sheniblog-EM-SSLC Worksheets - Questions and Answers - John P A

5) ABCD is a quadrilateral .AC = CD = AD,∠BAD = 120◦, ∠B = 90◦ ,The perpendicular distance
from D to the diagonal AC is 12cm .

a) What is the length of AC?
b) What are the angles of triangle ABC ?
c) What are the length AB and BC
d) Find the area of triangle ABC.
e) Find the area of triangle ADC.
f) Find the area of the quadrilateral ABCD

a) DE = 12, AE = √12

3

AC = 2× 12 = √24
sqrt3 3

b) ∠B = 90◦, ∠A = 120 − 60 = 60◦, ∠C = 30◦

c) AB = √12 , BC = 12.

3

d) Area ABC = 1 × 12 × √12 = √72
2 3
3

e) Area ADC = 1 AC × DE = 1√44
2 3

f) Area ABCD= √72 + 1√44 = 2√16
3 3
3

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2020-21 Academic year Works

Mathematics X
Trigonometry

46
Concepts

a) There are some special right triangles. The diagonal of a square makes two right triangles of angles

45◦, 45◦, 90◦

b) If the si√de opposite to 45◦ is 1 then the side opposite to 90◦ will be √ The sides are in the ratio
2.

1:1: 2

c) The altitude of an equilateral triangle makes two right triangles.The angles of these triangles are

30◦, 60◦, 90◦. to 30◦ is 1, the side opposite to 90◦ will be 2, side opposite tos 60◦ will be √
If the side opposite 3

Worksheet 46
1) In triangle ABC , the line AD is perpendicular to BC, AB = 12cm

a) What is the length of AD?
b) What is the length of AC?
c) What is the length of BC?
d) Calculate the area of triangle ABC?

a) △ADB is a 30◦ − 60◦ − 90◦ triangle.
∠A = 30◦. The side opposite to 90◦ is 12cm.
The side opposite to 3√0◦ is 6cm
BD = 6cm, AD = 6 3cm

b) △ADC √is a 45√◦ − 45◦√− 90◦ triangle.
AC = 6 3 × 2 = 6 6cm

√√
c) CD = 6 3cm, BC = 6 + 3cm

d) Area A = 1 × BC × AD = 1 × (6 + √ × √√ + √
2 2 6 3) 6 3=18 3(1 3)sq.cm

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2) In the fogure O is the centre of the circle. ∠BAC = 60◦,BC = 10 then

a) Draw the diametre from B which meet the circle at P
b) Draw triangle BP C, write the measure of ∠BP C?
c) What is the diametre of the circle? What is its radius?
d) Calculate the area of triangle BP C?

a) Picture is given below

Draw BP as in the figure. Draw P C

b) ∠BP C = 60◦. (Angle in the same arc)

c) △BP C is a 30◦ − 60◦ − 90◦ triangle.

Side opposite to 60◦ is 10cm

Side opposite to 30◦ is √10 cm

3

Side opposite to 90◦ is 2 × √10

3

Diametre = √20 cm, Radius = √10 cm

33

d) Area = 1 × BC × PC = √50 sq.cm
2 3

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3) O is the centre of the circle.If ∠ABC = 120◦,AC = 10then

a) The diametre from Ameet the circle at P . Mark the point P on the circle.
b) Draw triangle AP C ,write the measure of ∠AP C
c) What is the diametre and radius of the circle.
d) Calcualte the area of the triangle.

a) See the figure given below

b) ∠AP C = 180 − 120 = 60◦ (ABCP is cyclic)

c) In triangle ACP , side opposite to 60◦ is 10cm

AC = 10cm

Side opposite to 30◦ is √10 cm
3

AP = √20 cm. Diametre is = √20 cm , radius = √10 cm

3 33

d) Area = 1 × 10 × √10 = √50 sq.cm
2 3 3

4) In the figure AP is perpendicular to BC, ∠ACD = 150◦,∠BAP = 60◦

a) What is the length of AP and AB.
b) What is the measure of angle ACP ?

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c) What is the length of P C?
d) What is the area of the triangle ABC

a) △AP B is a 30◦ − 60◦ −√90◦ triangle.
Side opposite to 60◦ is 8 3cm

Side opposite to 30◦ is 8cm

Side opposite to 90◦ is 16cm.

AP = 8cm, AB = 16cm

b) ∠ACP = 180 − 150 = 30◦

√
c) AP = 8cm, P C = 8 3cm

d) Area = 1 × BC × AP = 1 √√
2 2 × 16 3 × 8 = 64 3 sq.cm

5) In the figure ABCDis a square .∠DP A = 30◦, ∠CP B = 45◦, P B = 4 cm then

a) What is the length of BC?
b) What is the length of AP ?
c) Find the area of the rectangle ?
d) Find the length of P Dand P C

a) △P BC is a 45◦ − 45◦ − 90◦ triangle.
BC = 4cm

b) △AP D is a 30◦ − 30√◦ − 90◦ triangle.
AD = 4cm, AP = 4 3cm.
√

c) AB = 4 + 4 3√cm,BC = 4cm √
Area = 4(4 + 4 3) = 16(1 + 3) sq.cm
√

d) P D = 8cm, P C = 4 2cm

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2020-21 Academic year Works

Mathematics X
Trigonometry

47

Concepts

a) A triangle can be scaled without altering its angles.While doing so length of the sides changes
keeping the ratio of the sides constant.This is what we have studied in similar triangles.

b) Unchaging angles and unchanging ratio of the sides make a special type of angle
measurements.These are known as trigonometric measurement of the angles.

c) We define trigonometric measurement of angles on the acute angles of a right triangle.
I√n 45◦ − 45◦ − 90◦right triangle the sides opposite to these angles are in the ratio1 : 1 :
2.Whatever be the size of the triangle this ratio remains unchanged.
√

d) The sides opposite to 30◦ − 60◦ − 90◦ angles of a right triangle are in the ratio1 : 3 : 2.

This ratio is independent of the size of the triangle.This leads to the measurement of angles,known

as sin, cos, tan.

e) In triangle ABC, A, B, Care the angles and a, b, c are the opposite sides. IfB = 90◦then

sin A = a , cos A = c , tan A = a
b b c

sin 30◦ = 21√, sin 60◦ = √ sin 45◦
3 √1
f) 2 , = 2 .

cos 30◦ = 3 , cos 60◦ = 21√, 3c,otsa4n5◦45=◦ √1 .
tan 30◦ = 2 2

√1 , tan 60◦ = =1
3

Worksheet 47

1) In triangle ABC,if ∠B = 90◦ ,sin A = 3 then
5

a) Draw a rough diagram

b) Write cos A and tan A

c) Write cos C and tan C

a) See the diagram

√
b) AB = 52 − 32 = 4

cos A = 4 , tan A = 3
5 4

d) cos C = 3 , tan C = 4 1
5 3

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2) If in triangle ABC, ∠B = 90◦ ,sin A = 0.8then

a) Draw a rough diagram.
b) Write cos A, tan A
c) Write cos C, tan C

a) See the diagram

b) sin A =√0.8 = 8
10

AB = 102 − 82 = 6

cos A = 6 = 0.6 , tan A = 8
10 6

d) cos C = 8 , tan C = 6
10 8

3) In triangle ABC , ∠B = 90◦ ,cos C = x

a) Draw a rough diagram.
b) Write sin C, tan C
c) Write cos A, sin A tan A

a) See the diagram

√√
b) AB = √1√211−−x2x2==√1 1 − x2
sin C = − x2

tan C = 1−x2
x

c) √ x =x
cos A = 1 − x2,sin A = 1

tan A = √x
1−x2

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4) In the figure AB = 10cm, the altitue from Ato BC is 8cm ,∠C = 45◦.

a) What is the length of BD?
b) Write sin B, cos B, tan B
c) Write the length of BC
d) Find the area of triangle ABC
e) Find the perimetre of triangle ABC.

√
a) BD = 102 − 82 = 6cm

b) sin B = 8
10
6
cos B = 10

tan B = 8
6

c) △ADC is a 45◦ − 45◦ − 90◦ triangle.
AD = CD = 8cm
BC = 6 + 8 = 14cm

d) Area = 1 × BC × AD = 1 × 14 × 8 = 56 sq.cm
2 2

√√
e) Perimetre = AB + AC + BC = 10 + 8 2 + 14 = 24 + 8 2cm

5) In the figure the side AB is perpendicular to BC, BP = 1, ∠BAP = ∠P AC = 30◦ then

a) What is the length of AB
b) What is the length of BC and P C

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a) △ABP is a 30◦ − 60◦ − 90◦ triangle . √
side opp√osite to 30◦ is 1. Therefore the side opposite to 60◦ is 3

AB = 3

b) △TSihdAeerBeofpCoproiesstiathee3t0so◦id3−e0◦o6p0isp◦o√−s3it9.e0t◦ot6ri0a◦ngisle√. 3

× √ = 3
3

BC = 3, P C = 3 − 1 = 2

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

48

Concepts

a) A triangle can be scaled without altering its angles.While doing so length of the sides changes
keeping the ratio of the sides constant.This is what we have studied in similar triangles.

b) Unchaging angles and unchanging ratio of the sides make a special type of angle
measurements.These are known as trigonometric measurement of the angles.

c) We define trigonometric measurement of angles on the acute angles of a right triangle.

I√n 45◦ − 45◦ − 90◦right triangle the sides opposite to these angles are in the ratio1 : 1 :
2.Whatever be the size of the triangle this ratio remains unchanged.
√
d) The sides opposite to 30◦ − 60◦ − 90◦ angles of a right triangle are in the ratio1 : 3 : 2.

This ratio is independent of the size of the triangle.This leads to the measurement of angles,known

as sin, cos, tan.

e) In triangle ABC, A, B, Care the angles and a, b, c are the opposite sides. IfB = 90◦then

sin A = a , cos A = c , tan A = a
b b c

sin 30◦ = 21√, sin 60◦ = √ sin 45◦
3 √1
f) 2 , = 2 .

cos 30◦ = 3 , cos 60◦ = 12√, 3c,otsa4n5◦45=◦ √1 .
tan 30◦ = 2 2

√1 , tan 60◦ = =1
3

Worksheet 48
1) In triangle ABC , AB = 8cm , BC = 12cm ,∠B = 120◦

a) What is the altitude from C to AB?
b) Find the area of triangle ABC.

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a) ∠ABP = 180 − 120 = 60◦

sin 60◦ = AP
AB
√
3 AP
2 =
12√
AP = 6 3cm

b) Area= 1 √√
2 × 8 × 6 3 = 24 3 sq.cm

2) In triangle ABC, ∠B = 90◦, ∠A = 40◦

a) What is the measure of ∠C
b) Compare sin A and cos C
c) Compare sin C and cos A?
d) Write the conclusion on comparing sin measure and cos measure of acute angles of a right triangle.

a) ∠C = 50◦

b) sin A = a
b
a
cos C = b

sin A = cos C

If A + C = 90◦, sin A = cos C and cos A = sin C always

c) sin C = c
b
c
cos A = b

sin C = cos A

d) sin of one acute angle is equal to cos of other acute angle.

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3) In triangle ABC, ∠B = 90◦ , tan A = 12 then
5

a) Find the tan measure of angle C

b) What is sin A

c) Find 1+sin A
1−sin A

a) tan C =√ 5
12

AC = 52 + 22 = 13

b) sin A = 12
13

c) 1+sin A = 1+ 12 = 25
1−sin A 13
12
1− 13

4) In triangle ABC, ∠B = 30◦, ∠C = 60◦, BC = 12cm

a) BC is perpendicular to DA,If DB = x then what is DC?
b) Write equations connecting △BDA and △CDA
c) Find x
d) What is the perpendicular distance from A to BC
e) Find the area of triangle ABC.

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a) CD = 12 − x

b) If AD = h

h = tan 30 = √1 ,h = √x √
x 3 √ 3 3(12

h = tan 60 = 3, h = − x)
12−x

c) √x √
= 3 × (12 − x)
3√ √
x = 3 × 3 × (12 − x)

x = 3(12 − x),4x = 36, x = 9

d) h= √x = √9 √
3 3 = 3 3cm

e) Area = 1 √√
2 × 12 × 3 3 = 18 3sq.cm

5) ABCD is a trapezium .∠A = 60◦, ∠B = 30◦, AB = 14cm , AD = 4cm

a) What is the diatance between parallel sides
b) What is the length of CD
c) Find the perimetre of the trapezium
d) Claculate the area of the trapezium.

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a) Draw DP and CQ perpendicular to AB
△AP D is a 30◦ − 60◦ − 90◦ triangle.
Side opposite to 90◦ i√s 4cm.
AP = 2cm, P D = 2 3cm.

b) △CQB √is a 30 − 60 − 9√0 trian√gle.
QC = 2 3cm, QB = 2 3 × 3 = 6cm

P Q = 14 − (2 + 6) = 6cm

CD = 6cm

√
c) CB = 4 3cm.
√√
perimetre = 14 + 4 + 6 + 4 3 = 24 + 4 3cm

d) Area = 1 × h × (a + b) = 1 √√
2 2 × 2 3 × (14 + 6) = 20 3sq.cm

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

49
Concepts
a) We can see a table of all trigonometric measures of angles ranging from 0◦ to 90◦in the text
book.You can learn a detailed table in the higher
b) When the angle measures increases from 0◦ to 90◦, sinmeasure measure increases from 0 to 1.
c) When angle measures increases from 0 to 90 cos measure decreases from 1 to 0
d) Maximum value of sin and cos is 1.
e) sinmeasure and cosmeasure become equal at the acute angle 45◦.
f) sin 0 = cos 90 = 0, sin 90 = cos 0 = 1
g) If the sum of two angles is 90◦then sin of one angle is equal to cos of other angle.
Example,sin 40◦ = cos 50◦

Worksheet 49
1) There is a regular octagon of side 4cm . A quadrilateral is shaded .

a) What is the measure of ∠C?
b) What is the length the rectangleABCF
c) Calculate the area of the coloured region.

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a) Angle sum of a polygon of side n is (n − 2) × 180

One angle = (8−2)×180 = 135◦
8

b) Look at the picture

Draw DP perpendicular toCF .Mark the point Q in the figure.

△DP C is a 45◦ − 45√◦ − 90◦ triangle.
CD = 4cm, P C = 2 2cm. √
√
Length√of the recta√ngle :CF = 2 2 + 4 + 2 2cm
4 + 4 2 = 4(1 + 2)cm

c) Area of CDQ is equal to area of AF Q.
Area of rectangle is√equal to area o√f coloured region.
Area = 4 × 4(1 + 2) = 16(1 + 2)

2) In the parallelogram ABCD, ∠A = 40◦, AB = 10cm , AD = 8cm

a) Draw a rough diagram
b) What is the diatance between the parallel sides AB, CD
c) Calculate the area of the parallelogram.
[sin 40◦ = 0.6428, cos 40◦ = 0.7660, tan 40◦ = 0.8391]

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a) See the diagram

b) sin 40◦ = DP
8

DP = 8 × 0.6428 = 5.1424cm

c) Area = AB × P D = 10 × 5.14 = 51.4sq.cm

3) In triangle ABC, ∠A = 25◦, AB = 14cm , AC = 18cm .
a) Draw a rough diagram
b) What is the altitude from C to AB
c) Calculate the area of the triangle ABC.

[sin 25◦ = 0.64226, cos 25◦ = 0.9063, tan 25◦ = 0.4553]

a) See the diagram

b) sin 25◦ = PC
18

P C = 0.64 × 18 = 11.52cm

c) Area 1 × 14 × 11.52 = 80.64 sq.cm
2

4) In the figure AB = AC = 12cm , ∠A = 140◦

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a) Write the measure of ∠B and ∠C
b) What is the altitude fron C to AB?
c) Find the area of triangle ABC
d) Find the perimetre of triangleABC

[sin 40◦ = 0.6428, cos 40◦ = 0.7660, tan 40◦ = 0.8391]

a) ∠B = ∠C = 180−140 = 20◦
2

b) Draw CP perpendicular to AB

In triangle CP A,∠A = 180 − 140 = 40◦

sin 40 = PC
12

P C = 12 × .64 = 7.68 cm

c) Area = 1 × 12 × 7.68 = 46.08 sq.cm
2

5) In triangle ABC, O is the centre of its circumcircle.One side AB = 24cm , opposite angle ∠C = 140◦

a) Draw a rough diagram .Draw the diametre from A which cut the circle at P . Join P B
b) What is the measure of ∠AP B?
c) Find the radius of the circle.
[sin 50◦ = 0.7660, cos 50◦ = 0.6420, tan 50◦ = 1.1918]

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a) See the diagram

b) ∠AP B = 180 − 140 = 40◦

c) Triangle ABP is a right triangle

sin 40◦ = 24
AP
24 24
AP = sin 40 = cos 50 = 37.5

Radius = 18.75

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

49

Concepts
a) We can see a table of all trigonometric measures of angles ranging from 0◦ to 90◦in the text

book.You can learn a detailed table in the higher
b) When the angle measures increases from 0◦ to 90◦, sin measure increases from 0 to 1.
c) When angle measures increases from 0 to 90 cos measure decreases from 1 to 0
d) Maximum value of sin and cos is 1.
e) sinmeasure and cosmeasure become equal at the acute angle 45◦.
f) sin 0 = cos 90 = 0, sin 90 = cos 0 = 1
g) If the sum of two angles is 90◦then sin of one angle is equal to cos of other angle.

Example,sin 40◦ = cos 50◦

Worksheet 49
1) Consider the following trigonometric measures of angles

sin 42◦, cos 78◦, sin 70◦, cos 14◦

a) Rewrite all these into equivalent sin measures
b) Find the largest and smallest among these
c) Write these in the ascending order.

a) sin 42◦ = sin 42◦
cos 78◦ = sin(90 − 78) = sin 12◦
sin 70◦ = sin 70◦
cos 14◦ = sin(90 − 14) = sin 76◦

b) Smallest is sin 12◦
Largest is sin 76◦.
So in the given measures , smallest is cos 78◦ and largest is cos 14◦

c) sin 12◦, sin 42◦, sin 70◦, sin 76◦
cos 78◦ < sin 42◦ < sin 70◦ < cos 14◦

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2) In the concentric circles, O is the centre and the radius of the outer circle is 12cm .The line AB , the chord
of the big circle touches the small circle at P .Also, OP is perpendicular to AB. If ∠P BO = 40◦ then

a) What is the radius of the small circle?
b) What is the length of the chord AB?
c) Find the area in between these two circles?

a) In △OP B, OP = sin 40◦
OB
OP = OB × sin 40◦ = 12 × 0.6428 = 7.71 cm

b) cos 40◦ = PB
12
P B = 12 × cos 40◦ = 12 × 0.7660 = 9.12cm

Length of the chord AB = 2 × 9.12 = 18.24cm

c) If R is the radius of the outer circle and r is the radius of the incircle ,
R2 = r2 + P B2

R2 − r2 = P B2 = 9.122
R2 − r2 = 83.17
Area = π(R2 − r2) = 83.17π.

3) ABCD is a trapezium.The line AB parallel to CD.Also, AB = 18cm , AD = 6cm , ∠A = 50◦then

a) What is the distance between the parallel lines AB and CD
b) What is the length of the side CD
c) Calcualte the area of the trapezium.

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a) Draw rough diagram . Draw DP perpendicular to AB

In △ADP , sin 50◦ = PD
6

P D = 6 × sin 50 = 6 × 0.7660 = 4.6 cm

b) cos 50◦ = AP
6

AP = 6 × cos 50 = 3.8 cm

P B = 18 − 3.8 = 14.2cm.

Length of CD = 14.2cm

c) Area = 1 × h(a + b) = 1 × 4.6 × (18 + 14.2) = 2.3 × 32.2 = 74.06
2 2

sq.cm

4) 40◦ angle is drawn with the line AB of length 8cm as an arm and the vertex at A. A line is drawn from B
perpendicular to the arm ,which is marked as BP .

a) What are the angle measures of the triangle AP B?
b) Find the length of the sidesAP and BP
c) Find the area of the triangle.

a) Draw a rough diagram as given below .

∠A = 40◦, ∠P = 90◦, ∠B = 50◦

b) sin 40◦ = PB
8

P B = 8 × sin 40 = 8 × .64 = 5.12cm

cos 40◦ = AP
8

AP = 8 × 0.76 = 6.08cm

c) Area = 1 × BP × AP = 1 × 5.12 × 6.08 = 15.56 sq.cm
2 2

5) a, b, c are the sides opposite to the angles A, B, C of triangle ABC.

a) Draw a rough diagram

b) Prove that area of the triangle is 1 bc sin A.
2

c) If two sides are 16cm and 10cm, the angle between them is 50◦, then find the area of the triangle.

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a) Look at the picture

b) sin A = PC = PC
AC b

P C = b sin A

Area = 1 × AB × PC = 1 × c × b × sin A = 1 bc sin A
2 2 2

c) Area = 1 × 16 × 10 × sin 50 = 60.8 sq.cm
2

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

50

Concepts
a) We can see a table of all trigonometric measures of angles ranging from 0◦ to 90◦in the text

book.You can learn a detailed table in the higher
b) When the angle measures increases from 0◦ to 90◦, sin measure increases from 0 to 1.
c) When angle measures increases from 0 to 90 cos measure decreases from 1 to 0
d) Maximum value of sin and cos is 1.
e) sinmeasure and cosmeasure become equal at the acute angle 45◦.
f) sin 0 = cos 90 = 0, sin 90 = cos 0 = 1
g) If the sum of two angles is 90◦then sin of one angle is equal to cos of other angle.

Example,sin 40◦ = cos 50◦

Worksheet 50

1) Find the value of the following without using trigonometric table.

a) sin 18◦
cos 72◦

b) cos 48◦ − sin 42◦

c) cos 38◦ cos 52◦ − sin 38◦ sin 52◦

a) sin 18◦ = cos(90−18) = cos 72◦ = 1
cos 72◦ cos 72◦ cos 72◦

b) cos 48◦ − sin 42◦ = sin(90 − 48)◦ − sin 42◦ = sin 42◦ − sin 42◦ = 0

c) cos 38◦ cos 52◦ − sin 38◦ sin 52◦
cos 38 sin 38 − sin 38 cos 38 = 0

2) sin3 A = sin A × sin A × sin A.
If sin A + sin B + sin C = 3then
a) What is the value of sin3 A + sin3 B + sin3 C?
b) What is the value of cos3 A + cos3 B + cos3 C?
c) What is the value of cos A + cos B + cos C?

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a) Maximum value of sin measure is 1.
Here it is given that sin A + sin B + sin C = 3
So, sin A or sin B or sin C cannot be less than 1.
If one of them is less than 1 , other should be greater than 1 to keep the
sum 3, which is not possible.
sin A = sin B = sin C = 1
sin3 A + sin3 B + sin3 C = 1 + 1 + 1 = 3

b) sin A = sin B = sin C = 1 ⇒ cos A = cos B = cos c = 0(Clear from
trigonometric table)
cos3 A + cos3 B + cos3 C = 0

c) cos A + cos B + cos C = 0

3) If BD = a, BC = b, AB = h, ∠ADB = 30◦, ∠ACB = 60◦then

√
Prove that h = ab

⋆ In △ABD ,tan 30◦ = h
a

√1 h (1)
= (2)
3a

⋆ In △ABC ,tan 60◦ = h √h
b 3=
b

⋆ Multiplying 1 and 2 √1 √ = h × h
√ ×3
3 ab
h2 = ab, h = ab

4) ABCD is a trapezium .ABis parallel to CD, AD = BC = 12cm , CD = 5cm .
If ∠A = 40◦then

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a) What is ∠B?
b) What is the diatance between the parallel sides ?
c) What is the length of AB
d) Find the area of the trapezium.

a) ∠B = 40◦ (ABCD is an isosceles trapezium)

b) Draw DP and CQ perpendicular to AB

sin 40◦ = DP
12

DP = 12 × 0.64 = 7.68 cm

c) cos 40◦ = AP
12
AP
0.76 = 12

AP = 12 × 0.76 = 9.12cm

QB = 9.12cm , P Q = 5cm

AB = 9.12 + 5 + 9.12 = 23.24cm

d) Area = 1 × h(a + b)
2
1
Area = 2 × 7.68 × 28.24 = 108.44 sq.cm

5) The diagonals of the rhombus ABCD intersect at O. One side is 10 cm ,
If ∠OAB = 20◦then

a) What is the measure of ∠AOB?

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b) What is the length of the diagonal AC?
c) What is the length of the diagonal BD?
d) Calculate the area of the rhombus ?

a) ∠AOB = 90◦

b) cos 20◦ = OA
AB

OA = 10 × 0.9397 = 9.39

AC = 18.76cm

c) sin 20 = OB
AB

OB = 10 × 0.3420 = 3.42cm

BD = 6.84cm

d) Area = 1 × d1 × d2 = 64.15 sq.cm
2

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

50

Concepts
a) We can see a table of all trigonometric measures of angles ranging from 0◦ to 90◦in the text

book.You can learn a detailed table in the higher
b) When the angle measures increases from 0◦ to 90◦, sin measure increases from 0 to 1.

c) When angle measures increases from 0 to 90 cos measure decreases from 1 to 0

d) Maximum value of sin and cos is 1.
e) sinmeasure and cosmeasure become equal at the acute angle 45◦.

f) sin 0 = cos 90 = 0, sin 90 = cos 0 = 1
g) If the sum of two angles is 90◦then sin of one angle is equal to cos of other angle.

Example,sin 40◦ = cos 50◦

Worksheet 50

1) a, b, c are the sides opposite to the angles ∠A, ∠B, ∠C of triangle ABC.

a) Draw a rough diagram of triangle its circumcircle and the diametre from B

b) If R is the radius of the circumcircle then prove that a = 2R
sin A

c) If ∠A = 40◦, a = 12cm then find the radius of the circumcircle.

a) Picture is given below

b) ∠A = ∠D(Angle in the same arc), ∠BCD = 90◦

sin A = sin D = BC
BD
a a
sin A = 2R ∴ sin A = 2R

c) 12 = 2R, 12 = 2R, R = 12 ÷2 = 9.33cm
sin 40◦ 0.6428 0.6428
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2) Radius of the circumcircle of a regular pentagon is 6cm.Consider the triangle formed by a side and two radii
at its ends.
a) Find the side of the pentagon.
b) What is the perpendicular distance from the circumcentre to the side.
c) Calculate the area of the pentagon.

a) Look at the figure

One angle of a regular pentagon is (5−2)×180 = 108◦
5
In the right triangle OP B, ∠B = 54◦

cos 54 = PB , PB = cos 54 × 6 = .58 × 6 = 3.48cm.
OB

Side of the pentagon AB = 2 × 3.48 = 6.96cm

b) sin 54 = OP
6

OP = 6 × sin 54 = 6 × 0.80 = 4.8cm

c) Area of triangle OAB = 1 × AB × OP = 1 × 6.96 × 4.8 = 16.7 sq.cm
2 2

Area of pentagon = 5 × 16.7 = 83.5 sq.cm

3 a, b, c are the sides of a triangle and its opposite angles A, B and C

a) Prove that a = b cos C + c cos B

b) If R is the radius of the circumcircle then prove that area of the triangle A = abc
4R

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a) Look at the picture

In triangle AP B , cos B = axc−b x⇒⇒x = c cos B a − x
In triangle AP C, cos C = b cos C =

c cos B + b cos C = x + a − x = a

a = b cos C + c cos B

b) In triangle AP B , sin B = h
c

h = c cos B

Area = 1 × a × h = 1 × a × c sin B
2 2
a
Again, sin A = 2R where R is the radius of the circumcircle.

Area of the traingle = 1 bc sin A or 1 ac sin B or 1 ab sin C
2 2 2
1 1 a abc
Area A = 2 × bc × sin A = 2 bc 2R = 4R

4) The chord of a circle with radius 7cm makes an angle 110◦ at its centre.

a) What is the length of the chord?
b) What is the perpendicular distance from centre to the chord.
c) Calculate the area of the triangle formed by the chord and the radii at its ends.

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a) sin 55 = PB
7

P B = 7 × sin 55 = 7 × 0.81 = 5.67 cm

AB = 2 × 5.67 = 11.34cm

b) cos 55 = OP
OB

OP = 7 × cos 55 = 7 × 0.57 = 3.99 cm

c) Area = 1 × AB × OP = 1 × 11.34 × 3.99 = 22.62sq.cm
2 2

5) There is a stetched string from the top of a flag post to a point on the ground.The point on the ground is at
the diatance 50m away from the foot of the flag-post.The string makes an angle 40◦ with the ground.

a) Draw a rough diagram.
b) Calculate the height of the flag post.
c) Find the length of the string.

a) diagram

b) h = tan 40◦
50
h = 50 × tan 40◦ = 50 × .83 = 41.5 m

c) cos 40 = 50
x
50 50
x = cos 40 = .76 = 65.78cm

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

51
Concepts
a) In the first figure the top of a flagpost is observed at an angle of elevation x.Angle of
elevation is the angle between the ray of vision and horizontal ground
b) When the object is viewed from the top of a tower, the angle become the angle of
depression. It is denoted by y in the second figure.

Worksheet 51
√
1) The height of a building is 100 3metre.The top of this building is observed from a point at the
distance 100m from the foot of the building.
a) Draw a rough diagram
b) What is the angle of elevation?
c) If the angle of elevation is 45◦ what is the distance from the foot of the building to the point
on the ground?

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a) Diagram

b) Let x be the a=ng1le00o1√f030ele=va√tio3n.
tan x = → x = 60◦
tan x = AB

√AC
3

c) If the angle of elevation is 45◦, AB = AC. The√diatance from the foot
of the building to the point on the ground = 100 3m

2) The string of a kite is 100m long and it makes an angle of 60◦ with the horizontal.
a) Draw a rough diagram
b) Find the height of the kite assumimg that there is no slack in the string.

a) Diagram.

b) sin 60◦ = h
100
√
3 h
2 = √
100√
2h = 100 3, h = 50 3m

3) The top of building is observed from a point at the diatance a and b from the foot of a the building
on either side .The angle of elevation are 30◦, 60◦.

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a) Draw a rough diagram.
√

b) If h is the height of the building , prove that h = ab
a) Diagram.

b) t√√a1133n×=30√◦ha ,3=√=ha3,ha=ta×nhb 60 = h
b

1 = h2 , h2 = ab, h
ab
b√
h = ab

4) The top of a building can be observed at an angle of elevation 45◦ from a point at some diatance
away from the foot of the building.When moved a diatance 20m towards the tower the angle of
elevation becomes 60◦.

a) Draw a rough diagram.
b) Form equations using the given conditions.
c) Calculate the height of the tower.
d) What is the distance from the foot of the tower to the point the top is observed.

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a) Diagram.

b) Let the height of the tower be h

tan 45◦= h , h = 1, h =√x + 20, x = h − 20.
tan x+20 = 3x.
√x+20
60◦ = h , 3= h , h
x x

√ √√
c) h =√ 3(h√− 20) = 3h√− 20 3
20 3 =√ 3h − h = √h( 3 − 1)
h= = 10(3 + 3)metre.
√20 3
3−1

d) x = h − 20 = 10(3 + 1.732) − 20 = 47.32 − 20 = 27.32m
Distances are 20 + x = 47.32metre , 27.32metre

5) The top and bottom of a 7 metre tall building can be seen at the angle of depression 45◦ and 60◦
from the top of a tower.

a) Draw a rough diagram.
b) Calculate the height of the tower.
c) What is the diatance between the tower and the building.

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a) Diagram.

⋆ Let AB = CD = d,tan 45◦ = x , 1 = x , x = d
d d

⋆ √ta3n√=60◦7+x=x 7+x √
+ x, 3x −
√d =7 x =7
, 3x x= = 9.6m

x( 3 − 1) = 7, √7 = 7
3−1 .73

Height of the tower x + 7 = 9.6 + 7 = 16.6m

d) d = x, Diatance = 9.6m

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

52
Concepts
a) In the first figure the top of a flagpost is observed at an angle of elevation x.Angle of
elevation is the angle between the ray of vision and horizontal ground
b) When the object is viewed from the top of a tower, the angle become the angle of
depression. It is denoted by y in the second figure.

Worksheet 52
1) From a point on the ground 40 metre away from the foot of the tower sees the top of the tower at

an angle of elevation 30◦ and sees the top of the water tank on the top of the tower at an angle
of elevation 45◦.

a) Draw a rough diagram.
b) Find the height of the tower.
c) Find the height of the water tank

1

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a) Diagram

b) In triangle ABD we have tan 45◦ = BD
AB
h+h1
1 = 40 , h + h1 = 40

tan 30◦ = BC , √1 = h
AB 3 40

h = √40 = 23.1m

3
Height of the tower 23.1m

c) 23.1 + h1 = 40, h1 = 40 − 23.1 = 16.9metre.

2) A man standing on the bank of a river sees the top of the tree at an angle of elevation 50◦.Steppimg
20 m back finds the angle to sees at an angle of elevation 30◦

a) Draw a rough diagram.
b) Find the width of the river.
c) Calcualte the height of the tree.

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a) Diagram

b) tan 50◦ = BC , tan 30◦ = BC
AB BD
h √1 h
1.19 = x , 3 = x+20

h = 1.19x, h = x√+20
3

1.19x = x√+20
√3
1.19 × 3x = x + 20, 1.19 × 1.73x = x + 20, x = 19.05m

Width of the river 19.05metre

c) When x = 19.05, h = 1.19x = 1.19 × 19.05 = 22.67metre.

3) A tree of height 12m is broken by the wind . The top struck the ground at an angle of 35◦ .

a) Draw a rough diagram
b) At what height from the bottom of the tree is broken by the wind?
c) Calculate the distance from the foot of the tree to the its tip on the on the ground.

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a) Diagram

b) sin 60◦ = AC
DC
√
23√= x
√√
12−√x
12 3 −√ 3x = 2x, 12 3 = (2 + 3)x
x = 24 3 − 36 = 5.56metre.

c) √tan 60◦ = x
3 DA
= x , = 5.56
DA 1.73 DA

DA = 5.56 = 3.2m
1.73

4) The shadow of a vertical tower on level ground increases by 10m when the altitude of the Sun
changes from the angle of elevation 45◦ to 30◦.
a) Draw a rough diagram
b) Calculate the height of the tower.

a) Diagram

b) tan 45 = AB , 1 = h , h = x
AC x
√1 h
tan 30 = = x+10
√3
x + 10 = 3h, h = 13.65metre.

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5) A tall building and a short building are standing on a level ground.The angle of elevation of the top
of the short building from the foot of the tall building is 30◦.
The angle of elevation of the top of the tall building from the foot of the short building is 60◦. The
tall building has height 50m.

a) Draw a rough diagram
b) What is the diatance between the buildings.
c) Calculate the height of the short building.

a) Diagram

b) tan 60◦ = 50 , √ = 50
x 3 x

x = √50 = 50 = 28.9m
3 1.73

The distance between the buildings = 28.9m

c) tan 30 = h , √1 = h √ 28.9 = 16.7m
x 3 28.9 , 3h = 28.9, h = 1.73

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

52
Concepts
a) In the first figure the top of a flagpost is observed at an angle of elevation x.Angle of
elevation is the angle between the ray of vision and horizontal ground
b) When the object is viewed from the top of a tower, the angle become the angle of
depression. It is denoted by y in the second figure.

Worksheet 52
1) The diatance between two buildings is 100metre.The height of one building is double the height of

other building.The top of the buildings can be seen at the angle of elevations 60◦ and 30◦ from a
point in between the buildings.

a) Draw a rough diagram
b) What is the diatance from the foot of the tall tower to the point of observation.
c) Claculate the height of the buildings. 1

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a) Diagram

b) t1Ltaae0nnt0A36−00Cx====12xhh0√0h,−a3√nxhd3,,x1B100=D00 −−2=hxx,2hh===√√√33×2h32x√3x

= 3x
2

200 − 2x = 3x, 5x = 200, x = 40m.

The point of observation is at the diatance 40m from tall building.

c) h= √ = √ √ √
3x 3×40 = 20 3m
2 2

He√ight of the small building is 20 3m and height of the tall building is
40 3m

2) The top of a 30 high building can be seen from a point at some diatance from the foot of the
building is at an angle of elevation 30◦. When the poit of observation is some distance closer to
the building the angle become 60◦

a) Draw a rough diagram

b) What is the distance from the foot of the building to the second point of observation.

c) What is the diatance between two points from which the top of the building is observed .

d) What is the diatance from the foot of the tower to the first point of observation.

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a) Diagram

b) In △ABP ,tan 60◦ = 30 , √ = 30 , y = √30 √
y 3 y 3 = 10 3m

The second point is at the distance 10 × 1.732 = 17.32m away from the

building.

c) tan 30 x=3+0yx,3+30y0√3 √
= = x + 17.3x = 30 3 − 17.3 = 34.66m
√1 = x + y
3
The distance between the points is 34.66m

√
d) Distance is x + y = 30 3 = 51.96m

3) A baloon , moving horizontally at the height 88.2m from the ground sees at an angle of elevation
60◦ from a point on the ground. After some time it can be seen at the angle 40◦ from the same
point
a) Draw a rough diagram.
b) Calculate the diatance moved by the baloon during this time.

a) Diagram

b) In △P AB, tan 60◦ = 88.2 , P A = 88.2 = 50.98m
PA 1.73
88.2 88.2
tan 40 = PC , PC = 0.83 = 106.2m

Distance moved by the baloon is 106.2 − 50.98 = 55.2metre.

4) A man sees the top and bottom of a hill at the angle of elevation 70◦ and angle of depression 30◦
from the top of a tower of height 10m

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a) Draw a rough diagarm.
b) What is the distance between the hill and the building
c) Calculate the height of the hill

a) Diagram

b) ∠ACB = 90 − 30 = 60◦√
, AB = 10 3
tan 60 = AB = 17.32m
10

c) In △CDE,tan 70 = ED
17.32
ED
2.7 = 17.32 , ED = 2.7 × 17.32 = 46.76m

Height of the hill = 46.76 + 10.0 = 56.76m

5) A man sees a car moving towards a building at uniform speed at an angle of depression 30◦ from
the top of the building. After 6 seconds the angle of drepression becomes 60◦.

a) Draw a rough diagram.
b) How long the car take to reach near the building?

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a) Diagram

⋆ tan 60◦ = h , h = √
x 3x

⋆ tan 30 = h , √1 = h
x+y 3 x+y

x + y = 3x, y = 2x.

The car takes 6 seconds to travel y distance. Since y = 2x, x = y . Since
2

the car travels equal distance in the equal intervel of time the car takes

3 seconds to travel y distance. That is the car takes 3 more seconds to
2

reach the building

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2020-21 Academic year Worksheets

Mathematics X
Trigonometry

52

Concepts √
2
⋆ The sides of a 45◦ − 45◦ − 90◦ triangle are in the ratio 1:1 :

⋆ The sides of a 30◦ − 60◦ − 90◦ triangle are in the ratio √ : 2
1: 3

⋆ The angle between the direction of vision and horizontal is called angle of elevation or angle
of depression.

Worksheet 52

1) A quadrilateral is colouded inside an equilateral triangle of side 12cm .The sides of the equilateral
triangles are divided into three parts by dots.

a) What is the area of △QRC?
b) What is the area of △P BS?
c) What is the area of △AP Q?
d) Calculate the area of the coloured quadilateral.

a) △QRC is√an equil√ateral triangle of side 4cm.Altitude 2 √ 3cm. Area A1 =
23 = 4 3sq.cm
1 × 4 ×
2

b) △P BS has two sides 8cm√and 4cm, included 12a×ng4le×640√◦.T3h=is is√a 30◦ −
60◦ − 90◦ triangle. P S = 4 3cm Area A2 = 8 3sq.cm

c) △AP Q is a 30◦ − 60◦ − 90◦ triangle. A3 = 1 × 4 × √ = √ sq.cm
2 43 83

d) Altitude of △ABC is √ Area √21 × 12 √ √= √ √
6 3cm1, ×√6 3 36 √3 sq.cm

Area of the shaded quadrilateral is 36 3 − (8 3 + 8 3 + 4 3 = 16 3

sq.cm

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2) Ois the centre of a circle of radius 4cm. ∠P AB = 20◦

a) What is the length of the side P B in triangle AP B?
b) What is the length of the side P A?
c) Calculate the area of △P AB

a) AB is the diametre of the circle.∠AP B = 90◦.

sin 20◦ = PB = PB
AB 8

P B = 8 × 0.34 = 2.72cm

b) cos 20◦ = PA
8

P A = 8 × 0.93 = 7.44cm

c) Area = 1 × PA × PB = 10.11 sq.cm
2

3) A man sees a boat approaching the shore at the angle of depression 30◦ from the top of a light
house . After 6 seconds the angle of depression becomes 60◦.

a) Draw a rough diagram
b) How long will the boat takes to reach the shore .
c) If the speed of the boat is 25kilometre per hour, what is the distance from the shore to the

second point of observation.

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a) Diagram

b) △ABC is a 30◦ − 60◦ − 90◦ triangle.IF AB = h, AC √
= 3h

△ABD is a 30◦ − 60◦ − 90◦ triangle.AD = √h
√ 3
3h
CD = AC − AD = − √h = √2h
33

AD : CD = 1 : 2

The boat takes 6 minutes to reach C to D.So the boat takes 3 minutes

to reach he shore from D

c) If the speed 25 km per hour, , CD = speed × time . That is CD =

25 × 6 = 2.5km.
60

The diatance from A to D is 1.25km

4) A man sees the top of a tower at angle of elevation 40◦ from the top of a building of height
15metre. He sees the top of the tower at the angle 70◦ from the foot of the building.

a) Draw a rough diagram .
b) Calculate the height of the tower.
c) What is the distance between tower and building?

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a) Diagram

b) Let AC = y ,DE = x

tan 40◦ = x
y
x
0.83 = y , x = 0.83y

tan 70 = x+15 , 2.7 = x+15 = 0.83y+15
y y y

2.7y = 0.83y + 15, 2.7y − 0.83y = 15, 1.8y = 15, y = 8.02m,x =

0.83y = 6.65m. Height of the tower = 6.65 + 15 = 21.65m

c) Distance = 8.02m

5) A man sees the top of a light house of height 100m at the angle of elevation 60◦. After 2 minutes
the angle becomes 45◦.

a) Draw a rough diagram.
b) Calculate the distance between the positions of obaservation.
c) Calculate the speed of the boat.

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