x-CH5 Trigonometry

Trignomerty 5

Q6) A regular pentagon is drawn with all its vertices on a circle

of radius 15 centimetres. Calculate the length of the sides

of this pentagon.

Ans) Given r = 15cm

Let the side of the hexagon be ‘a’ .

Since given polygon is a pentagon O
360 °
∠ POQ = 5 = 72° 15 cm

We know ,

Length of a chord of x° Pa Q
central angle x° 2
= 2r Sin

∴ a= 2r Sin 72 ° = 2r Sin36°
2

= 2 × 15 × 0.5878

= 17.634

ie, Length of the sides of the pentagon = 17.634 cm

Assignment
T.B Page 114

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 59 13 /11 / 2020
Answer of last class assignment
To view class

Ans) We know,

Each angle of equilateral triangle is 60°. 8 cm 60°
Let ‘r’ be the circumradius. 8 cm
Then, each side = 2r Sin 60°

Given each side is 8 cm.

∴ 2r Sin 60° = 8 60° 60°
8 cm
2r × √3 =8

2

r × √3 = 8

r= 8 = 8 × √3 = 8 √3 = 8× 1.732 ≈ 4.62cm
√3 3
√3 √3 3

Ans) i) Given ∠A = 70°, A
70°
So, BC = 2r Sin 70°

Given, BC = 4 cm

∴ 2r Sin 70° = 4 50° 60°
4
2r = sin70 ° B 4 cm C

2r = 4 ≈ 4 ≈ 4.26
0.9397 0.94

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

ie, Diameter of circle ≈ 4.26 cm

ii) Other two sides of circle are AB & AC

AB = 2r Sin 60° AC = 2r Sin 50°

= 4.26 × 0.8660 = 4.26 × 0.7660
≈ 3.69 cm ≈ 3.26 cm

Another measure

In the given figure Δ ABC, Δ ADE, Δ AFG, Δ AHI I
are similar triangles. H

That is their corresponding angles G
F
are equal and the corresponding

sides are in proportion. E

C

Consider ∠A

AB D
Opposite Side BC DE FG HI
Sin A = Hypotenuse = AC = AE = AG = AI

Cos A = Adjacent Side = AB = AD = AF = AH
Hypotenuse AC AE AG AI

Let us see another ratio:

The number got by dividing the opposite side of an angle by

the adjacent side in right triangles is called the C
tangent of the angle and is shortened as ‘tan’.

Tan A = Opposite Side of ∠A Opposite Side
Adjacent Side of ∠A

A Adjacent Side B

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Here there are three right triangles of different size, but with

the same acute angle ‘A’ at the lower left C

CC
4.6188
3.7528

2.8868

30° B 30° 30° B
A5 A 6.5 B A8

Opposite Side = BC Opposite Side = BC Opposite Side = BC
Adjacent Side AB Adjacent Side AB Adjacent Side AB
2.8868 3.7528 4.6188
= 5 = 6.5 = 8

= 0.5774 = 0.5774 = 0.5774

In the above three triangles the angles are equal, but the sides
are different and the ratio between opposite side and adjacent
side of ∠A is always same.
Tan 30° = 0.5774

The value of the tangent (and the sine and cosine) depends on
the size of the angle, not on the size of the triangle.

Q1) In Δ ABC ,∠C = 90° ,∠B = 50° , BC = 3cm , find AC. A
C
Ans) Opposite Side AC
Adjacent Side BC
Tan 50° = =

∴ AC = BC × Tan 50°

= 3 × 1.1918 (Given BC = 3cm)

= 3.5754

≈ 3.6 cm 50°
B 3 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

Q2) Consider the given picture.
Width of each step is 20cm
and stairway makes an
angle 35° with the floor,
Find how high the man in
the picture is standing.

Ans)

From figure

CB = 20 + 20 + 20 = 60 cm

Given ∠C = 35° 20 cm A
Opposite Side AB 20 cm B
Tan35° = = Adjacent Side = CB 20 cm

∴ AB = CB × tan35° C 60 cm
= 60 × 0.7002

= 42.01
≈ 42 cm

Thus the height is about 42 centimetres.

Assignment

Q1) In a right triangle ABC, ∠C = 65° , AB = 5cm. Find the
A
length of side BC.

5 cm 65°
B C

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Q2) From the top of a vertical post, a rope is stretched and tied
to the ground 3 m away from the bottom of the post. The
rope makes an angle 50° with the ground. Find the height
of the post.

C

Rope
Post

50°

A Ground B

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 60 16 /11 / 2020

To view class

Answer of last class assignment
Q1) In a right triangle ABC, ∠C = 65° , AB = 5cm.AFind the

length of side BC.

Ans) Tan65° = AB
BC
5 cm
BC × tan65° = AB B
AB
BC = tan 65 ° 65°
C
BC = 5
2.1445
5
≈ 2.14 ≈ 2. 34cm

Q2) From the top of a vertical post, a rope is stretched and tied

to the ground 3 m away from the bottom of the post. The

rope makes an angle 50° with the ground. Find the height
C
of the post.

Ans) BC
AB
Tan50° =

∴ BC = AB × tan50° Rope
Post
= 3 × 1.1918

= 3.5754

50°

A 3m B
Ground
So, height of the post ≈ 3.58 meters

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

T.B Page 117
Q1) One angle of a rhombus is 50° and the larger diagonal is 5

centimetres. What is its area?
Ans) In a rhombus, diagonals bisect each other at right angles.

and diagonals bisect the angles of a rhombus.

So, D C

PA = PC = 5 = 2.5 d2 d1
2
P
∠PAB = ∠PAD = 50 ° = 25° 2.5cm
2 2.5cm

Consider Δ APB,

Tan 25° = PB = PB A B
PA 2.5

∴ PB = 2.5 × tan 25°

= 2.5 × 0.4663 =1.16575 = 1.165cm

BD = 2 × PB = 2 × 1.165 = 2.33 cm

ie, The two diagonals are AC = 5 cm , BD = 2.33cm

Area of rhombus = 1 × d1 × d2
2

= 1 × 5 × 2.33
2

= 2.5 × 2.33

= 5.825
≈ 5.83 cm2

Q2) A ladder leans against a wall, with its foot 2 metres away
from the wall and the angle with the floor 40°. How high
is the top end of the ladder from the ground?

Ans) In the figure AC represents length of ladder and BC
represents the height of wall.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Given BC = 2 m A
Wall
Tan 40° = AB B
BC

∴ AB = BC × tan 40° Ladder

AB = 2 × 0.8391 40°
C 2m
= 1.6782
≈ 1.7 Ground

∴ Height of the top end of the ladder from the ground = 1.7m

A working model

This working model help you to
find different values of
trigonometric ratios.

Click to view

Assignment

Q) Find the values of tan 30°, tan 45°, tan 60° without using the
trigonometric table.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 61 17 /11 / 2020

To view class

Answer of last class assignment
Q) Find the values of tan 30°, tan 45°, tan 60° without using the

trigonometric table.

Ans) * Considering right triangle with angles 30°, 60°, 90°

we have

Tan 30° = Opposite Side = 1 30° 2
Adjacent Side √3 60°
√3

Tan 60° = Opposite Side = √3 = √3
Adjacent Side
1

1

* Considering right triangle with angles 45°, 45°, 90°
we have

Tan 45° = Opposite Side = 1 =1 45° √2
Adjacent Side 1 1

45°
1

Q4)

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Ans) In the figure , let OB = x and let BD = DF = FH = y

Tan 40° = AB = AB
OB x

∴ AB = x tan 40°

Tan 40° = CD = CD
OD x+y

∴ CD = (x + y) tan 40°

CD = x tan 40° + y tan 40°

Tan 40° = EF = EF
OF x+2 y

∴ EF = (x + 2y) tan 40°

EF = x tan 40° + 2y tan 40°

Tan 40° = GH = GH
OH x+3 y

∴ GH = (x + 3y) tan 40°

GH = x tan 40° + 3y tan 40°

Sequence of height of lines is

x tan 40°, x tan 40° + y tan 40° , x tan 40° + 2y tan 40° ,

x tan 40° + 3y tan 40° ,........

This is an arithmetic sequence with first term ‘ x tan 40° ’ and
common difference ‘ y tan 40° ’.

Q5)

Ans) Draw AP perpendicular to BC.

Given BC = 6 cm

Let BP = x , then CP = 6 − x

Consider Δ BPA ,

Tan 40° = AP = AP
BP x

∴ AP = x tan 40° ............(1) x 6−x

6

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Consider Δ CPA ,

Tan 65° = AP = AP
CP 6−x

∴ AP = (6 − x) tan 65° ............(2)

From equations (1) & (2) we have,

x tan 40° = (6 − x) tan 65°

x tan 40° = 6 tan 65° − x tan 65°

x tan 40° + x tan 65° = 6 tan 65°

x ( tan 40° + tan 65°) = 6 tan 65°
6 tan 65 °
x= tan 40 ° +tan 65 °

= 6×2.1445
0.8391+2.1445
12.867
= 2.9836

= 4.3126

So, AP = x tan40° = 4.3126 × 0.8391 = 3.6187≈ 3.62
1
Area of triangle = 2 × BC × AP
= 1
2 3

× 6 × 3.62

= 3 × 3.62

= 10.86 cm2

Q3)

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

Ans) Given ABCDE is a regular A

pentagon with side 30 cm. 30cm 108° 30cm
Consider ΔACD

It is an isosceles triangle. B 108° E

Draw AP perpendicular to CD. R Q
30
So ,CP = PD = 2 = 15cm 30cm 30cm

Each angle of a regular pentagon

is 108°. 108° D
Consider Δ AED
It is an isosceles triangle. C 15cm P 15cm
30cm

Draw EQ perpendicular to AD. EQ bisects ∠E.
108 °
Since ∠E =108° , ∠AEQ = ∠DEQ = 2 = 54°

In Δ EQD, ∠QDE = 180° − ( 90° + 54° ) = 180° − 144° = 36°

In Δ APD, ∠ADP = 108° − 36° = 72°

Consider Δ EQD, QE
QE 30
Cos 54° = DE = ∴ QE = 30 × Cos 54°

= 30 × 0.5878

QD QD ≈ 17.63 cm
QE 17.63
Tan 54°= = ∴ QD = 17.63 × Tan 54°

= 17.63 × 1.3764
≈ 24.27 cm

∴ Length of small rectangle = 24.27 cm

Breadth of small rectangle = 17.63 cm

Consider Δ APD,
AP AP
Tan 72° = PD = 15 ∴ AP = 15 × Tan 72°

= 15 × 3.0777

≈ 46.17 cm

∴ Length of large rectangle = 46.17 cm

Breadth of large rectangle = 15 cm

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 5

Assignment

Q1) One angle of a rhombus is 140° and the smallest diagonal is
6cm. What is its area ?

Q2) In Δ ABC, AB = 8cm, ∠A =50°, ∠B = 70°
i) Find the perpendicular distance fro C to AB.
ii) Find the area of the triangle.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 1

Chapter -5
Trigonometry

Online class – 62 19 /11 / 2020
Answer of last class assignment
To view class

Q1) One angle of a rhombus is 140° and the smallest diagonal is

6cm. What is its area ?

Ans) DC
ABCD is the given

rhombus. 3cm O
Given , 3cm

BD = 6cm

So, 6 = 3 cm , A B
OB = OD = 2
∠BOC = 90° Diagonals of a rhombus
bisect each other at right
angles

Consider Δ BOC
OC OC
Tan 70° = OB = 3 ∴ OC = 3 × Tan 70°

AC = 2 × OC = 2 × 3 × Tan 70°
1
Area of rhombus = 2 × BD × AC

= 1 ×6× 2× 3 × Tan 70°
2

= 18 × 2.7475

= 49.455 cm2

Q2) In Δ ABC, AB = 8cm, ∠A =50°, ∠B = 70°
i) Find the perpendicular distance from C to AB.
ii) Find the area of the triangle.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 2

Ans)

i) Given AB = 8cm

Let AD = x, then DB = 8 − x

Draw CD perpendicular to AB.

Consider Δ ADC ,

Tan 50° = CD = CD
AD x

∴ CD = x tan 50° ............(1) x 8 −x
8 cm
Consider Δ BDC ,

Tan 70° = CD = CD
DB 8− x

∴ CD = (8 − x) tan 70° ............(2)

From equations (1) & (2) we have,

x tan 50° = (8 − x) tan 70°

x tan 50° = 8 tan 70° − x tan 70°

x tan 50° + x tan 70° = 8 tan 70°

x ( tan 50° + tan 70°) = 8 tan 70°
8 tan 70°
x= tan 50° + tan 70°

= 8×2.7475
1.1918+ 2.7475
21.98
= 3.9393

= 5.5796 ≈ 5.58 cm

So, CD = x tan 50°

= 5.58 × 1.1918

= 6.65 cm

ie, Perpendicular distance from C to AB = CD = 6.65 cm

ii) Area of triangle = 1 × AB × CD
2
1 × 84× 6.65
= 2

= 4 × 6.65

= 26.6 cm2

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 3

Distances and Heights

Raised View
Angle of elevation

Straight View

Usually our line of vision is parallel to the ground. To see
things a bit high, we have to raise (elevate) this.

The angle between straight view and raised view is called
the angle of elevation.

Q1) A man standing 20 metres away from the foot of a tree sees

its top at an elevation of 40° . His height is 1.7 metres.

What is the height of the tree? (Tan 40° =0.84) A
Ans) In the rough sketch,

BE represents height of man

AD represents height of tree.

Consider Δ ACB,
AC
Tan 40° = BC

Tan 40° = AC 40° C
20 20 m D
B
∴ AC = 20 × tan 40° 20 m
1.7m
= 20 × 0.84 = 16.8 meters
E

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 4

AD = AC + CD = 16.8 + 1.7 = 18.5 m
∴ Height of the tree = 18.5 m

Straight View
Angle of depression

Lowered View

When we stand at a high place, we have to lower (depress)
our sight to see things below.

The angle between straight view and lowered view is called
the angle of depression .

Q) A man, 1.8 metres tall, stands on top of a light house 25

metres high and sees a ship at sea at a depression of 35°.

How far is it from the foot of the light house? B
35° 1.8 m
Ans) In the rough sketch, E
55° C
BC represents height of man,

CD represents height of lighthouse,

AD represents distance of the ship 25 m
from the lighthouse.

A D
Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

11..77mm

Trignomerty 5

In △ ADB , given BC= 1.8, CD = 25

So, BD = BC + CD = 1.8 + 25

= 26.8 m
Given , ∠ EBA = 35° , So, ∠ ABD = 90°− 35° = 55°

Consider Δ ADB, AD
BD
Tan 55° = AD
Tan 55° = 26.8

∴ AD = 26.8 × tan 55°

= 26.8 × 1.4281

= 38.27 meters

∴ Distance of the ship from the lighthouse = 38.27 meters

Q2) A boy standing on the top of a building of height 10 meters,

sees the top of a tower at an elevation of 30° and its bottom

at an angle of depression 60° . Find the height of the tower

and distance of the tower from the building .
E
Ans)

60° In the rough sketch,

BA represents height of building,

A 30° x D CE represents height of tower,
60° AD represents distance of the
tower from the building.

10 m From figure,

BA = CD = 10 m

30° Given ∠DAC = 60°

BC So, ∠ DCA =180°−(90°+ 60°) = 30°

Consider Δ ADC, the angles are 30°, 60°, 90° 30° , 60° , 90°
Let AD = x , then CD = √3x 1 : √3 : 2
x √3x

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trignomerty 6

But CD = 10 ∴ √ 3 x = 10
10
x=
√3
10
ie, AD = meters
√3

Given, ∠ DAE = 30°,
So, ∠AED = 180°−(90°+ 30°) = 60°

Consider Δ ADE, the angles are 30°, 60°, 90°

Since AD = x , 30° , 60° , 90°
1: √3 : 2
DE = x x x
√3
√3
10
= ÷√3
√3
10 1
= ×
√3 √3
10
= 3 √3 × √3 = 3

= 3.33 meters

∴ Height of the tower = CE

= CD + DE

= 10 + 3.33

= 13.33 meters

Distance of the tower from the building = AD
10
= meters
√3

Assignment
T. B Page 122

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 1

Chapter -5
Trigonometry

Online class – 63 20 /11 / 2020

Answer of last class assignment To view class
T. B Page 122

Ans) Draw a rough figure as shown. C
In the figure,
BC represents the height of tree, Tree
AB represents the length of the shadow,

In △ ABC, ∠A =40° , AB = 18 m

BC 40° Shadow
AB
Tan 40° = A 18 m B

Tan 40° = BC
18

∴ BC = 18 × tan40°

= 18 × 0.84 = 15.12 m

So, Height of the tree = 15.12 m

Clinometer
Clinometer is an instrument used for measuring angles of
slope , elevation, or depression of an object with respect to
gravity's direction.

How to make clinometer ? , watch this video

How to use clinometer ? , watch this video

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 2

Q1) In the given figure ∠B =90° , AC = 50 metres ,∠DCB = 60° ,

∠DAB = 30° . Find the length of BC and BD .

Ans) D
Consider △ACD

∠DCB is an exterior angle.
In any triangle,

the outer angle at a vertex

is equal to the sum of the

inner angles at the other

two vertices. A 50 m C xB
∴ ∠ADC + ∠DAC =∠DCB

So, ∠ ADC = ∠ DCB −∠ DAC

= 60° − 30° = 30°

Here ∠ADC = ∠DAC , So △ACD is an isosceles triangle.
Sides opposite to equal angles are equal.

∴ AC = CD = 50 metres

Consider Δ DBC, the angles are 30°, 60°, 90° 30° , 60° , 90°
Let BC = x , then BD = √3x , CD =2x 1 : √3 : 2
x √3x 2x

We have CD = 50 m ,

So, 2x = 50
50
∴x= 2 = 25

ie. BC = 25 metres

BD = √3x

= √3 × 25 = 25√3 metres

Q2) Manu 1.5m tall, sees the top of a tree at a distance, at
an angle of elevation 40° . Stepping back 18 meters he sees
its top at an elevation of 20°. What is the height of
the tree? ( Sin 40° = 0.64)

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 3

Ans) Draw a rough figure as shown. A
In the figure,

AG represents height of tree,

DE = CF= BG =1.5m represents

height of Manu,

Given, DC = 18 m, ∠ADC= 20° , D 18 m C B
∠ACB = 40°
1.5 m 1.5 m 1.5 m

E F G

∠ACB is an outer angle of ∆ACD.
In any triangle, the outer angle at a vertex is equal
to the sum of the inner angles at the other two vertices.

∴ ∠ CAD + ∠ADC =∠ACB

∠ CAD = ∠ ACB −∠ADC
So, ∠CAD = 40°− 20° = 20°

Here ∠ADC = ∠DAC , So △ACD is an isosceles triangle.
Sides opposite to equal angles are equal.

∴ AC = CD = 18metres

Consider ∆ABC, AB
AB 18
Sin 40° = AC =

So, AB = 18 × sin40 °

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 4

= 18 × 0.64
= 11.52 m
∴ AG = AB + BG
= 11.52 + 1.5 = 13.02 m
So, Height of the tree = 13.02 m

Q3) A boy 1.5 m tall, sees the top of a D

hill at a distance at an angle of

elevation 20°. Stepping back a

distance of 100 m , he sees it at A

an angle of elevation 10°. Find

the height of the hill.

(sin20° = 0.342)

Ans)

Draw a rough figure as shown. D

In the figure,

DG represents the height of hill.

AE = BF = CG = 1.5 m represents

height of the boy.

Given AB = 100 m ,∠DAB= 10° , A 100 m B C
∠DBC = 20° 1.5 m
1.5 m 1.5 m

∠DBC is an outer angle of ∆ABD. E F G

In any triangle, the outer angle at a vertex is equal

to the sum of the inner angles at the other two vertices.

∴ ∠ADB = 20°− 10° = 10°

Here ∠ADC = ∠DAC , So ∆ADB is an isosceles triangle.
Sides opposite to equal angles are equal.

∴ AB = DB = 100 m

Consider ∆DCB,
DC DC
Sin 20° = DB = 100

So, DC = 100 × sin20 °

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 5

= 100 × 0.342
= 34.2 m
∴ DG = DC + CG
= 34.2 + 1.5 = 35.7 m
So, Height of the hill = 35.7 m

Assignment
Q) 1.5 m tall Manu is standing at the edge of a river sees the

top of the tree on the opposite edge at an angle of elevation
30°. Stepping back 50 m, he sees its top at an angle of
elevation 15°. Find the height of the tree and width of the
river.

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 1

Chapter -5
Trigonometry

Online class – 64 23 /11 / 2020

To view class

Answer of last class assignment

Q) 1.5 m tall Manu is standing at the edge of a river sees the

top of the tree on the opposite edge at an angle of elevation

30°. Stepping back 50 m, he sees its top at an angle of

elevation 15°. Find the height of the tree and width of the

river.

Ans) Draw a rough figure as shown.

In the figure,

CG represents the height of the

tree, Tree

DF = AE = BG = 1.5 m represents 50 m

height of the boy. 1.5 m 1.5 m 1.5 m

AB represents the width of the

river.
∠CAB is an outer angle of ∆DAC.

So, ∠DCA = 30°− 15° = 15°

∆DAC is an isosceles triangle.

So, DA = CA = 50m

Consider ∆ABC, angles are 30° , 60° , 90°

So sides are in the ratio 1 : √3 : 2 30° , 60° , 90°
Since CA = 50m , CB = 25, AB = 25√3
1 : √3 : 2
∴ CG = CB + BG 25 25√3 50

= 25 + 1.5 = 26.5 m

So, Height of the tree = 26.5 m

Width of the river = 25√3 = 25 × 1.732 =43.3m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 2

T.B Page 122
Q2)When the sun is at an elevation of 35° , the shadow of a tree

is10 metres. What would be the length of the shadow of the
same tree, when the sun is at an elevation of 25° ?
(tan 65° = 2.14, tan 35° = 0.70 )

)25° )35° 10m

Ans)

Draw a rough figure as shown.

DC represents the height of tree

65° BC represents the length of the
shadow at an elevation of 35°,

AC represents the length of the

shadow at an elevation of 25°.

10m

Consider △ BCD, given ∠B =35° & BC= 10 m
DC
Tan 35° = 10

∴ DC = 10 × tan35°

= 10 × 0.70 = 7.0m

Consider △ ACD, OR DC 7
∠ADC = 90° − 25° = 65° AC AC
Tan 25° = =

Tan 65° = AC = AC AC × Tan 25° = 7
DC 7
7 7
∴ AC = 7 × tan65° ∴ AC = tan 25 ° = 0.46

= 7 × 2.14 = 14.98≈15 = 15

So, length of the shadow at an elevation of 25° = 15 m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 3

Q3)From the top of an electric post, two wires are stretched to

either side and fixed to the ground, 25 metres apart. The

wires make angles 55° and 40° with the ground. What is the

height of the post?

Ans) Draw a rough figure as

shown.

AB is the post, AC and BD

are the wires stretched to h
the ground.

Height of post = AB = h x 25 m 25 - x
Let CB = x , then BD = 25 − x

In ΔABC In ΔABD
h h
Tan 55° = x Tan 40° = 25 − x

1.43 = h 0.84 = h
x 25 − x

h = 1.43 × x.....(1) h = 0.84(25 −x).....(2)

Comparing equations (1) & (2)

1.43 × x = 0.84(25 −x)

1.43 x = 21 – 0.84 x

1.43 x + 0.84 x = 21

2.27 x = 21 21.00 2100
21 2.27 227
x= 2.27 = = = 9.25 m

h = 1.43 x = 1.43 × 9.25 = 13.2275 ≈ 13.23 m

So, height of the post = 13.23 m

Assignment

T.B Page122

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 1

Chapter -5
Trigonometry

Online class – 65 24 /11 / 2020

To view class

Answer of last class assignment
T.B Page 122
Q4) A 1.5 metre tall boy saw the top of a building under

construction at an elevation of 30° . The completed building
was 10 metres higher and the boy saw its top at an elevation
of 60° from the same spot. What is the height of the
building?
10 m

60°

)30°

1.5 m

Ans) Draw a rough figure as given below. F

AB represents the boy, CF represents 10 m

the height of the building. E

AB = CD = 1.5m 10 m
In Δ BDE , ∠D = 90° , ∠EBD = 30°
∠BEF is an outer angle of this triangle,
So, ∠BEF =90° + 30° = 120°.

In Δ BEF ,∠BFE =180° −( 120°+ 30°) = 30° B D
1.5 m
Δ BEF is isosceles triangle. 1.5 m
C
Given EF = 10m , So , BE= 10m A

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 2

In Δ BDE three angles are 30°, 60°, 90° 30° , 60° , 90°

So, sides are in the ratio 1 : √3 : 2 1 : √3 : 2
10 5 10
Since BE =10, DE = 2 =5

∴ Height of the building = CF = CD + DE + EF

= 1.5 + 5 + 10 = 16.5 m

Q5) A 1.5 metre tall man, standing at the foot of a tower, sees
the top of a hill 40 metres away at an elevation of 60° .
Climbing to the top of the tower, he sees it at an elevation of
50°.Calculate the heights of the tower and the hill.

50°

1.5 m

60°
1.5 m

40 m

Ans) Draw a rough figure as given below.

D BA is the tower, DC is the hill, AF = BE

represents the man.

Given AC = 40m , So, FG = EH = 40m
DH
E In Δ DHE, Tan 50° = 40
1.5 m
40 m H ∴ DH = 40 × Tan 50°
B

= 40 × 1.20 = 48
DG
In Δ DGF, Tan 60° = 40

40 m G ∴ DG = 40 × Tan 60°
40 m C
F = 40 × 1.732 = 69.28
1.5 m
≈ 69.3 m
A

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 3

So, Height of the hill = DC = DG + GC
= 69.3 + 1.5
= 70.8 m

Height of tower = AB = FE = GH
= DG − DH
= 69.3 − 48 = 21.3 m

Q5)A man 1.8 metre tall standing at the top of a telephone
tower, saw the top of a 10 metre high building at a
depression of 40° and the base of the building at a
depression of 60° . What is the height of the tower? How
far is it from the building?

40°
60°

10 m

Ans) Draw a rough figure as given below.

A F BG represents the tower, CD
1.8 m represents the building, AG represents

G

h the man .
Given CD = 10m , So, BE = 10m

Let BC = x, then ED = x

E x D Let AE = h
10 m x
In Δ AED, h
B x
10 m Tan 40° =

C ∴ h = x tan 40°........(1)

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi

Trigonomerty 4

In Δ ABC , h+10
x
Tan 60° =

∴ h + 10 = x tan 60°........(2)

Substituting equation (1) in equation (2)

x tan40° + 10 = x tan60°

10 = x tan60° − x tan40°

10 = x ( tan60° – tan40° )

10 = x ( 1.7321 − 0.8391)

10 = x × 0.893

x × 0.893 = 10
10
∴ x= 0.893

= 10000
893

= 11.23
∴ Distance between the tower and the building = 11.23 m

h = x tan 40°.
= 11.23 × 0.8391
= 11.23 × 0.84 = 9.43

Height of the tower = BG = BE + EG
= 10 + ( h – 1.8 )
= 10 + 9.43 – 1.8
= 19.43 – 1.8
= 17.63 m

Cecilia Joseph , St. John De Britto’s A.I.H.S , Fortkochi