= p h h2 – p2 h2 [ By inserting h under square root] = p h h2 h2 – p2 h2 [ By separating h2 for numerators] = sin θ 1 – sin2 θ [ By using trigonometric ratios] Example-7 In the ∆ABC, ∠ABC = 90°, ∠ABC = θ, AC = h cm, AB = b cm and BC = p cm, prove that: sin2 θ + cos2 θ = 1 Solution: Here, in the ∆ABC, ∠ABC = 90°, ∠ABC = θ, AC = h cm, AB = b cm and BC = p cm, To prove: sin2 θ + cos2 θ = 1 Now, we have sin θ = p h, cos θ = b h and tan θ = p b LHS = sin2 θ + cos2 θ = p2 h2 + b2 h2 = p2 + b2 h2 = h2 h2 [ By Pythagorean theorem, p2 + b2 = h2 ] = 1 Hence, sin2 θ + cos2 θ = 1 B C A b cm h cm p cm Trigonometry Trigonometry 400 Allied The Leading Mathematics-9 Trigonometry 401
PRACTICE 20.1 Keeping Skill Sharp 1. Given the triangle ABC, ∠B = 90°, (a) Which side is the base for the angle A? (b) Which side is the base for the angle C? (c) Which side is the base for the angle B? 2. (i) ∆PQR is a right-angled triangle at Q. In each of the following, write down the following trigonometric ratios: sin P, cos R and tan P. (a) PQ = QR = 1, PR = 2 (b) PQ = 5, QR = 12, PR = 13 (c) PQ = 2, PR = 4, QR = 2 3 (ii) Given ∆LMN, ∠ M = 90o , (a) sin N = 15 17, h = 5; find p. (b) cos N = 8 17, b = 3; find h. (c) tan L = 8 15, p = 4; find b. 3. From the following figures, find the values of sin θ, sin a, cos a and tan θ: (a) (b) (c) (d) 4. (a) If sin θ = 4 5, find the values of cos A and tan A. (b) If cos A = 5 13, find the values of sin A and tan A. (c) If tan A = 12 5 , find the values of sin A and cos A. 5. Find the value of; (a) sin2 A + cos2 A (b) 1 – cos2 θ (c) sin A ÷ cosA (d) sec2 θ – tan2 θ (e) cosec2 A – cot2 A (f) 1 + tan2 θ 6. Prove that: (a) sec2 A – tan2 A = 1 (b) 1 – sin2 θ = cos2 θ (c) cot A = cos A sin A (d) sec2 θ – 1 = tan2 θ (e) cosec A . sin A = 1 (f) cos A = 1 sec A 7. If sin A = 4 5, prove that: (a) cosec2 A – cot2 A = 1 (b) 1 – cos2 A = sin2 A (c) 1 + tan2 A = sec2 A B C A A B α D C θ Q R S P θ α M L K N α θ X Y W Z α θ Trigonometry Trigonometry 402 Allied The Leading Mathematics-9 Trigonometry 403
8. If tan A = 5 13, prove that: (a) sec A. cos A = 1 (b) cosec2 A – cot2 A = 1 (c) sin2 A + cos2 A = 1 9. From the following figures, find the values of sin θ, cos θ and tanθ: (a) (b) (c) (d) 10. From the following figures, find the values of sin θ, cos θ and tan θ: (a) (b) 11. (a) Express sin θ into cos θ. (b) Express sin A into tan A. 12. (a) A vertical pole of 16 ft is supported by a wire of length 20 ft. What is the distance between the bottom of the pole and base of wire? If the wire makes an angle θ with the ground, find the values of sin θ, cos θ and tan θ. (b) A 25 m tall tower castes 60 m long shadow. How long wire will need to support the tower ? If the wire makes an angle φ with the ground, find the values of sin θ, sec θ and cot θ. 1. (a) BC (b) AB (c) None 2. (i) (a) 1 2 , 1 2 , 1 (b) 12 13, 12 13, 12 5 (c) 3 2 , 3 2 , 3 (ii) (a) 75 17 (b) 51 8 (c) 15 2 3. (a) AD BD, BD BC, CD BC, AD AB (b) QR PR, PR PS, RS PS, QR PQ (c) KN MN, LM MK, KL MK, KN KM (d) YZ XY, WX XY, WX XY, YZ XZ 4. (a) 3 5 , 4 3 (b) 12 13, 12 5 (c) 12 13, 5 13 5. (a) 1 (b) sin2 θ (c) tan A (d) 1 (e) 1 (f) sec2 θ 9. (a) 5 13, 12 13, 5 12 (b) 5 13, 12 13, 5 12 (c) 3 5 , 4 5 , 3 4 (d) 3 5 , 4 5 , 3 4 10. (a) 2 3 , 7 3 , 2 7 (b) 3 5 , 4 5 , 3 4 12. (a) 1 – cos2 θ (b) tan A 1 + tan2 A Answers A B C 4 cm D 3 cm 12 cm θ θ Q R S P 26 cm 8 cm 6 cm A B D 24 cm C 20 cm 7 cm θ K L M θ N 29 cm 15 cm 21 cm B C D A 4 cm 3 cm θ 3 cm Q S R T P 12 cm 12 cm 3 cm θ 4 cm Trigonometry Trigonometry 402 Allied The Leading Mathematics-9 Trigonometry 403
20.2 Trigonometric Ratios of Standard Angles At the end of this topic, the students will be able to: ¾ find the trigonometric ratios of the standard angles 0°, 30°, 45°, 60° and 90°. Learning Objectives In an equilateral triangle ∠A = ∠B =∠ C = 60° AD⊥ BC, then ∠BAD = ∠CAD = 30° In right-angled ΔABC, ∠C = 90° and AC = BC then ∠A = ∠B = 45° Hence within the limit 0° ≤ θ ≤ 90°, the angles 30°, 45° and 60° occur frequently. In this list we include 0° and 90°. In this lesson, we will learn the trigonometric ratios of the special angles 0°, 30˚, 45˚, 60˚, and 90˚ and how to use them to find exact values of trigonometric expressions without a calculator. Ratios of 60° and 30° In ΔABC, AB = BC = CA = 2a (suppose). Then ∠A = ∠B = ∠C = 60°. Draw AD⊥BC. Then ∠BAD = ∠CAD = 30° and BD = DC = a. Now, in ΔABD, ∠B = 60°, AB = 2a, BD = a. By Pythagoras’ theorem, AB2 = AD2 + BD2 or, AD2 = AB2 – BD2 = (2a)2 – (a)2 = 3a2 ∴ AD = 3 a. Now, taking ∠ABD = 60° as the angle of reference, we have sin 60° = AD AB = 3a 2a = 3 2 = 0.866; cos 60° = BD AB = a 2a = 1 2 = 0.5; tan 60° = AD BD = 3a a = 3 = 1.732 Again, taking ∠BAD = 30° as the angle of reference, we have sin 30° = BD AB = a 2a = 1 2 = 0.5; cos 30° = AD AB = 3a 2a = 3 2 = 0.866; tan 30° = BD AD = a 3a = 1 3 = 0.577 Trigonometric Ratios of 45° In ΔABC, ∠B = 90° and AB = BC = a (suppose), then ∠A = ∠C = 45° (Why?) Now, AC2 = AB2 = BC2 = a2 + a2 = 2a2 ∴ AC = 2 a Now, taking ∠A = 45° as the angle of reference, we have sin 45° = BC AC = a 2a = 1 2 = 0.707; cos 45° = a 2a = 1 2 = 0.707; tan 45° = BC AB = a a = 1 B C A B D A C B a D a 2a 2a A C 2a B C A a a Trigonometry Trigonometry 404 Allied The Leading Mathematics-9 Trigonometry 405
Trigonometric Ratios of 0° In ΔABC, ∠C = 90° and ∠B = θ be very small When θ = 0° (nearly) then AC = 0, and AB = BC Hence, taking ∠B = θ = 0° as the angle of reference, sin 0° = AC AB = 0 AB = 0; cos 0° = BC AB = BC BC = 1; ten 0° = AC BC = 0 BC = 0 Trigonometric Ratios of 90° In ΔABC, ∠C = 90° and ∠B = θ (near to 90°). When θ = 90° nearly. Then, BC = 0 and AB = AC Now, taking ∠B = θ = 90° as the angle of reference, we have sin 90° = AC AB = AC AC = 1; cos 90° = BC AB = 0 AB = 0; tan 90° = AC BC = AC 0 = Undefined/infinity (∞). Now, making a table of the values of trigonometric ratios of certain angles 0°, 30°, 45°, 60° and 90°, we have Angle θ 0° 30° 45° 60° 90° Rule Ratio 0 4 1 4 2 4 3 4 4 4 sin θ 0 1 2 1 2 3 2 1 cos θ 1 3 2 1 2 1 2 0 tan θ 0 1 3 1 3 ∞ Example-1 Evaluate: (a) sin 60° (1 + cos 60°)(b) sin2 30° + cos2 30° (c) 2 tan 30° 1 + tan2 30° Solutions: Here, (a) sin 60° (1 + cos 60°) = 3 2 1 + 1 2 = 3 2 2 + 1 2 = 3 2 × 3 2 = 3 3 4 . (b) sin2 30° + cos2 30° = 1 2 2 + 3 2 2 = 1 4 + 3 4 = 4 4 = 1. (c) 2 tan 30° 1 + tan2 30° = 2 1 3 1 + 1 3 2 = 2 3 1 + 1 3 = 2 3 3 + 1 3 = 2 3 × 3 4 = 2 3 × 3 × 3 4 = 3 2 . θ B C A B C a θ A Trigonometry Trigonometry 404 Allied The Leading Mathematics-9 Trigonometry 405
Example-2 Verify that: (a) sin 60° = 2 sin 30°.cos 30° (b) cos 60° = cos2 30° – sin2 30° (c) sin 90° = sin 30°.cos 60° + cos 30°.sin 60° Solution: Here, (a) LHS = sin 60° = 3 2 and RHS = 2 sin 30° cos 30° = 2 . 1 2 . 3 2 = 3 2 Hence, LHS = RHS. Proved. (b) LHS = cos 60° = 1 2, and RHS = cos2 30° – sin2 30° = 3 2 2 – 1 2 2 = 3 4 – 1 4 = 3 – 1 4 = 2 4 = 1 2 Hence, LHS = RHS. Proved. (c) LHS = sin 90° = 1, and RHS = sin 30° . cos 60° + cos 30° . sin 60° = 1 2 . 1 2 + 3 2 . 3 2 = 1 4 + 3 4 = 3 + 1 4 = 4 4 = 1. Hence, LHS = RHS. Proved. Example-3 Prove that: (a) cos 60o = 1– 2sin2 30o (b) tan 60° – tan 30° 1 + tan 60°.tan 30° = tan 30o (c) 1 + tan 30° 1 – tan 30° = 1 + sin 60° 1 – sin 30° Solutions: Here, (a) LHS = cos 60o = 1 2 RHS = 1 – 2 sin2 30° = 1 – 2 × 1 2 2 = 1 – 2 × 1 4 = 1 – 1 2 = 2 – 1 2 = 1 2 Hence, LHS = RHS. Proved. (b) LHS = tan 60° – tan 30° 1 + tan 60°. tan 30° = 3 – 1 3 1 + 3 × 1 3 = 3 – 1 3 1 + 1 = 2 3 2 = 2 3 × 1 2 = 1 3 = tan 30o = RHS. proved. (c) LHS = 1 + tan 30° 1 – tan 30° = 1 + 1 3 1 – 1 3 = 3 + 1 3 3 – 1 3 = 3 + 1 3 × 3 3 – 1 = 3 + 1 3 – 1 = 3 + 1 3 – 1 × 3 + 1 3 + 1 Trigonometry Trigonometry 406 Allied The Leading Mathematics-9 Trigonometry 407
= 3 + 2 3 + 1 3 – 1 = 4 + 2 3 2 = 2(2 + 3) 2 = 2 + 3 Again, RHS = 1 + sin 60° 1 – sin 30° = 1 + 3 2 1 – 1 2 = 2 + 3 2 2 – 1 2 = 2 + 3 2 × 2 1 = 2 + 3 Hence, LHS = RHS. Proved. Example-4 If A and B are acute angles, sin (A – B) = 1 2 and cos (A + B) = 1 2, find the values of A and B. Solution: Here, sin (A – B) = 1 2 or, sin (A – B) = sin 30o and so A – B = 30° .............. (i) cos (A + B) = 1 2 or, cos (A + B) = cos 60o and so A + B = 60° ............. (ii) By adding the eqn (i) and (ii), we get A – B = 30° A + B = 60o 2A = 90o ∴ A = 45o ......................... (iii) Using the eqn (iii) in (ii), we get 45° + B = 60° or, B = 60° – 45° Hence, A = 45° and B = 15°. Example-5 The hypotenuse of a right-angled triangle is 8 cm and one of its acute angle is 30°. Find the remaining angle and sides. Solution: Let ABC be the given triangle with hypotenuse AC = 8 cm and ∠ACB = 30°. To find AB, BC and ∠CAB. Since, A + C = 90° or, A + 30° = 90°, So, A = 90° – 30° = 60° By definition, AB AC = sin 30° or, AB 8 = 1 2 So, AB = 1 2. 8 = 4 cm. Also, BC AC = sin 60° or, BC 8 = 3 2 . ∴ BC = 8 × 3 2 = 4 3 cm. 30° C 8 cm B A Trigonometry Trigonometry 406 Allied The Leading Mathematics-9 Trigonometry 407
PRACTICE 20.2 Keeping Skill Sharp 1. Without using tables or calculator, evaluate: (a) sin 30o + cos 30o + tan 30o (b) sin 0° + cos 0° + tan 0° (c) 2sin 30°. cos 30° (d) 2 tan 30° 1 – tan2 30° (e) sin 30°.cos 45° + cos 30°.sin 45° (f) cos 30°.cos 45° – sin 30°.sin 45° 2. Prove the following relations: (a) sin 45°.cos 30° – cos 45°. sin 45° = 6 – 2 4 (b) sin 60°.cos 30° – cos 60°. sin 30° = 1 2 (c) 2 tan 30° 1 – tan2 30° = 3 (d) cos 60° = 2cos2 30° – 1 (e) tan 60° – tan 30° 1 + tan 60°.tan 30° = 1 3 (f) sin 30°.cos 60° + sin 30°.cos 60° = sin 90° 3. (a) If sin θ = 1 2, find cos θ and tan θ. (b) If 2sin θ = 3, find cos θ and tan θ. (c) If cos θ = 1, find sin θ and tan θ. (d) If sin θ = cos θ, find sin θ and cos θ. 4. Verify that: (a) tan 60° = 2 tan 30° 1 – tan2 30° (b) cos 60° = 2 cos2 30° – 1 (c) cos 90° = cos 30°.cos 60° – sin 30°.sin 60° (d) 1 – sin 60° cos 60° = – 1 – tan 30° 1 + tan 30° 5. (a) The hypotenuse of a right-angled triangle is 8 cm and one of its side angles is 60°. Find the remaining angle and sides. (b) The hypotenuse of a right-angled triangle is 8 2 cm and one of its side angles is 45°. Find the remaining angle and sides. 2. (a) 5 + 3 2 3 (b) 1 (c) 3 2 (d) 3 (e) 1 + 3 2 2 (f) 3 – 1 2 2 3. (b) 3 2 , 1 3 (b) 1 2 , 3 (c) 0, 0 (d) 1 2 , 1 2 5. (a) 30°, 4 3 cm, 4 cm (b) 45°, 8 cm, 8 cm Answers Project Work Draw a right-angled triangle and label it. Identify the special name of its sides by taking reference angle. Also, draw the table of the values of the trigonometric ratios of the standard angles 0°, 30°, 45°, 60° and 90° with rules. Trigonometry Trigonometry 408 Allied The Leading Mathematics-9 Trigonometry 409
CONFIDENCE LEVEL TEST - VII Unit VII : Trigonometry Class: 9, The Leading Maths Time: 45 mins. FM: 20 Attempt all the questions. 1. In the figure, PQR is a right-angled triangle with ∠PQR = 90° and ∠PRQ = 30°. (a) Write the value of cos 30°. [1] (b) Prove: sin2 30° + cos2 30° = 1 [1] (c) Find the length of QR. [1] (d) For PQ = PR, how much the value of ∠PRQ is increased or decreased? Calculate it. [1] 2. On the basis of the given triangle, answer the following questions. (a) Examine the length of QR. [1] (b) Examine the length of PQ. [1] (c) Interpret the value of tan θ. [1] (d) Prove that: sin2 30° = 1 – cos2 30° [1] 3. In the right angled triangle ACB, ∠ABC = θ be the reference angle. (a) Write the trigonometric ratio of sin θ. [1] (b) If θ = 60°, and BC = 1 unit, find the value of AC. [1] (c) Prove that tan2 60° + cos2 60° = 13 4 [1] (d) For what value of θ, AB is double of AC? [1] 4. (a) Write the values of tan 30° and sin 30°. [1] (b) Write the value of 1 + tan 30° 1 – tan 30°. [1] (c) Write the value of 1 + sin 60° 1 – sin 30°. [1] (d) Are the values obtained from the numbers (b) and (c) equal ? [1] Best of Luck 30° P 1 unit R Q P 13 cm 4 3 cm cm Q O q R A B C q Trigonometry Trigonometry 408 Allied The Leading Mathematics-9 Trigonometry 409
Additional Practice – VII 1. In the given, figure ∠ABC = ∠ACD = 90°, ∠ACB = α and ∠ADC = θ. Also, AD = 13 cm, BC = 4 cm and CD = 12 cm is given. (a) Which trigonometric ratio is equal with AC AB ? (b) Find the length of AC. (c) Find the value of sin q. (d) If ABC is a right-angled isosceles triangle, what would be the value of α in degree? Write with reason. 2. In the given, figure ∠ACB = ∠BDC = 90° and ∠CAB = θ. Also, BD = 3 cm, CD = 4 cm and AC = 12 cm is given. (a) Which trigonometric ratio is equal with BC AB? (b) Find the length of BC. (c) Find the value of tan θ. (d) Is the value of sin θ = 5 13? Why? 3. The hypotenuse of a right-angled triangle is 8 cm and one of its acute angle is 30°. (a) Write the relation of Pythagoras theorem. (b) Find the length of BC. (c) Prove that sin2 30° + cos2 30° = 1 (d) For what angle of ∠ABC, AC is equal to BC? 4. In the right-angled triangle ACB, the value of sin θ is 3 6. (a) Write the trigonometric ratio of sin θ. (b) Find the value of cos θ. (c) Is the opposite side of reference angle shorter than other? Give reason. (d) Prove that: sin2 θ + cos2 θ = 1. 5. In a right-angled triangle PQR, the ∠PRQ = α is the referenced angle. (a) Which trigonometric ratio is denoted by the ratio of PQ PR ? (b) What is the value of PQ PR 2 + QR PR 2 ? Find it. (c) If α = 45° and PR = 10 unit, find the length of base RQ. (d) If PR = 2PQ, what would be the value of α in degree? 1. 5 cm, 3 5 , 45° 2. 5 cm, 5 12 3. 4 3, 45° 4. 3 2 5. 5 2, 30° Answers A B C 13 cm 12 cm 4 cm D q a A B 4 cm 3 cm 12 cm C D q 30° A 8 cm B C 3 cm q A 6 cm B C P R Q a Trigonometry Trigonometry 410 Allied The Leading Mathematics-9
PB Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 411 I. PRACTICE QUESTION SET According to Latest Specification Grid Class: 9, The Leading Maths Time: 3 hrs. FM: 75 Attempt all questions. 1. If universal set U = {x: x is a natural number, x ≤ 10}, A = {1, 2, 4, 6}, B = {4, 5, 6, 7, 8} and C = {8, 9, 10}. Answer the given questions. (a) How many elements does (A ∪ C) contains? [1] (b) Show the given information in Venn diagram. [1] (c) What type of sets are A and C? [1] (d) Show that n(A ∪ B) = n(A) + n(B) – (A ∩ B) [3] 2. The marked price of a motorbike is Rs. 350,000 and 13% VAT is levied after allowing 5% discount. (a) Write the formula to find the VAT amount. [1] (b) How much should a customer have to pay to buy motorbike? Find it. [2] (c) If 8% discount was allowed instead of 5%, how much should have to pay by the customer? Find it. [2] 3. The monthly salary of an employee working in a shop is Rs. 25,000. He receives 1.5% commission on the total sales more than Rs. 15,00,000. (a) State the definition of commission. [1] (b) If the total sales of items in a month is Rs. 25,00,000, find the commission amount. [1] (c) How much should be the sales of item in the next month in order to make his total income Rs. 49,000? [2] 4. Rambhajan bought 500 share out of 50000 shares sold by a hydropower company. When the company distributed 25% of its net profit, he received Rs. 43850 as his dividend in a year, (a) Write the formula to calculate the dividend rate. [1] (b) What is the profit of 1 share? [1] (c) Find the net profit of the company. [2] 5. The sides of a triangular field are 51 m, 25 m and 74 m. Answer the given questions. (1 Aana = 31.80 m2 ) (a) Write the formula to calculate the area of a scalene triangle. [1] (b) Calculate the area of the field. [2] (c) Find the area of the field in Aana. [1] (d) If one Aana of the land costs Rs. 7,50,000 then find the total cost of the field. [1]
412 Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 413 6. Study the given figure and answer the following: (a) Calculate the lateral surface area of the solid. [2] (b) Find the volume [2] 7. A cylindrical tank has diameter 2.8 m and height 3 m. (a) Write the formula to calculate the curved surface area of cylinder. [1] (b) How many liters of water is required to fill the tank completely? [3] 8. Answer the following questions. (a) Write the formula to find the general term of AP. [1] (b) Find the common difference of given sequence 5, 9, 13, 17, ... [1] (c) The fourth term of GP whose first term is 2 is 54. Find its common ratio. [2] 9. Factorize the given algebraic expression given below. (a) x4 + x2 y2 + y4 [3] (b) x2 – 12x – 28 + 16y – y2 [3] 10. From the given algebraic expressions x2 – 9, x3 – 27 and x2 + x – 12, answer the following questions. (a) What are the factors of x2 – 9? [1] (b) What are the factors of x3 – 27? [1] (c) Find the LCM and HCF of given expression and represent in Venn diagram. [3] 11. Solve the given problem (a) Simplify: 7x + 2 – 9 × 7x 8 × 7x [2] (b) Simplify: xl xm l 2 + lm + m2 × xm xn m2 + mn + n2 × xn xl n2 + nl + l 2 [3] 12. Answer the given questions. (a) Find the values of x and y. [2] (b) Draw two triangles PQR and produce QR to X then examine 'the exterior angle of a triangle is equal to the sum of two opposite interior angle' experimentally? Also write the conclusion of above experiment. [3] A A' B B' C C' 20 cm 10 cm A D B E C 60° 65° x° y°
412 Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 413 13. Study the given figure and answer the given questions. (a) Name the two parallel line. [1] (b) Write any one condition of similarity for two triangles. [1] (c) Find the value of AP. [3] 14. In the given figure, BO = 8 cm and OD = x + 3 cm. (a) Write the relation of diagonals of a parallelogram. [1] (b) Find the value of x. [1] (c) Look at the given parallelogram PQRS and answer the following: (i) Prove that: ∆PSR ≅ APQR. [3] (ii) Write the relation between triangle PSR and parallelogram PQRS. [1] 15. On the basis of the given triangle, answer the following questions. (a) Calculate the length of BC. [1] (b) Find the length of OC. [1] (c) Interpret the value of cos q. [1] (d) Prove that: sin2 60° = 1 – cos2 60° [1] 16. From the table given below, answer the following questions. Marks 25 30 35 40 45 50 Frequency 4 7 10 14 3 2 (a) Find the the mean of the above data. [1] (b) Find median its. [1] (c) Compute the mode of the given data. [1] Best of Luck P A B 3 cm 6 cm 4 cm 12 cm Q R A D B O C P R Q S A 13 cm 5 cm 3 cm B O q C
414 Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 415 III. Project and Practical Work Evaluation Sheet Practical Recorded Form School’s Name and Address: ………………………………………………………. Academic Year: 208 …. Subject: The Leading Mathematics Class: 9 RN Student’s Name Conducting experimental and project work based on main learning achievements: (8) Conducting and documenting experimental and project work: (5) Drawing - Naming - Character Description - Spotting and Oral Examination: (3) Total Marks Remarks 123456789101112 …………………… …………………….. ………………………. Signature of student Signature of parents Signature of class teacher II. Classroom Individual Form Student’s Activities Form Subject: The Leading Mathematics Class: 9 Student’s Name: …………………………............................................ Roll No. ……. SN Unit Chapter/ Lesson Practice/ Exercise Participation (3 Marks) (1.5) (1.5) Attendance Class work Home work Classroom Activities 12345678910111213 …………………… …………………….. ………………………. Signature of student Signature of parents Signature of class teacher
414 Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 415 III. Project and Practical Work Evaluation Sheet Practical Recorded Form School’s Name and Address: ………………………………………………………. Academic Year: 208 …. Subject: The Leading Mathematics Class: 9 RN Student’s Name Conducting experimental and project work based on main learning achievements: (8) Conducting and documenting experimental and project work: (5) Drawing - Naming - Character Description - Spotting and Oral Examination: (3) Total Marks Remarks 123456789101112 …………………… …………………….. ………………………. Signature of student Signature of parents Signature of class teacher
416 Allied The Leading Mathematics-9 Allied The Leading Mathematics-9 PB IV. Internal Evaluation Form Practical Recorded Form for 25 Marks School’s Name and Address: ………………………………………………………. Academic Year: 208 …. Subject: The Leading Mathematics Class: 9 RN Student’s Name Participation (Attendence + Class work): (3) Project and Practical Work: (16) Terminal Exams: (6) Total Marks Percent % Grade Grade point Result Remarks First (3) Second (3) 123456789101112 …………………… …………………….. ………………………. Signature of student Signature of parents Signature of class teacher