CBSE Class 10 Mathematics Introduction to Trigonometry MCQs Set 07

Question. In triangle ABC, right angled at B, If tan A = 4/3 , then the value of cosC is
(a) 3/5
(b) 4/5
(c) 1
(d) none of the options

Answer: B

Question. In ΔOPQ, right-angled at P, OP = 7 cm and OQ – PQ = 1 cm, then the values of sin Q.
(a) 7/25
(b) 24/25
(c) 1
(d) none of the options

Answer: A

Question. In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm, then the value of sin P is
(a) 5/13
(b) 12/13
(c) 3/5
(d) 4/5

Answer: B

Question. If 4tan θ = 3 then (cos2θ – sin2θ) = ?
(a) 4/25
(b) 7/25
(c) 1
(d) 11/25

Answer: B

Question. If 3cot θ = 2, then the value of 4sin𝜃 −3cos𝜃/2sin𝜃 +6cos𝜃 is
(a) 3/2
(b) 1/2
(c) 1/3
(d) none of the options

Answer: C

Question. If 7tan θ = 4, then the value of 7sin𝜃 − 3cos𝜃 / 7sin𝜃 + 3cos𝜃 is
(a) 1/7
(b) 5/7
(c) 3/7
(d) 5/14

Answer: A

Question. The value of 2(sin²45°+ cot²30°) – 6(cos²45°— tan²30°) is
(a) 6
(b) 3
(c) 2
(d) 4

Answer: B

Question. The value of 2(sin⁴30°+ cos⁴60°) - (tan²60°+ cot²45°) + 3cosec²60° is
(a) 1/2
(b) √3/2
(c) 1/4
(d) none of the options

Answer: C

Question. If tan²45° – cos²30° = x sin45° cos45° then x = ?
(a) -2
(b) 2
(c) − 1/2
(d) 1/2

Answer: D

Question. If tan (A – B) = 1/√3 and tan (A + B) = √3, then the value of A and B, respectively are
(a) 45° and 15°
(b) 30° and 15°
(c) 45° and 30°
(d) none of the options

Answer: A

Question. (1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?
(a) 0
(b) 1
(c) 2
(d) -1

Answer: C

Question. (sec A + tan A) (1 – sin A) = ?
(a) sec A
(b) sin A
(c) cosec A
(d) cos A

Answer: D

Question. If A + B = 90°, cot B = 3/4 then tanA is equal to;
(a) 3/4
(b) 4/3
(c) 1/4
(d) 1/3

Answer: A

Question. If cos 𝜃 = 1/2, sin 𝜑 = 1/2 then value of 𝜃 + 𝜑 is
(a) 30°
(b) 60°
(c) 90°
(d) 120°

Answer: C

Question. The value of 2sin²30° – 3cos²45° + tan²60° + 3sin²90° is
(a) 1
(b) 5
(c) 0
(d) none of the options

Answer: B

Question. sin𝜃 - 2sin³𝜃 / 2cos³𝜃 − cos = ?
(a) sin θ
(b) cos θ
(c) tan θ
(d) cot θ

Answer: C

Question. 2tan 30° / 1 + tan 230° = ?
(a) sin60°
(b) cos60°
(c) tan60°
(d) sin30°

Answer: A

(CASE STUDY BASED QUESTIONS)

1. Raj is an electrician in a village. One day power was not there in entire village and villagers called Raj to repair the fault. After thorough inspection he found an electric fault in one of the electric pole of height 5 m and he has to repair it. He needs to reach a point 1.3m below the top of the pole to undertake the repair work.

Based on the above situation, answer the following questions:

Question. when the ladder is inclined at an angle of α such that √3tanα + 2 = 5 to the horizontal, then find the angle α?
(a) 45°
(b) 30°
(c) 60°
(d) none of the options

Answer: C

Question. How far from the foot of the pole should he place the foot of the ladder? (Use 3 = 1.73)
(a) 2.89 m
(b) 2.14 m
(c) 3 m
(d) none of the options

Answer: B

Question. In the above situation, find the value of sin𝛼cos (a/2) – cos𝛼sin (𝛼/2)
(a) 0
(b) 1
(c) 1/2
(d) none of the options

Answer: C

Question. In the above situation if BD = 3 cm and BC = 6 cm. Find α
(a) 45°
(b) 30°
(c) 60°
(d) none of the options

Answer: B

Question. Given 15 cotα=8, find sinα.
(a) 17/15
(b) 17/15
(c) 17/8
(d) 15/17

Answer: D

2. ‘Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The ‘Skysails’ technology allows the towing kite to gain a height of anything between 100 metres – 300 metres. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a ‘telescopic mast’ that enables the kite to be raised properly and effectively.
Based on the following figure related to sky sailing, answer the questions:

Based on the above situation, answer the following questions:

Question. In the given figure, if √3 tan2θ – 3 = 0, where θ is acute angle, then find the value of θ.
(a) 45°
(b) 30°
(c) 60°
(d) none of the options

Answer: B

Question. What should be the length of the rope of the kite sail in order to pull the ship at the angle (calculated above) and be at a vertical height of 300 m?
(a) 300 m
(b) 400m
(c) 500 m
(d) 600 m

Answer: D

Question. What should be the distance BC in order to pull the ship at the angle (calculated above) and be at a vertical height of 300 m?
(a) 300√3 m
(b) 400√3 m
(c) 500√3 m
(d) 600√3 m

Answer: A

Question. If BC = 100 m, θ = 60°, then AB is
(a) 100√3 m
(b) 200√3 m
(c) 500√3 m
(d) 300√3 m

Answer: A

Question. If the length of the rope, AC = 200m and θ = 30°, then the vertical height, AB is
(a) 300 m
(b) 400m
(c) 100 m
(d) 200 m

Answer: C

3. Kings and queens used the Tower in times of trouble to protect their possessions and themselves. Arms and armour were made, tested and stored here until the 1800s. The Tower also controlled the supply of the nation’s money. All coins of the realm were made at the Tower Mint from the reign of Edward I until 1810. Kings and queens also locked away their valuables and jewels at the Tower and even today, the Crown Jewels are protected by a garrison of soldiers. A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°

Based on the above situation, answer the following questions:

Question. Find the height of the tower in the above situation.
(a) 10√3 m
(b) 15√3 m
(c) 25√3 m
(d) 30√3 m

Answer: B

Question. If the above tower casts a shadow of length 2√3 m on the ground when the sun's elevation is 60°. The height of the tower is
(a) 4√3 m
(b) 3 m
(c) 12 m
(d) 6 m

Answer: D

Question. The angle of elevation of the top of a tower from a point on the ground 30 m away from the foot of the tower is 30°. The height of the tower is
(a) 30 m
(b) 20 m
(c) 10√3 m
(d) 10√2 m

Answer: C

Question. In the above situation if AB = 30 m and AC = 60 m. Find ∠A.
(a) 45°
(b) 60°
(c) 30°
(d) none of the options

Answer: B

Question. If 2cos 3θ = 1 then θ = ?
(a) 10°
(b) 15°
(c) 30°
(d) 20°

Answer: D

4. The Circus Arts Program is one of the most popular activities at Camp Lohikan. It brings a level of excitement and enthusiasm to the camp experience that can't be found at home or in school. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground such that the angle made by the rope with the ground level is 30°

Question. Find the height of the pole in the above situation.
(a) 10√3 m
(b) 15√3 m
(c) 15 m
(d) 10 m

Answer: D

Question. In the above situation, if AB = 5 m and ∠ ACB = 30° then find the length of the side BC.
(a) 5√3 m
(b) 5 m
(c) 10 m
(d) 6 m

Answer: A

Question. In the above situation, if the height of pole is 3 m and the length of rope is 6 m then find ∠ACB
(a) 45°
(b) 60°
(c) 30°
(d) none of the options

Answer: C

Question. Find the value of 2 tan² 45° + cos² 30° – sin² 60°
(a) 0
(b) 2
(c) 1
(d) none of the options

Answer: B

Question. If √3 tan 2θ − 3 = 0 then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer: B

5. Kite flying is also a major part of Makar Sankranti, although the states of Gujarat and Rajasthan indulge in this with a lot more enthusiasm. Makar Sankranti is a major harvest festival celebrated in India and is dedicated to the Sun God, Surya. It is the first major festival to be celebrated in India and usually takes place in January, this year the festival will be celebrated on January 14. Aditya flying a kite at a height of 60 m above the ground. He attached the string to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°.

Question. In the above situation, find the length of the string, assuming that there is no slack in the string.
(a) 50√3 m
(b) 40√3 m
(c) 50√2 m
(d) 100 m

Answer: B

Question. The string of a kite is 100 m long and it makes an angle of 60° with the horizontal. If these is no slack in the string, the height of the kite from the ground is
(a) 50√3 m
(b) 100√3 m
(c) 50√2 m
(d) 100 m

Answer: A

Question. A kite is flying at a height of 30 m from the ground. The length of string from the kite to the ground is 60 m. Assuming that there is no slack in the string, the angle of elevation of the kite at the ground is
(a) 45°
(b) 60°
(c) 90°
(d) 30°

Answer: D

Question. The value of (sin²30° – sec²60° + 4cot²45°) = ?
(a) 4
(b) 2
(c) 1/4
(d) 1

Answer: C

Question. If 2sin 2θ = √3 then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer: A