Worksheet on Trigonometry for Class 10

Learn how Triangles reveal amazing mathematical relationships with our grade 10 printable trigonometry practice questions. These practice material explores angle and side connections. Students learn sine, cosine, and tangent ratios through practical applications. Each exercise connects abstract concepts to real-world problem-solving. These activities build foundation skills for engineering and science careers. Download a free class 10 trigonometry worksheet PDF for mathematical discovery.

Solved Questions on Trigonometry

1. If sin A : cos A = 3 : 9, then what is the value of the given trigonometric expression?

a) 1
b) 2
c) 0
d) 3

Answer: d) 3

Explanation:

2. If cot A + tan A = p and sec A − cos A = q, then what is the value of the given expression?

a) 0
b) 1
c) 2
d) − 2

Answer: b) 1

Explanation:

3. If sin A = cos A/√3, then what is the value of sin2 A + sec2 A + sin A sec A + 2 cosec A?

a)
b)
c)
d)

Answer: a)

Explanation: 

4. A woman on a cliff observes a boat at an angle of depression of 30° which is approaching the shore to the point immediately beneath the observer with a uniform speed. 18 minutes later, the angle of depression of the boat was found to be 60°. What is the total time taken by the boat to reach the shore?

a) 9 minute
b) 25 minutes
c) 27 minutes
d) 28 minutes

Answer: c) 27 minutes

Explanation: The two positions of the boat at the two instants are P and S and let the speed of the boat be x m/min, h be the height of the cliff and R be the location of the woman.

Thus, the figure is shown as:

The boat takes 18 minutes to reach point S from point P.

We know that: Distance = Speed × Time                        

Thus, the distance covered by the boat from point P to point S is:

PS = x × 18

PS = 18x

Let the boat take t time to reach the shore, then the distance covered from point S to point Q is:

SQ = xt

Now, apply the trigonometric ratio in the triangle SQR,

tan 60° = ^{Perpendicular}⁄_Base

In the triangle SQR, QR is the perpendicular and SQ is the base whose lengths are QR = h

SQ = xt

→ tan 60° = ^h⁄_xt
→ √3 = ^h⁄_xt
→ h = xt√3 … (1)

Now, apply the trigonometric ratio in the triangle PQR,

tan 30° = ^{Perpendicular}⁄_Base

In the triangle PQR, QR is perpendicular and PQ is the base whose lengths are QR = h

PQ = PS + SQ = 18x + xt

→ tan 30° = $\frac{\frac{h}{}}{\frac{\mathrm{18x\; +\; xt}}{}}$

→ $\frac{\frac{1}{}}{\frac{\mathrm{\surd 3}}{}}$ = $\frac{\frac{h}{}}{\frac{\mathrm{18x\; +\; xt}}{}}$

Compare the values h from the equation (1) and equation (2),

→ xt√3 = $\frac{\frac{\mathrm{x(18\; +\; t)}}{}}{\frac{\mathrm{\surd 3}}{}}$
→ xt√3 × √3 = x(18 + t)
→ 3xt = x(18 + t)
→ 3t = 18 + t
→ 3t − t = 18
→ 2t = 18
→ t = ¹⁸⁄₂

t = 9 mins

Thus, the boat takes 9 minutes to reach the shore from the point S.

∴ Total time taken by the boat to reach the shore Q from point P
= 18 + 9 = 27 minutes

5. Simplify the following trigonometric expression:

a) 2cos θ (1 + sec θ)
b) sec θ (1 + sec θ)
c) cos θ (1 + sec θ)
d) 2sec θ (1 + sec θ)

Answer: d) 2sec θ (1 + sec θ)

Explanation: