CBSE Class 10 Mental Maths Trigonometry And Its Applications Worksheet - Free Printable

Let's solve each problem step-by-step from the given worksheet on Trigonometry and Its Applications.

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1. If $ x = \sin^2\theta $ and $ y = \cos^2\theta + 1 $, then find the value of $ x + y $.

We are given:
- $ x = \sin^2\theta $
- $ y = \cos^2\theta + 1 $

So,
$$
x + y = \sin^2\theta + (\cos^2\theta + 1) = \sin^2\theta + \cos^2\theta + 1
$$

Using the identity: $ \sin^2\theta + \cos^2\theta = 1 $

$$
x + y = 1 + 1 = \boxed{2}
$$

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2. If $ \sec^2\theta(1 + \sin\theta)(1 - \sin\theta) = k $, then find the value of $ k $.

We know:
- $ \sec^2\theta = \frac{1}{\cos^2\theta} $
- $ (1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta = \cos^2\theta $

So,
$$
k = \sec^2\theta \cdot (1 - \sin^2\theta) = \frac{1}{\cos^2\theta} \cdot \cos^2\theta = 1
$$

$$
\boxed{k = 1}
$$

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3. Find the value of $ \sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ $

Use identity:
$$
\sin A \sin B - \cos A \cos B = -\cos(A + B)
$$

But let’s use another approach:

Note that:
- $ \sin 70^\circ = \cos 20^\circ $
- $ \cos 70^\circ = \sin 20^\circ $

So,
$$
\sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = \sin 20^\circ \cos 20^\circ - \cos 20^\circ \sin 20^\circ
$$

$$
= \sin 20^\circ \cos 20^\circ - \cos 20^\circ \sin 20^\circ = 0
$$

$$
\boxed{0}
$$

Alternatively, using identity:
$$
\sin A \sin B - \cos A \cos B = -\cos(A + B)
$$

Let $ A = 20^\circ, B = 70^\circ $, then:
$$
= -\cos(90^\circ) = -0 = 0
$$

Same result.

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4. If $ \tan\theta \tan 45^\circ = 1 $, then find the value of $ \theta $.

We know $ \tan 45^\circ = 1 $, so:
$$
\tan\theta \cdot 1 = 1 \Rightarrow \tan\theta = 1
$$

So $ \theta = 45^\circ $ (principal value)

$$
\boxed{\theta = 45^\circ}
$$

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5. Find the value of $ \sin^2 10^\circ + \sin^2 80^\circ $

Note: $ \sin 80^\circ = \cos 10^\circ $, since $ \sin(90^\circ - x) = \cos x $

So,
$$
\sin^2 10^\circ + \sin^2 80^\circ = \sin^2 10^\circ + \cos^2 10^\circ = 1
$$

$$
\boxed{1}
$$

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6. If $ \cos A = \frac{3}{5} $, then find the value of $ \tan^2 A - \sec^2 A $.

We know:
- $ \cos A = \frac{3}{5} $
- So adjacent = 3, hypotenuse = 5 → opposite = $ \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 $

So:
- $ \sin A = \frac{4}{5} $
- $ \tan A = \frac{\sin A}{\cos A} = \frac{4/5}{3/5} = \frac{4}{3} $
- $ \tan^2 A = \left(\frac{4}{3}\right)^2 = \frac{16}{9} $
- $ \sec A = \frac{1}{\cos A} = \frac{5}{3} $
- $ \sec^2 A = \left(\frac{5}{3}\right)^2 = \frac{25}{9} $

Now:
$$
\tan^2 A - \sec^2 A = \frac{16}{9} - \frac{25}{9} = \frac{-9}{9} = \boxed{-1}
$$

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7. If $ \sin 2A = \cos 3A $, then find the value of $ A $.

We use identities:
- $ \sin 2A = 2\sin A \cos A $
- $ \cos 3A = 4\cos^3 A - 3\cos A $

But better to use angle relations.

Let’s write:
$$
\sin 2A = \cos 3A
\Rightarrow \sin 2A = \sin(90^\circ - 3A)
$$

So,
$$
2A = 90^\circ - 3A \quad \text{(or other general solutions)}
\Rightarrow 2A + 3A = 90^\circ \Rightarrow 5A = 90^\circ \Rightarrow A = 18^\circ
$$

Check:
- $ \sin(2 \times 18^\circ) = \sin 36^\circ $
- $ \cos(3 \times 18^\circ) = \cos 54^\circ $
- But $ \sin 36^\circ = \cos(90^\circ - 36^\circ) = \cos 54^\circ $ → Yes!

So it works.

$$
\boxed{A = 18^\circ}
$$

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8. If $ \tan\theta - \tan(90^\circ - \theta) = 0 $, then find the value of $ \theta $.

We know:
- $ \tan(90^\circ - \theta) = \cot\theta $

So:
$$
\tan\theta - \cot\theta = 0 \Rightarrow \tan\theta = \cot\theta
\Rightarrow \tan\theta = \frac{1}{\tan\theta}
\Rightarrow \tan^2\theta = 1 \Rightarrow \tan\theta = \pm 1
$$

So $ \theta = 45^\circ $ or $ 135^\circ $, etc.

But in standard range $ 0^\circ < \theta < 90^\circ $, $ \theta = 45^\circ $

$$
\boxed{\theta = 45^\circ}
$$

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9. Complete the following:

> The angle nearer to altitude is _________ than the angle away from the altitude.

In trigonometry, especially in height and distance problems, the angle of elevation (nearer to altitude) is greater than the angle away from the altitude.

But more accurately:
The angle nearer to the altitude (i.e., closer to vertical) is larger than the one farther from it.

So, the correct word is: greater

$$
\boxed{\text{greater}}
$$

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10. In the figure, find the area of $ \triangle ABC $ in which $ \angle ABC = 90^\circ $, $ \angle ACB = 45^\circ $, and $ BC = 8 $ cm.

Given:
- $ \angle B = 90^\circ $
- $ \angle C = 45^\circ $
- So $ \angle A = 45^\circ $ → triangle is right-angled isosceles

So $ AB = BC $? Wait — check angles:

- $ \angle B = 90^\circ $
- $ \angle C = 45^\circ $
- So $ \angle A = 45^\circ $

Thus, $ \angle A = \angle C \Rightarrow AB = BC $

Wait! $ \angle A = 45^\circ $, $ \angle C = 45^\circ $ → sides opposite equal angles are equal

Opposite to $ \angle A $ is $ BC $, opposite to $ \angle C $ is $ AB $

So $ AB = BC = 8 $ cm

Now, area of right triangle:
$$
\text{Area} = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 8 = \boxed{32} \text{ cm}^2
$$

Wait — but $ AB $ and $ BC $ are legs? Let's label properly.

From the diagram (Fig. 1):
- Right angle at $ B $
- $ BC = 8 $ cm
- $ \angle C = 45^\circ $

So in triangle $ ABC $, right-angled at $ B $, $ \angle C = 45^\circ $ → so $ \angle A = 45^\circ $

So it's a 45-45-90 triangle → legs are equal

Legs: $ AB $ and $ BC $

So $ AB = BC = 8 $ cm

Then area:
$$
\frac{1}{2} \times AB \times BC = \frac{1}{2} \times 8 \times 8 = \boxed{32} \text{ cm}^2
$$

✔ Correct.

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11. Name the angle of depression in the figure.

Look at Fig. 2, there are two parts:

- Left figure: Cloud at P, observer at Q, horizontal line QR, line from Q to P.
- Right figure: Bird at B, horizontal line BA, line BC down to ground point C.

In right figure (bird), the bird sees point C below it.

Angle of depression is the angle between the horizontal and the line of sight downward.

So from point B, horizontal is BA, line of sight is BC.

So angle between BA and BC is $ \angle ABC $, but wait — direction?

Actually, angle of depression is $ \angle ABC $ only if measured from horizontal down.

But from the diagram, it's clear: the angle between horizontal $ BA $ and the line $ BC $ is the angle of depression.

So the angle of depression is $ \angle ABC $?

Wait — no. Point B is the bird, A is to the right (horizontal), C is down.

So line BC goes down. Angle between horizontal (BA) and BC is $ \angle ABC $, but we must identify the actual angle.

Actually, the angle of depression is the angle between the horizontal and the line of sight, measured downward.

So from point B, horizontal is BA, line of sight is BC → angle between them is $ \angle ABC $, but we need to confirm.

Wait — in the diagram, point A is to the right, C is below and to the right.

So vector BA is horizontal right, vector BC is diagonal down-right.

So the angle between BA and BC is the angle of depression.

Hence, the angle of depression is $ \boxed{\angle ABC} $?

Wait — no! $ \angle ABC $ is at B, between points A-B-C.

Yes — so $ \angle ABC $ is the angle between horizontal (BA) and the line of sight (BC).

So yes, angle of depression is $ \angle ABC $.

But wait — in some diagrams, they label it as $ \angle BAC $, but here it's clearly $ \angle ABC $.

But let’s double-check:

- Horizontal line from B is BA
- Line of sight to C is BC
- So angle between them is $ \angle ABC $

Yes.

So answer: $ \boxed{\angle ABC} $

But wait — is it possible they mean $ \angle ABC $ is the angle at B between A and C?

Yes — and since it's formed by horizontal and downward line, it is indeed the angle of depression.

So:
$$
\boxed{\angle ABC}
$$

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✔ Final Answers Summary:

1. $ x + y = \boxed{2} $
2. $ k = \boxed{1} $
3. $ \sin 20^\circ \sin 70^\circ - \cos 20^\circ \cos 70^\circ = \boxed{0} $
4. $ \theta = \boxed{45^\circ} $
5. $ \sin^2 10^\circ + \sin^2 80^\circ = \boxed{1} $
6. $ \tan^2 A - \sec^2 A = \boxed{-1} $
7. $ A = \boxed{18^\circ} $
8. $ \theta = \boxed{45^\circ} $
9. The angle nearer to altitude is greater than the angle away from the altitude.
10. Area of $ \triangle ABC = \boxed{32} \text{ cm}^2 $
11. Angle of depression = $ \boxed{\angle ABC} $

Let me know if you'd like explanations for any specific question in more detail!