Maharashtra SSC Geometry HOTS Questions with Answers – Trigonometry | Class 10 Board Exam 2027

Asked in: March 2024 board paper (asked again as a 1-mark question in July 2024)

\(\sin\theta \times \operatorname{cosec}\theta = \,?\)

Asked in: July 2024 and March 2026 board papers

If \(\sec\theta = \dfrac{25}{7}\), complete the following activity to find the value of \(\tan\theta\).

Activity: We have, \(1 + \tan^2\theta = \sec^2\theta\)

\(\therefore 1 + \tan^2\theta = \text{____}\)

\(\therefore \tan^2\theta = \dfrac{625}{49} - \text{____}\)

\(\therefore \tan^2\theta = \dfrac{625 - 49}{49}\)

\(\therefore \tan^2\theta = \dfrac{\text{____}}{49}\)

\(\therefore \tan\theta = \text{____}\)

Asked in: March 2025 board paper (same type with \(\sin\theta = \dfrac{20}{29}\) in June 2026)

If \(\sin\theta = \dfrac{11}{61}\), find the value of \(\cos\theta\) using trigonometric identity.

Asked in: March 2026 board paper (asked as an activity in March 2024)

Prove that: \(\sin^2\theta + \cos^2\theta = 1\).

Asked in: March 2025 board paper (same type, 80 m and 45° for a church, in June 2026)

An observer at a distance of 10 m from a tree looks at the top of the tree; the angle of elevation is 60°. To find the height of the tree complete the activity. (\(\sqrt3 = 1.73\))

[Figure: AB = h is the tree, B its foot; C is the observer with BC = 10 m; \(\angle BCA = 60^\circ\).]

Activity: \(\tan\theta = \dfrac{\text{____}}{BC}\) ...(I) \(\tan 60^\circ = \text{____}\) ...(II)

\(\dfrac{AB}{BC} = \sqrt3\) (from (I) and (II)) \(AB = BC \times \sqrt3 = 10\sqrt3\)

AB = 10 × 1.73 = ____ \(\therefore\) height of the tree is ____ m.

Asked in: July 2025 board paper

A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60° with the horizontal. Find the height of the tree.

Asked in: July 2025 board paper

If \(\tan\theta = 1\), then find the value of \(\dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta}\) by completing the following activity.

Activity: \(\tan\theta = 1\) (given), but \(\tan\text{____} = 1\) \(\therefore \theta = \text{____}\)

\(\therefore \dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \dfrac{\sin 45^\circ + \cos 45^\circ}{\sec 45^\circ + \operatorname{cosec} 45^\circ} = \dfrac{\dfrac{1}{\text{____}} + \dfrac{1}{\sqrt2}}{\sqrt2 + \text{____}} = \dfrac{\dfrac{2}{\sqrt2}}{\text{____}} = \dfrac{1}{\text{____}}\)

Asked in: March 2025 board paper

Taking θ = 30°, verify the following trigonometric identities:

(i) \(\sin^2\theta + \cos^2\theta = 1\) (ii) \(1 + \tan^2\theta = \sec^2\theta\) (iii) \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\)

Asked in: June 2026 board paper

α is an acute angle. If \(3\sin\alpha - 4\cos\alpha = 0\), then find the values of \(\tan\alpha\) and \(\sec\alpha\).

Asked in: March 2026 board paper

Two eagles are flying in the sky at a height of \(10\sqrt3\) m above the ground. A boy standing at a point on the ground looks towards the eagles making angles of elevation of 60° and 30° respectively. Find the distance between these two eagles.

Asked in: July 2024 board paper

To find the width of the river, a man observes the top of a tree on the opposite bank making an angle of elevation of 60°. When he moves 24 metre backward from the bank and observes the same top of the tree, his line of vision makes an angle of elevation of 30°. Find the height of the tree and width of the river. (\(\sqrt3 = 1.73\))

Asked in: March 2024 board paper

If \(\dfrac{1}{\sin^2\theta} - \dfrac{1}{\cos^2\theta} - \dfrac{1}{\tan^2\theta} - \dfrac{1}{\cot^2\theta} - \dfrac{1}{\sec^2\theta} - \dfrac{1}{\operatorname{cosec}^2\theta} = -3\), then find the value of θ.

\(\cos^2\theta\,(1 + \tan^2\theta) = \) ____ .

If \(\sec\theta = 2\), where θ is acute, then θ = ____ .

If \(\cot\theta = \dfrac{40}{9}\), find \(\tan\theta\).

Find the value of \(\sec^2\theta - \tan^2\theta + \sin^2\theta + \cos^2\theta\).

If \(5\sec\theta - 13 = 0\), find the value of \(\tan\theta\) (θ acute).

Prove that: \(\cot\theta + \tan\theta = \operatorname{cosec}\theta\,\sec\theta\).

From the top of a 30 m high lighthouse, the angle of depression of a boat is 30°. Find the distance of the boat from the foot of the lighthouse. (\(\sqrt3 = 1.73\))

Prove that: \((\sec\theta - \cos\theta)(\cot\theta + \tan\theta) = \tan\theta\,\sec\theta\).

If \(\tan\theta = \dfrac{7}{24}\), find the values of all the other trigonometric ratios of θ (θ acute).

Prove that: \(\sqrt{\dfrac{1 - \sin\theta}{1 + \sin\theta}} = \sec\theta - \tan\theta\).

If \(\sin\theta + \sin^2\theta = 1\), show that \(\cos^2\theta + \cos^4\theta = 1\).

A kite is flying with a string 60 m long. The string makes an angle of 60° with the level ground. Assuming the string is straight, find the height of the kite above the ground. (\(\sqrt3 = 1.73\))

From the top of a building 20 m high, the angle of elevation of the top of a tower is 30° and the angle of depression of its foot is 45°. Find the distance between the building and the tower, and the height of the tower. (\(\sqrt3 = 1.73\))

If \(\sec\theta + \tan\theta = p\), show that \(\sin\theta = \dfrac{p^2 - 1}{p^2 + 1}\).

Two poles of equal height stand on either side of a road 80 m wide, exactly opposite each other. From a point on the road between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the distance of the point from each pole. (\(\sqrt3 = 1.73\))