Every question from past board papers on Trigonometry, newest paper first. Open any question for its full worked answer.
Q.2 (A)(ii) If \(\sin\theta = \dfrac{20}{29}\), then complete the following activity to find \(\cos\theta\).
Activity : \(\sin^2\theta + \cos^2\theta = 1\)
\(\left(\text{____}\right)^2 + \cos^2\theta = 1\)
\(\dfrac{400}{841} + \cos^2\theta = 1\)
\(\therefore \cos^2\theta = 1 - \text{____}\)
\(\cos^2\theta = \text{____}\)
Taking square roots of both sides
\(\therefore \cos\theta = \text{____}\)
Q.2 (B)(v) A person is standing at a distance of 80 m from a Church looking at its top. The angle of elevation is of 45º. Find the height of the Church.
Q.5 (ii) \(\alpha\) is acute angle. If \(3\sin\alpha - 4\cos\alpha = 0\), then find the values of \(\tan\alpha\) and \(\sec\alpha\).
Q.2 (A)(2) If \(\sec\theta = \dfrac{25}{7}\), find the value of \(\tan\theta\).
Solution :
1 + tan2 θ = ____
∴ 1 + tan2 θ = ____2
∴ tan2 θ = \(\dfrac{625}{49}\) – ____
= \(\dfrac{625 - 49}{49}\)
∴ tan2 θ = \(\dfrac{576}{49}\)
∴ tan θ = ____
Q.2 (B)(5) Prove that : sin2 θ + cos2 θ = 1.
Q.5 (2) Two eagles are flying in the sky at height of \(10\sqrt{3}\) m above the ground. A boy standing at a point on the ground looks towards the eagles making angles of elevation of 60º and 30º respectively. Find the distance between these two eagles.
Q.2 (A)(3) An observer at a distance of 10 m from tree looks at the top of the tree, the angle of elevation is 60º. To find the height of tree complete the activity. (\(\sqrt3\) = 1.73)
Activity :
In the figure given above, AB = h = height of tree, BC = 10 m, distance of the observer from the tree.
Angle of elevation (θ) = ∠BCA = 60º
tan θ = ____ / BC .................. (I)
tan 60º = ____ .................. (II)
\(\dfrac{AB}{BC} = \sqrt3\) (From (I) and (II))
AB = BC × \(\sqrt3\) = \(10\sqrt3\)
AB = 10 × 1.73 = ____
∴ height of the tree is ____ m.
Q.2 (B)(5) If \(\sin\theta = \dfrac{11}{61}\), find the values of cos θ using trigonometric identity.
Q.5 (2) Taking θ = 30º to verify the following Trigonometric identities :
(i) \(\sin^2\theta + \cos^2\theta = 1\)
(ii) \(1 + \tan^2\theta = \sec^2\theta\)
(iii) \(1 + \cot^2\theta = \operatorname{cosec}^2\theta\).
Q.1 (B)(4) If \(\sin\theta = \dfrac{7}{25}\), then find the value of \(\operatorname{cosec}\theta\).
Q.3 (A)(1) 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 \(\tan ____ = 1\)
\(\therefore \theta = ____\)
\(\therefore \dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \dfrac{\sin 45^\circ + \cos 45^\circ}{\sec 45^\circ + \operatorname{cosec} 45^\circ}\)
\(= \dfrac{\dfrac{1}{____} + \dfrac{1}{\sqrt2}}{\sqrt2 + ____}\)
\(= \dfrac{\dfrac{2}{\sqrt2}}{____}\)
\(\dfrac{\sin\theta + \cos\theta}{\sec\theta + \operatorname{cosec}\theta} = \dfrac{1}{____}\)
Q.3 (B)(3) A storm broke a tree and the treetop rested 20 m from the base of the tree, making an angle of 60º with the horizontal. Find the height of the tree.
Q.1 (B)(1) In \(\Delta ABC\), \(\angle B = 90^\circ\), \(\angle C = 30^\circ\), AC = 12 cm, then find AB.
Q.1 (B)(3) Find the value of \(\sin\theta \times \operatorname{cosec}\theta\).
Q.2 (A)(2) If \(\sec\theta = \dfrac{25}{7}\), complete the following activity to find the value of \(\tan\theta\).
Activity :
We have, \(1 + \tan^2\theta = \sec^2\theta\)
\(\therefore 1 + \tan^2\theta =\) ____
\(\therefore \tan^2\theta = \dfrac{625}{49} -\) ____
\(\therefore \tan^2\theta = \dfrac{625 - 49}{49}\)
\(\therefore \tan^2\theta = \dfrac{\text{____}}{49}\)
\(\therefore \tan\theta =\) ____
Q.4 (3) 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 meter backward from 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. (\(\sqrt{3}\) = 1.73)
Q.1 (A)(2) \(\sin\theta \times \operatorname{cosec}\theta = \,?\)
Q.2 (A)(2) Find the value of \(\sin^2\theta + \cos^2\theta\).
Solution :
In Δ ABC, ∠ABC = 90º, ∠C = θ.
AB2 + BC2 = ____ ............ (Pythagoras theorem)
Divide both sides by AC2
\(\dfrac{AB^2}{AC^2} + \dfrac{BC^2}{AC^2} = \dfrac{AC^2}{AC^2}\)
∴ \(\left(\dfrac{AB}{AC}\right)^2 + \left(\dfrac{BC}{AC}\right)^2 = 1\)
But \(\dfrac{AB}{AC}\) = ____ and \(\dfrac{BC}{AC}\) = ____
∴ \(\sin^2\theta + \cos^2\theta\) = ____
Q.4 (1) \(\dfrac{1}{\sin^2\theta} - \dfrac{1}{\cos^2\theta} - \dfrac{1}{\tan^2\theta} - \dfrac{1}{\cot^2\theta} - \dfrac{1}{\sec^2\theta} - \dfrac{1}{\operatorname{cosec}^2\theta} = -3\), then find the value of θ.
