A ladder leans against a wall making an angle of 45 ° with the ground. It reaches 3 - 2 m higher when the angle with the ground is 60 ° . How long is the ladder?

Given, B E = A D = Ladder, ∠ B E C = 45 ° , A B = 3 - 2 m and ∠ A D C = 60 ° Let B C = x . So, A C = A B + B C = x + 3 - 2 m In △ B C E , sin 45 ° = B C B E ⇒ x B E = 1 2 ⇒ B E = 2 x In △ A D C , sin 60 ° = A C A D ⇒ sin 60 ° = A C B E ⇒ x + 3 - 2 2 x = 3 2 ⇒ x + 3 - 2 x = 3 2 ⇒ 3 - 2 = 3 2 x - x ⇒ 3 - 2 = 3 x - 2 x 2 ⇒ 3 - 2 = 3 - 2 x 2 ⇒ x = 2 Now, B E = 2 x = 2 × 2 = 2 Therefore, the length of the ladder is 2 m .

A ladder leans against a wall making an angle of 45 ° with the ground. It reaches 3 - 2 m higher when the angle with the ground is 60 ° . How long is the ladder?

Given, B E = A D = Ladder, ∠ B E C = 45 ° , A B = 3 - 2 m and ∠ A D C = 60 ° Let B C = x . So, A C = A B + B C = x + 3 - 2 m In △ B C E , sin 45 ° = B C B E ⇒ x B E = 1 2 ⇒ B E = 2 x In △ A D C , sin 60 ° = A C A D ⇒ sin 60 ° = A C B E ⇒ x + 3 - 2 2 x = 3 2 ⇒ x + 3 - 2 x = 3 2 ⇒ 3 - 2 = 3 2 x - x ⇒ 3 - 2 = 3 x - 2 x 2 ⇒ 3 - 2 = 3 - 2 x 2 ⇒ x = 2 Now, B E = 2 x = 2 × 2 = 2 Therefore, the length of the ladder is 2 m .

From two points P and Q at distances of a and b respectively from the base of a tower and in the same straight line with it, the angles of elevation of the top of the tower are complementary. Prove that the height of the tower is a b .

Let the height of the tower be A B . Q is a point at the distance b away from the base B of the tower and P is a point at the distance a away from point B in the same straight line. The angles of elevation of the top A of the tower from points P and Q are complementary. Since the angles are complementary if one angle is θ then the other is 90 ° - θ . In △ A Q B , tan θ = A B B Q tan θ = A B b ....(1) In △ A P B , tan 90 ° - θ = A B P B cotθ = AB b [Since tan 90 ° - θ = cotθ ] 1 tan θ = AB b [As we know that cotθ = 1 tanθ ] tan θ = b A B ....(2) From equation (1) and (2) AB a = b A B AB 2 = a b AB = ± a b Since height cannot be negative, therefore, the height of the tower is a b . Hence proved that the height of the tower is a b .

The horizontal distance between two poles is 15 m . The angle of depression of the top of the first pole as seen from the top of the second pole is 30 ° . If the height of the second pole is 24 m , find the height of the first pole. [Use 3 = 1 . 73 ]

Let A B and C D be two poles which are 15 m apart. The second pole C D is 24 m in height. Let the height of first pole A B be h m . We draw C E ∥ D B ∴ E B = C D = 24 m E C = B D = 15 m In right-angled ∆ A E C , we have tan ⁡ 30 ° = A E E C ∵ tan ⁡ θ = P B ⇒ 1 3 = A E 15 [ ∵ tan ⁡ 30 ° = 1 3 and E C = 15 m ] ⇒ A E = 15 3 × 3 3 = 15 3 3 = 5 3 m = 5 × 1 . 73 = 8.65 m Now, A B = A E + E B = 8.65 + 24 = 32.65 Thus, the height of first pole is 32.65 m

In the adjoining figure, find the value of x .

We have, We know that, in a right-angled triangle Hypotenuse 2 = perpendicular 2 + base 2 In △ A B E , A E 2 = 3 2 + 4 2 A E 2 = 9 + 16 = 25 ⇒ A E = 5 cos 40 ° = base hypotenuse = A B A E ⇒ cos 40 ° = 4 5 ....... ( i ) Now in △ C D E , cos 40 ° = C D C E = x 15 ⇒ 4 5 = x 15 ( Using ( i ) ) ⇒ x = 4 5 × 15 = 12 Hence, the value of x is 12 .