


15.
On the same side of a tower, two objects are located. Observed from the top of the tower, their angles of depression are 45° and 60°. If the height of the tower is 150 m, the distance between the objects is:
Answer & Solution
Answer: (D)63.5 m
Solution: Let AB be the tower and C and D be the objects. Then, AB = 150 m, ∠ACB = 45° and ∠ADB = 60° AB/AD = tan 60° = √3 => AD = AB/√3 = 150/√3 m AB/AC = tan 45 ° = 1 => AC = AB = 150 m CD = (AC  AD) CD = (150  150/√3 ) m = ( (150 (√3  1))/√3 * √3/√3 ) m = 50 (3  √3 ) m CD = (50 * 1.27) m = 63.5 m 16.
A man on the top of a vertical observation tower observes a car moving at a uniform speed coming directly towards it. If it takes 12 minutes for the angle of depression to change from 30° to 45°, how soon after this will the car reach the observation tower?
Answer & Solution
Answer: (C)16 min. 23 sec.
Solution: Let AB be the tower and C and D be the two positions of the car. Then, ∠ACB = 45°, ∠ADB = 30°. Let AB = h, CD = x and AC = y. AB/AC = tan 45° = 1 h/y = 1 => y = h AB/AD = tan 30° = 1/√3 => h/(x + y) = 1/√3 => x + y = √3 h x = (x + y)  y = √3 h  h = h(√3  1) Now, h(√3  1) is covered in 12 min. So, h will be covered in ( 12/(h (√3  1)) * h) = 12/((√3  1)) min = 1200/73 min = 16 min. 23 sec. 17.
The top of a 15 metre high tower makes an angle of elevation of 60° with the bottom of an electric pole and angle of elevation of 30° with the top of the pole. What is the height of the electric pole?
Answer & Solution
Answer: (C)10 metres
Solution: Let AB be the tower and CD be the electric pole. Then, ∠ACB = 60°, ∠EDB = 30° and AB = 15 m Let CD = h. Then, BE = (AB  AE) = (AB  CD) = (15 h) AB/AC = tan 60° = √3 => AC = AB/√3 = 15/√3 And, BE/DE = tan 30° = 1/√3 => DE = (BE * √3 ) = √3 (15  h) AC = DE => 15/√3 = √3 (15  h) => 3h = (45  15) => h = 10 m AC = DE => 15/√3 = √3 (15  h) => 3h = (45  15) => h = 10 m 