# pl help me in solving the following questions1) A lamp-post stands at the center of a circular park.Let P and Q be two points on the boundary such that PQ subtends an angle 90 deg at the foot of the lamp-post and the angle of elevation of the top of the lamp-post from P is 30 deg. If PQ=30m, then find the height of the lamp-post.2) A person in a helicopter flying at a height of 700m, observes 2 objects lying opposite to each other on either bank of a river. The angles of depression of the 2 objects are 30 deg and 45 deg. Find the width of the river.3)A straight highway leads to the foot of a tower. A man standinng on the top of the tower spots a van at an angle of depression of 30 deg. The van is approaching the tower with uniform speed. After 6 min the angle of depression of the van is 60 deg. How many minutes will it take for the van to reach the tower?4)From the top of a 60m tall tower, the angles of depression of the top and the bottom of a building are 30deg and 60deg respectively. Find the height of the building.5)The angle of elevation of a hovering helicopter as seen from a point 45m above a lake is 30deg. and the angle of depressionof its reflection in the lake as seen from the same point and at the same time, is 60deg. Find the distance of the helicopter from the surface of the lake.

Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
8 years ago
Solution:
1. Since the distance between P and Q is 30, so if we take r as the radius of the circle, then, using trigonometry we would get$r^{2}+r^{2}=30\Rightarrow r=\sqrt{15}$. Thus the required height is equal to$\sqrt{15}tan(30^{\circ})$.
2. The width of the river would be equal to:$\frac{700}{tan(30^{\circ})}+\frac{700}{tan(45^{\circ})}$.
3. For the two positions, the angle of depressions are 30 and 60 degrees respectively. So, if we assume the height of the tower to be h, and x is the distance covered in 6 mins and z be the required distance to be covered. Hence, we get:$\frac{h}{x+z}=\frac{1}{\sqrt{3}}, \frac{h}{z}=\sqrt{3}\Rightarrow \frac{2h}{\sqrt{3}}$is the distance covered in 6 mins. Thus the required distance would be covered in 3 mins.
4. Let us assume x to be the depth of the top of the building and h be the height of the building. Also the distance of the tower from the building be y. Hence, we have:$y=\frac{x+h}{tan\left ( 60^{\circ} \right )}=\frac{60}{\sqrt{3}}$. Therefore the required height would be equal to:$60-\frac{60}{\sqrt{3}}tan\left ( 30^{\circ} \right )=60-20=40$.
5. Let the distance of the helicopter from the point 45 m high be x. Let the horizontal dstance of the helicopter from the point of observation be z. Hence, we have:$z=\frac{45}{tan\left ( 60^{\circ} \right )}$. Hence, the required height is equal to:$45+\frac{45}{tan\left ( 60^{\circ} \right )}tan\left ( 30^{\circ} \right )=60$.
Thanks & Regards
Sumit Majumdar,