Question

Angles of elevation of the top of a 20 m high pole from two points in a line are 30 and 60 find the distance between the points​

Answer 1

Answer:

23.094 m

Step-by-step explanation:

Here, a png file is attached.

length of the pole AB = 20 m

elevation of point A from point C = 60°   ∠ACB = 60°

elevation of point A from point D = 30°   ∠ADB = 30°

Now, $tan\angle ACB= \frac{AB}{BC}$  ►  $BC= \frac{AB}{tan\angle ACB} = \frac{20}{tan60} = \frac{20}{\sqrt3}$  m

Again,$tan\angle ADB= \frac{AB}{BD}$  ► $BD= \frac{AB}{tan\angle ADB} = \frac{20}{tan30} = 20\sqrt3$ m

Finally, distance between the points​ C & D = CD

= BD-BC

=( $20\sqrt3-\frac{20}{\sqrt{3}}$ ) m

= 23.094 m

Answer 2

Let the height of pole be PQ=h m.

And QB=d be the distance between B and foot of pole.

tan600=dh

⇒d=3h

tan300=100−dh

⇒−3h+100=3h

⇒100=h(3+31)

⇒h=253 m