Q. 94.2( 5 Votes )
The angle of elevation of the top of an unfinished tower at a distance of 75 m from its base is 30°. How much higher must the tower be raised so that the angle of elevation of its top at the same point may be 60°? [Take √3 = 1.732.]
In the above figure, let DB be the unfinished tower. Let C be the point on the ground, 75 m from the base B from which the angle of elevation of the top D is 30°. Let A be the point to which the tower must be raised such that the angle of elevation of it from C becomes 60°. Join B and C. We get two triangles ABC and BCD with right angle at B. We need to find the excess height that is an AD. We have, BC = 75 m, ∠BCD = 30° and ∠ACB = 60°. To find the AD, we will find AB and BD from the triangles ABC and BCD respectively using the trigonometric ratio tan and then subtract BD from AB.
Now, from ∆BCD,
Again, from ∆ABC,
Now, we have got both AB and DB. Then we are asked to find the AD.
Hence the answer is 86.6 m.
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