Question
42. (4) Here ( A B ) and ( D C ) are given poles and ( B C ) is distance between poles.
Now by Trigonometric ratio in figure in ( triangle A B C ) and ( Delta D C B )
( tan theta=frac{D C}{B C}, tan left(90^{circ}-thetaright)=frac{A B}{B C} )
By sign of Trigonometric function ( tan left(90^{circ}-thetaright)=cot theta )
( tan theta=frac{D C}{B C}, cot theta=frac{A B}{B C} )
( tan theta=frac{9}{B C}, cot theta=frac{16}{B C} )
By Trigonometric ratio ( cot theta=frac{1}{tan theta} )
( tan theta=frac{9}{B C}, frac{1}{tan theta}=frac{16}{B C} )
On multiplying, ( tan theta times frac{1}{tan theta}=frac{9}{B C} times frac{16}{B C} )
( B C^{2}=9 times 16 Rightarrow B C=12 )

42. The distance between the two poles of length 16 m and 9 m, is X m. If two angles of elevation of their respective top from the 83 bottom of the other are complementary to each other than the value X is RRB NTPC (Phase 1) 2016 (1) 10m (2) 15m (3) 16m (4) 12m
Solution
