9. The angle of elevation of the top of an unfinished tower at a dista

1.	9. The angle of elevation of the top of an unfinished tower at a distance of75 m from its base is 30°. How much higher must the tower be raisedso that the angle of elevation of its top at the same point maybe 60°?[Take 3 = 1.732.]on the ton
Answer» Answer: 31.8 M (APPROX) Step-by-step EXPLANATION: Cos 30 = B/H $\frac{ \sqrt{3} }{2} = \frac{b}{h}$ $hypotenuse = \frac{75 \times 2}{ \sqrt{3} }$ h = 150/1.732 Sin30 = p/h $\frac{1}{2} = p \div \frac{150}{1.732}$ $\frac{1}{2} = \frac{p \times 1.732}{150}$ 1 = p × 1.732x 2 /150 1 = p × 3.464/150 p = 150/3.464 Now, p= 43.2(approx) Let x HEIGHT is increased. So, Sin 60 = $\frac{p + x}{h}$ $\frac{ \sqrt{3} }{2} = (p + x) \div \frac{150}{1.732}$ $\frac{ \sqrt{3} }{2} = \frac{(p + x) \times 1.732}{150}$ √3 and 1.732 will GET cancelled out. 1/2 = (p+x)/150 150/2 = p+x 75 = p+x 75-43.2= x 75 - 43.2 = x x = 31.8 m HOPE IT HELPS