Question

9. 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.]
on the ton​

Answer 1

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