Question

8. When the angle of elevation of the sun
increases from 30° to 60°, the length of the
shadow of a tower reduces by 20 m. Find the
height of the tower.

Answer 1

Given :

• When the angle of elevation of the sun increases from 30° to 60°, the length of the shadow of a tower reduces by 20 m.

To find :

• Height of the tower.

Solution :

According to the attachment :-

• AB = Height of the tower.
• BC = Shadow of the tower.

Here,

• $\angle\:ADB = 60\degree$
• $\angle\:ACB=30\degree$
• DC = 20 m

Consider,

• BD = x m

Then,

BC =( x+20 ) m

∆ ABD is a right triangle.

→ In case of ∆ ABD,

$\implies\sf{\dfrac{AB}{BD}=tan60\degree}$

$\implies\sf{\dfrac{AB}{x}=\sqrt{3}}$

$\implies\sf{AB=x\sqrt{3}............(1)}$

∆ ABC is a right triangle.

→ In case of∆ ABC,

$\implies\sf{\dfrac{AB}{BC}=tan30\degree}$

$\implies\sf{\dfrac{x\sqrt{3}}{x+20}=\dfrac{1}{\sqrt{3}}\:[put\:AB=x\sqrt{3}\: from\:eq(1)]}$

$\implies\sf{3x=x+20}$

$\implies\sf{3x-x=20}$

$\implies\sf{2x=20}$

$\implies\sf{x=10}$

Now put x=10 in eq(1)

$\implies\sf{AB=x\sqrt{3}}$

$\implies\sf{AB=10\sqrt{3}}$

Therefore, the height of the tower is 10√3 m.

Answer 2

$\large{\boxed{\bold{\bold{\purple{\sf{Mark\:me\:Brainliest}}}}}}$