Question

2
A
A 15 metres high tower casts a shadow of 24 metres long at a certain
24 metres long at a certain time and at the
same time, a telephone pole casts a shadow 16 metres long. Find the neig
telephone pole.
PLEASE EXPLAIN HOW TO PROVE THE TRIANGLES SIMILAR?. I NEED TO KNOW THIS ONLY. BY PROVIND 2 ANGLES EQUAL.​

Answer 1

Question :

A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the

same time, a telephone pole casts a shadow 16 metres long. Find the height

telephone pole.

Solution :

In question we have given the height of tower 15 m and height of telephone pole 16 m.

The tower casts a shadow of 24 m.

(Refer the attachment for fig.)

Assume a triangle ABC such that -

• AB = 15 m (height of tower)
• BC = 24 m (shadow of tower)
• CE = 16 m (shadow of telephone pole)

We have to find the height of telephone pole.

We know that.. both tower and telephone pole make an angle of 90° with the ground.

Now,

In ΔABC

=> tan 90° = AB/BC

=> tan 90° = 15/24 ____ (eq 1)

Similarly,

In ΔDEC

=> tan 90° = DE/CE

=> tan 90° = DE/16 ____ (eq 2)

On comparing (eq 1) & (eq 2) we get,

=> 15/24 = DE/16

Cross multiply them

=> 15 × 16 = 24 × DE

=> 240 = 24 × DE

=> DE = 10 m

∴ Height of tower is 10 m.

Answer 2

Answer:

$\tt\huge{Solution:}$

Let BC = 15 m be the tower and its shadow AB is 24 m.

At that time ∠CAB = 8, Again, let EF = h be a telephone pole and its shadow DE = 16 m.

At the same time ∠EDF = 8 Here, ΔASC and ΔDEF both are right angled triangles.

•°• Height of telephone pole is 10m.