Question

A 8m 40cm high vertical pole cast a shadow 4m 80cm. Find at the same time
a)The height of a pole which casts a shadow 4m long.
b)The length of the shadow cast by another pole 12m 25cm high.​

Answer 1

See the photo attached for solution

Answer 2

Given :-

• Height of the pole (P) = 8m 40 cm
• Length of the shadow (B) = 4m 80 cm

Solution:-

In Right angled Δ ABC formed,

• Height of the pole, Perpendicular, P = 8.40 m
• Length of the shadow, Base, B = 4.80 m

Since,

$\because \rm \tan \theta = \frac{P}{B}$

Therefore,

$\therefore \mathrm{\tan \theta = \frac{8.40}{4.80}}$

$\implies \mathrm{\tan \theta = \cancel{\frac{8.40}{4.80}}}$

$\implies \boxed{\bold{\mathrm{\tan \theta = \frac{3}{2}}}}$

(a) To find :- The height of a pole which casts a shadow 4m long.

Length of the shadow, Base, B = 4m

Since,

$\because \rm \tan \theta = \frac{P}{B}$

Therefore,

$\therefore \rm \tan \theta = \frac{P}{4}$

Hence, the sun is casting the same angle of elevation of light, so angle of suspension of shadow formed of all poles is equals, i.e., $\it \tan \theta$ is equal in both poles.

So,

$\because \rm \tan \theta = \frac{3}{2}$

Therefore,

$\therefore \rm \frac{3}{2} = \frac{P}{4}$

By Cross Multiplication,

$\implies \rm 2 \times P = 4 \times 3$

$\implies \rm P = \frac{4 \times 3}{2}$

$\implies \rm P = \frac{\cancel{4} \times 3}{\cancel{2}}$

$\implies \rm P = 2 \times 3$

$\implies \rm \bf P = 6 m$

Hence, the height of the pole that casts that shadow is 6 m.

(b) To find :- The length of shadow cast by the another pole 12 m 25 cm high.

Solution :-

Length of the pole, P = 12.25 m

Since,

$\because \rm \tan \theta = \frac{P}{B}$

Therefore,

$\therefore \rm \tan \theta = \frac{12.25}{B}$

Also,

$\because \rm \tan \theta = \frac{3}{2}$

So,

$\therefore \rm \frac{3}{2} = \frac{12.25}{B}$

By Cross multiplication,

$\implies \rm 3 \times B = 2 \times 12.25$

$\implies \rm B = \frac{2 \times 12.25}{3}$

$\implies \rm B = \frac{24.50}{3}$

$\implies \rm B = \cancel{\frac{24.50}{3}}$

$\implies \rm \bf B = 8.16 m$

Hence, the length of the shadow cast by that pole is 8.16 m or 8m 16cm.