Question

A pole of height 10m casts a shadow of 2m long on the ground. At the same time a tower casts a shadow of 50m on the ground. Then find the height of the tower.

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\therefore{\text{ Height\:of\:tower=250\:m}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• In the given question information given about a pole of height 10 metre cast a 2 metre long shadow on the ground at the same time a tower cast a 50 metre long shadow on ground.

• We have to find the height of Tower.

$\green{\underline \bold{Given : }} \\ :\implies \text{Height \: of \: pole = 10 \: m} \\ \\ :\implies \text{Shadow \: of \: pole = 2 \: m} \\ \\ :\implies \text{Shadow \: of \: tower = 50 \: m} \\ \\ \red{\underline \bold{To \: Find : }} \\ :\implies \text{Height \: of \: tower = ?}$

• According to given question :

$\bold{For \: pole : } \\ :\implies tan \: \theta = \frac{p}{b} \\ \\ :\implies tan \: \theta = \frac{Height \: of \: pole} {Shadow \: of \: pole} \\ \\ :\implies tan \: \theta = \frac{10}{2} \\ \\ :\implies tan \: \theta = 5 \\ \\ \bold{For \: tower : } \\ :\implies tan \: \theta = \frac{Height \: of \: tower}{Shadow \: of \: tower} \\ \\ :\implies 5= \frac{H}{50} \\ \\ :\implies \: H=50\times 5 \:m\\ \\ \green{:\implies H= 250 \: m}$

Answer 2

Question :--- A pole of height 10m casts a shadow of 2m long on the ground. At the same time a tower casts a shadow of 50m on the ground. Then find the height of the tower. ?

Formula used :--

→ when Time is same angle of Elevation and Angle of Depression of sun will be same all over ...

→ Tan@ = Perpendicular/Base

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Solution :---

✪✪ Case 1 ✪✪

❁❁ Refer To Image First .. ❁❁

in ∆ABC , we have ,

→ AB = Pole = 10m

→ BC = shadow of Pole = 2m .

→ Angle ACB = @

From this we get,

→ Tan@ = Perpendicular/Base

→ AB/BC = 10/2 = 5 ------------- Equation (1)

_____________________________

✪✪ Case 2 ✪✪

❁❁ Refer To Image First .. ❁❁

in ∆DEF , we have ,

→ DE = Building = Let h m.

→ EF = shadow of Building = 50m .

→ Angle ACB = @ { as Time is same }.

From this we get,

→ Tan@ = Perpendicular/Base

→ DE/EF = h/50 ----------------- Equation (2)

_____________________________

Putting value of Tan@ From Equation (1) in Equation (2) now , we get,

→ 5 = h/50

cross- Multiplying we get,

→ 5*50 = h

→ h = 250m.

Hence, Height of Building will be 250m at the same Time..

_____________________________

since, Time is same in both case, we can do this in simple way also ,, { or we can say Time is constant .}

→ when shadow is 2m, Height is = 10m

→ when shadow is 1m, Height is = (10/2)m

→ when shadow is 50m , Height is = (10/2) * 50 = 250m.

So,, Height of Building will be 250m at the same Time..

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