Question

the Shadow of the tower standing on a level plane is found to be 50 m longer when sun's evaluation is 30 degree Celsius then when is is 60 degree find the height of the tower
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Answer 1

We have:-

• The shadow of the tower standing on a level plane is found to be 50 m longer when sun's evaluation is 30° Celsius then when is is 60°.

To FinD:-

• Height of the tower$?$

Solution:-

Let the height of the tower AB = h

When sun's elevation is 30°, length of the shadow will be BC = 50 + x.

When sun's elevation is 60°, length of the shadow will be PB = x.

As per the question, the shadow increases by 50 when the elevation of the sun increases from 30° to 60°. We need to find the height of the tower where angle B = 90°.

In ∆ABC,

⇒ tan 30° = AB / BC

⇒ tan 30° = h / 50 + x

⇒ 1/√3 = h / 50 + x

⇒ h√3 = 50 + x ----------(1)

In ∆ABP,

⇒ tan 60° = AB / BP

⇒ √3 = h / x

⇒ h = x√3 ---------(2)

Now putting the value of h in equation (2),

⇒ x√3 = 50 + x

⇒ x√3 - x = 50

⇒ x(√3 - 1) = 50

⇒ x = 25(√3 + 1)

Then,

⇒ h = 25√3(√3 + 1)

Now considering √3 = 1.732

⇒ h = 25 × 1.732 × 2.732

⇒ h = 118.23 m (approx.)

Hence:-

The required height of the tower is 118.23 m or in the form of roots, 25√3(√3 + 1) m.

Answer 2

Answer:

$\huge \bf \: required \: answer$

Let the height of the tower AB = h

Sun's elevation = 30°, Therefore length of the shadow will be BC = 50 + x.

sun's elevation = 60°, length of the shadow will be PB = x.

Now,

$\sf \bold in \: \triangle \: abc \:$

$\sf \tan(30 \degree) = \dfrac{ab}{bc}$

$\sf \: \tan(30 \degree) = \dfrac{h}{ x + 50}$

• tan30⁰ = 1/√3

$\sf \: \dfrac{1}{ \sqrt{3} } = \dfrac{h }{50+x}$

$\sf \: h \sqrt{3} = 50 + x$

$\sf \bold {in \triangle \: abp}$

$\sf \: tan 60° = \dfrac{ab}{b}$.

$\sf \: √3 = \dfrac{h}{x}$

$\sf \: h = x√3 ---(2)$

Now putting the value of h in equation (2),

$\sf \: x√3 = 50 + x \\ \sf \: x√3 - x = 50$

$\sf \: x(√3 - 1) = 50 \\ \sf x = 25(√3 + 1)$

$\sf \: h \: = 25 \sqrt{3} + (\sqrt{3} + 1)$

As we know that

√3 = 1.732

$\sf \: h = 25 × 1.732 × 2.732$

$\sf h = 118.23 m (approx)$