A statue 1.6 m tall stands on a pedestal

Asked by admin @ 17/08/2021 in Math viewed by 618 People

A statue 1.6 m tall stands on the top of a pedestal. From a point on the ground the angle of elevation of the top of the statue is 60 degree and from the same point the angle of elevation of the top of the pedestal is 45 degree. Find the height of the pedestal.

Answered by admin @ 17/08/2021

$\huge\sf\pink{Answer}$

☞ Height of the pedestal is 0.8(√3+1) m

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$\huge\sf\blue{Given}$

✭ Height of statue is 1.6 m

✭ From group angle of elevation is 60°

✭ From the same point the angle of elevation to the top of the pedestal is 45°

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$\huge\sf\gray{To \:Find}$

◈ The height of the pedestal?

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$\huge\sf\purple{Steps}$

So we shall use the trigonometric function,tan θ

$\underline{\boxed{\sf tan \theta = \dfrac{Opposite}{Adjacent}}}$

In ∆BCD

➝ $\sf tan\theta = \dfrac{Opposite}{Adjacent}$

➝ $\sf tan 45^{\circ} = \dfrac{BC}{CD}$

$\sf \bigg\lgroup tan \ 45^{\circ} = 1\bigg\rgroup$

➝ $\sf 1 = \dfrac{BC}{CD}$

➝ $\sf CD = BC \:\:\: -eq(1)$

In ∆ACD

➳ $\sf tan \theta = \dfrac{Opposite}{Adjacent}$

➳ $\sf tan \ 60^{\circ} = \dfrac{AC}{CD}$

➳ $\sf tan \ 60^{\circ} = \dfrac{AB+BC}{CD}$

$\sf \bigg\lgroup tan \ 60^{\circ} = \sqrt{3} \bigg\rgroup$

➳ $\sf \sqrt{3} = \dfrac{AB+BC}{CD}$

➳ $\sf CD\sqrt{3} = AB+BC$

➳ $\sf BC\sqrt{3} = 1.6+BC$

➳ $\sf BC\sqrt{3} -BC = 1.6$

➳ $\sf BC(\sqrt{3}-1) = 1.6$

➳ $\sf BC = \dfrac{1.6}{\sqrt{3}-1}$

➳ $\sf \orange{BC = 0.8(\sqrt{3}+1)}$

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