Question

the angle of elevation of the top of a hill is 30° from a point on the ground. on walking 1 km walking towards the hill, angle is found to be 45°. calculate the height of the hill​

Answer 1

$\huge\sf\pink{Answer}$

☞ Height of the hill is 1.37 km

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$\huge\sf\blue{Given}$

✭ Angle of elevation at the top of a hill is 30° from a point on the ground

✭ On walking 1 km towards the hill,the angle of elevation becomes 45°

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

◈ The height of the hill?

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$\huge\sf\purple{Steps}$

So here In ∆ACB

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

➝ $\sf tan \ 45^{\circ} = \dfrac{AB}{x}$

«« $\sf tan \ 45^{\circ} = 1$ »»

➝ $\sf 1 = \dfrac{AB}{x}$

➝ $\sf x = AB \:\:\: -eq(1)$

In ∆ADC

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

➳ $\sf tan \ 30^{\circ} = \dfrac{AB}{DB}$

➳ $\sf tan \ 30^{\circ} = \dfrac{AB}{1+x}$

«« $\sf tan \ 30^{\circ} = \dfrac{1}{\sqrt{3}}$ »»

➳ $\sf \dfrac{1}{\sqrt{3}} = \dfrac{AB}{1+x}$

➳ $\sf 1(1+x) = AB(\sqrt{3})$

➳ $\sf 1+x = AB\sqrt{3}$

➳ $\sf 1+AB = AB \sqrt{3} \:\:\ \because eq(1)$

➳ $\sf 1 = AB\sqrt{3} - AB$

➳ $\sf 1 = AB(\sqrt{3}-1)$

➳ $\sf \dfrac{1}{\sqrt{3}-1} = AB$

➳ $\sf \dfrac{1}{1.732-1} = AB$

➳ $\sf AB = \dfrac{1}{0.732}$

➳ $\sf \orange{AB = 1.37 \ km}$

$\sf \star\: Diagram \: \star$

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