An observer -3 m tall is 3 m away from the pole 2 -3 m high- What is the angle of elevation of the top- -2 MARKS-

Concept: 1 Mark Application: 1 Mark Height of the pole that is above man's height = nbsp; nbsp;2√(3) nbsp;√(3) =√(3) m Hence, AB = nbsp;√(3)m BC = 3m ...

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Standard IX

Mathematics

Calculating Heights and Distances

An observer √...

Question

An observer $\sqrt{3}$ m tall is 3 m away from the pole 2 $\sqrt{3}$ m high. What is the angle of elevation of the top? [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Height of the pole that is above man's height = 2 $\sqrt{3}$ - $\sqrt{3}$ = $\sqrt{3}$ m

Hence, AB = $\sqrt{3}$ m

BC = 3m

Hence, $tan$ C = $\frac{A B}{B C}$ = $\frac{\sqrt{3}}{3}$ = $\frac{1}{\sqrt{3}}$

$⟹$ C, angle of elevation = $30^{\circ}$

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An observer √3m tall is 3 m away from the pole 2√3m high. What is the angle of elevation of the top? [2 MARKS]

Concept: 1 Mark Application: 1 Mark Height of the pole that is above man's height = 2√3 - √3 =√3 m Hence, AB = √3m BC = 3m Hence, tan C = ABBC = √33 = 1√3 ⟹ C, angle of elevation = 30∘

An observer $\sqrt{3}$ m tall is 3 m away from the pole 2 $\sqrt{3}$ m high. What is the angle of elevation of the top? [2 MARKS]

Concept: 1 Mark
Application: 1 Mark

Height of the pole that is above man's height = 2 $\sqrt{3}$ - $\sqrt{3}$ = $\sqrt{3}$ m

Hence, AB = $\sqrt{3}$ m

BC = 3m

Hence, $tan$ C = $\frac{A B}{B C}$ = $\frac{\sqrt{3}}{3}$ = $\frac{1}{\sqrt{3}}$

$⟹$ C, angle of elevation = $30^{\circ}$