The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45-and-30- Find the distance between the two objects- -Take -3- - 1-732-73-2 metres-107-5 meters-150 metres-200 metres -

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Angle of Depression

The angles of...

Question

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be $45^{\circ} a n d 30^{\circ}$ . Find the distance between the two objects. (Take $\sqrt{3} = 1.732$ )

73.2 metres

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107.5 meters

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150 metres

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200 metres

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Solution

The correct option is A
73.2 metres

Let C and D be the objects and CD be the distance between the objects.

In $Δ A B C$ , $t a n 45^{\circ}$ = $\frac{A B}{A C}$ = 1

$A B = A C = 100 m$

In $Δ A B D$ , $t a n 30^{\circ}$ = $\frac{A B}{A D}$
$A D \times \frac{1}{\sqrt{3}} = 100$
$A D = 100 \times \sqrt{3} = 173.2 m$
$C D = A D - A C = 173.2 - 100$
$C D = 73.2 m$

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The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be 45∘ and 30∘. Find the distance between the two objects. (Take √3=1.732)

The correct option is A 73.2 metres Let C and D be the objects and CD be the distance between the objects. In ΔABC, tan 45∘ = ABAC = 1 AB=AC=100 m In ΔABD, tan 30∘ = ABAD AD×1√3=100 AD=100×√3=173.2m CD=AD−AC=173.2−100 CD=73.2 m

The angles of depression of two objects from the top of a 100 m hill lying to its east are found to be $45^{\circ} a n d 30^{\circ}$ . Find the distance between the two objects. (Take $\sqrt{3} = 1.732$ )