The height of a hill is 150 m. From the top of the hill, the angle of depression of two objects lying towards the east of the hill is $45^{\circ} a n d 30^{\circ}$ . Find the distance between the objects.

$\sqrt{3} = 1.732$

$109.8 m$

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$107.5 m$

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$150 (\sqrt{3} - 1) m$

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$200 m$

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Solution

The correct options are
A
$109.8 m$

C
$150 (\sqrt{3} - 1) m$

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

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

$A B = A C = 150 m$

$\frac{A B}{A D} = t a n 30^{\circ}$

$A D \times \frac{1}{\sqrt{3}} = 150$

$A D = 150 \times \sqrt{3}$

$C D = A D - A C = 150 (\sqrt{3} - 1)$

$C D = A D - A C = 259.8 - 150 = 109.8 m$

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The height of a hill is 150 m. From the top of the hill, the angle of depression of two objects lying towards the east of the hill is 45∘ and 30∘. Find the distance between the objects. √3=1.732

The correct options are A 109.8 m C 150(√3−1) m Let C and D be the objects and CD be the distance between the objects. ABAC=tan 45∘=1 AB=AC=150 m ABAD=tan 30∘ AD×1√3=150 AD=150×√3 CD=AD−AC=150(√3−1) CD=AD−AC=259.8−150=109.8 m

The height of a hill is 150 m. From the top of the hill, the angle of depression of two objects lying towards the east of the hill is $45^{\circ} a n d 30^{\circ}$ . Find the distance between the objects.

$\sqrt{3} = 1.732$