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- What is the distance between the two objects- Take -3 - 1-732

Let C and D be the objects and CD be the distance between the objects. In Δ ABC, tan 45^∘ = AB/AC = 1 AB = AC = 100 m In nbsp;Δ ABD, tan 30^∘ = AB/AD AD × ...

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Standard X

Mathematics

Calculating Heights and Distances

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 $45^{\circ} and 30^{\circ}$ . What is the distance between the two objects? (Take $\sqrt{3} = 1.732$ )

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Solution

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

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

$A B = A C = 100 m$

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

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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∘. What is the distance between the two objects? (Take √3=1.732)

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 $45^{\circ} and 30^{\circ}$ . What is the distance between the two objects? (Take $\sqrt{3} = 1.732$ )