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In the given figure, AB is a tower and two objects C and D are located on the ground on the same side of AB. When observed from the top B of the tower, their angles of depression are 45° and 60° respectively. Find the distance between the objects, if the height of the tower is 180 m.

In the given figure, AB is a tower and two objects C and D are located on the ground on the same side of AB. When observed from the top B of the tower, their angles of depression are 45° and 60° respectively. Find the distance between the objects, if the height of the tower is 180 m. Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

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Answer

Considering right angled △ABC, we get

⇒tan⁡45∘=PerpendicularBase=ABAC⇒1=180AC⇒AC=180 m\Rightarrow \tan 45^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{AC} \\[1em] \Rightarrow 1 = \dfrac{180}{AC} \\[1em] \Rightarrow AC = 180 \text{ m}

Considering right angled △ADB, we get

⇒tan⁡60∘=PerpendicularBase=ABAD⇒3=180AD⇒AD=1803⇒AD=103.92 m.\Rightarrow \tan 60^{\circ} = \dfrac{\text{Perpendicular}}{\text{Base}} = \dfrac{AB}{AD} \\[1em] \Rightarrow \sqrt{3} = \dfrac{180}{AD} \\[1em] \Rightarrow AD = \dfrac{180}{\sqrt{3}} \\[1em] \Rightarrow AD = 103.92 \text{ m.}

Distance between two objects (CD) = CA - DA = 180 - 103.92 = 76.08 m

Hence,the distance between two objects = 76.08 meters.

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