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Mathematics

From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, the distance between the ships is:

  1. h(tan⁡α+tan⁡β)h(\tan \alpha + \tan \beta)

  2. (htan⁡αtan⁡βtan⁡α+tan⁡β)\Big(\dfrac{h \tan \alpha \tan \beta}{\tan \alpha + \tan \beta}\Big)

  3. (h(tan⁡α+tan⁡β)tan⁡αtan⁡β)\Big(\dfrac{h(\tan \alpha + \tan \beta)}{\tan \alpha \tan \beta}\Big)

  4. (h(cot⁡α+cot⁡β)cot⁡αcot⁡β)\Big(\dfrac{h(\cot \alpha + \cot \beta)}{\cot \alpha \cot \beta}\Big)

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Answer

From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be α and β. If the height of the lighthouse be h metres and the line joining the ships passes through the foot of the lighthouse, the distance between the ships is: Volume And Surface Area of solid RSA Mathematics Solutions ICSE Class 10.

Let AB(h) be the height of the lighthouse.

Let P and Q be the two ships on opposite sides of the lighthouse.

Let BP = x meters and BQ = y meters.

In triangle ABP,

⇒tan⁡α=hx⇒x=htan⁡α.\Rightarrow \tan \alpha = \dfrac{h}{x} \\[1em] \Rightarrow x = \dfrac{h}{\tan \alpha}.

In triangle ABQ,

⇒tan⁡β=hy⇒y=htan⁡β.\Rightarrow \tan \beta = \dfrac{h}{y} \\[1em] \Rightarrow y = \dfrac{h}{\tan \beta}.

The total distance between the ships is = x + y

=htan⁡α+htan⁡β=h(1tan⁡α+1tan⁡β)=h(tan⁡β+tan⁡αtan⁡αtan⁡β)=h(tan⁡α+tan⁡β)tan⁡αtan⁡β.= \dfrac{h}{\tan \alpha} + \frac{h}{\tan \beta} \\[1em] = h \Big( \dfrac{1}{\tan \alpha} + \dfrac{1}{\tan \beta} \Big) \\[1em] = h \Big( \dfrac{\tan \beta + \tan \alpha}{\tan \alpha \tan \beta} \Big) \\[1em] = \dfrac{h(\tan \alpha + \tan \beta)}{\tan \alpha \tan \beta}.

Hence, option 3 is the correct option.

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