Question

Answer the above question.

No spam ❎
​

Answer 1

$\huge{ \underline{ \overline{ \mid{ \rm \pink{Answer - }} \mid}}}$

$\underline{ \bold{Given - }}$

Angles of depression of two ships on the opposite sides are $\alpha \: and \: \beta$

And the height of lighthouse is h metres.

$\underline{ \bold{To \: Prove - }}$The distance between the ships is $\frac{h( \tan \alpha + \tan \beta ) }{ \tan \alpha \tan\beta }$

$\underline{ \bold{Proof - }}$

For figure, refer to the above attachment.

Now, Let AB be the lighthouse and C and D be the positions of the two ships. Then AB =h metres.

Clearly,

$\angle ACB = \alpha \: and \angle ADB = \beta$

Let AC=$x \: and \: AD = y$

From right $\triangle CAB$

$\frac{AC }{AB } = \cot \alpha \implies \: \frac{x}{h} = \cot\alpha \\ \implies \: x = h \cot \alpha \: \: \: ....(1)$

From right $\triangle DAB$

$\frac{AD }{AB } = \cot\beta \implies \: \frac{x}{h} = \cot\beta \\ \implies \: y = h \cot\beta \: \: \: \: ......(2)$

Adding the corresponding sides of (1)and (2),

$x + y = h( \cot\alpha + \cot \beta ) \\ \implies \: x + y = h( \frac{1}{ \tan\alpha } + \frac{1}{ \tan\beta } ) \\ \implies \: x + y = \frac{h( \tan\alpha + \tan\beta ) }{ \tan\alpha \tan \beta }$

Hence the distance between the ships is $\frac{h( \tan \alpha + \tan\beta ) }{ \tan\alpha \tan\beta }$

Answer 2

$\huge\orange{\boxed{\bold{Solution}}}$

________________________________________

Let AB be the lighthouse of height h metres.

Let AC = x and AD = y

In ∆CAB,

$\frac{AB}{AC} = \tan( \alpha) \\ \\ \tan( \alpha ) = \frac{h}{x} \\ \\ = > x = \frac{h}{ \tan( \alpha ) }$

In ∆DAB,

$\tan( \beta ) = \frac{h}{y} \\ \\ y = \frac{h}{ \tan( \beta ) }$

Distance between the ships = x + y

$= ( \frac{h}{ \tan( \alpha ) } + \frac{h}{ \tan( \beta ) } ) \\ \\ h \times( \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha) + \tan( \beta )} )$

_________________________________________

[tex]\huge\mathbb\blue{THANK\:YUH!}[/tex]