Question

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Answer 1

Solution:-

Given:-

=> ∠Q = 90⁰

=> YZ = h ( we assume height of the light house be h )

=> AB = x ( given)

Now take ∆BYZ

∠ = 2a

$\rm \: \tan 2a = \dfrac{p}{b} = \dfrac{h}{BZ}$

So now

$\rm \: BZ = \frac{h}{ \tan2a} \: \: \: \: \: \: ........(i)$

In ∆YZC

$\rm \: \tan3a = \dfrac{h}{ZC}$

$\rm \: ZC = \frac{h}{ \tan3a } \: \: \: \: \: \: \: .....(ii)$

In ∆AYZ

$\rm \: \tan a = \dfrac{h}{ZA}$

Now we can write

ZA = ZC + CB + BA

$\rm \: \tan a = \dfrac{h}{ZC + CB + BA }$

Now value of

BA = x ( given)

$\rm \: \tan a = \dfrac{h}{ZC + CB + x }$

$\rm \: ZC + CB + x = \frac{h}{ \tan a }$

Now we can write as

ZB = ZC + CB

=> CB = ZB - ZC

$\rm \: ZC +ZB - ZC + x = \frac{h}{ \tan a }$

$\rm \: \cancel{ZC }+ZB - \cancel{ZC} + x = \frac{h}{ \tan a }$

We get

$\rm \: ZB + x = \frac{h}{ \tan a }$

$\rm \: \to \: ZB = \dfrac{h}{ \tan2a}$

Now

$\rm \: \dfrac{h}{ \tan2a} + x = \frac{h}{ \tan a }$

Taking lcm

$\rm \: \dfrac{h + x \tan2a }{ \tan2a} = \frac{h}{ \tan a }$

Using cross multiplication

$\rm \tan \: a(x \tan2a + h) = h \tan2a$

$\rm \: \to x \tan a \: \tan2a + h \tan \: a = h \tan2a$

$\rm \: \to x \tan a \: \tan2a = h \tan2a -h \tan \: a$

$\rm \to\: x \tan a \: \tan2a = h (\tan2a - \tan \: a )$

$\to \rm \: h = \frac{x \tan \: a \tan2a }{ \tan2a - \tan a }$

Answer 2

Hii

Refer to the attachment ✌️