The angle of elevation of a cloud from a point h metre above a lake is $\theta$. The angle of depression of its reflection in the lake is $45^\circ$. The height of the cloud is

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KollegeApply · Accepted Answer

Let AB be the surface of the lake and p be the point of observation. So AP=h . The given situation can be represented as, Here,C is the position of the cloud and C' is the reflection in the lake. Then .CB=C'B Let PM be the perpendicular from P on CB. Then $\angle$CPM=$\theta$ and .$\angle$C'PM=$45^\circ$ LetCM=x , PM=y , thenCB=x+h and C'B=x+h Here, we have to find the height of cloud. So we use trigonometric ratios. In , ΔCMP $\Rightarrow$ tan $\theta$=(CM)/(PM) $\Rightarrow$ tan$\theta$=x/y $\Rightarrow$ y= x/tan$\theta$ (1) Again in ΔPMC', $\Rightarrow$ tan $45^\circ$=CM/PM $\Rightarrow$1=(C'B+BM)/(PM) $\Rightarrow$1=(x+h+h)/y $\Rightarrow$y=x+2h $\Rightarrow$ x/tan$\theta$=x+2h [using (1)] $\Rightarrow$ (x-x tan$\theta$)/tan $\theta$=2h $\Rightarrow$ x=(2h tan $\theta$)/((1-tan $\theta$)) Now, $\Rightarrow$ CB=h+x $\Rightarrow$ CB=h+(2h tan $\theta$)/(1-tan $\theta$) `$\Rightarrow$ CB=(h(1-tan$\theta$)+2h tan$\theta$)/(1-tan $\theta$) $\Rightarrow$ CB= (h(1+tan $\theta$))/(1-tan $\theta$)=h tan($45^\circ$+ $\theta$ ) [tan (A+B)=(tan A+tan B)/(1-tan A tan B)] ` [Solution diagram: 63555_solution_1.png]

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