Question

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Answer 1

Answer:-

• In the figure A is the point $a$ metre above the lake, and B is horizontally $d$ metre away from A. The cloud is at point C which is $x$ metre vertically above B, so that the cloud is at $a+x$ metre above the lake. Probably its reflection will be

⭐ Required Diagram ⭐

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,0){\line(1,0){60}}\multiput(0,-30)(8,0){8}{\line(1,0){4}}\multiput(60,30)(0,-60){2}{\circle*{2}}\multiput(0,10)(60,0){2}{\put(-1,-0.5){ a }\put(0,3){\vector(0,1){7}}\put(0,-3){\vector(0,-1){7}}}\put(60,27){\vector(0,1){3}}\put(60,23){\vector(0,-1){3}}\put(58.8,24){ x }\put(0,20){\line(6,1){60}}\put(0,20){\line(1,0){60}}\put(0,20){\line(6,-5){60}}\put(55.25,-16){ a+x }\put(60,-12){\vector(0,1){12}}\put(60,-18){\vector(0,-1){12}}\multiput(0,0)(2,0){31}{\qbezier(0,0)(-0.5,-0.5)(-1,-1)}\qbezier(15,20)(15,21)(14.5,22.2)\qbezier(5.5,20)(5.5,18)(3.75,17)\put(17.5,20.75){ \alpha }\put(6.75,16){ \beta }\put(-5,20){ A }\put(62,20){ B }\put(62,30){ C }\put(62,-30){ D }\put(30,15){ d }\end{picture}$

From triangle ABC,

$\longrightarrow\tan\alpha=\dfrac{x}{d}$

$\longrightarrow d=\dfrac{x}{\tan\alpha}\quad\quad\dots(1)$

From triangle ABD,

$\longrightarrow\tan\beta=\dfrac{2a+x}{d}$

$\longrightarrow d=\dfrac{2a+x}{\tan\beta}\quad\quad\dots(2)$

From (1) and (2),

$\longrightarrow\dfrac{x}{\tan\alpha}=\dfrac{2a+x}{\tan\beta}$

$\longrightarrow\dfrac{\tan\beta}{\tan\alpha}=\dfrac{2a+x}{x}$

Subtracting 1,

$\longrightarrow\dfrac{\tan\beta}{\tan\alpha}-1=\dfrac{2a+x}{x}-1$

$\longrightarrow\dfrac{\tan\beta-\tan\alpha}{\tan\alpha}=\dfrac{2a}{x}$

$\longrightarrow x=\dfrac{2a\tan\alpha}{\tan\beta-\tan\alpha}$

Hence the height of the cloud in metre is,

$\longrightarrow h=a+x$

$\longrightarrow h=a+\dfrac{2a\tan\alpha}{\tan\beta-\tan\alpha}$

$\longrightarrow h=\dfrac{a\tan\alpha+a\tan\beta}{\tan\beta-\tan\alpha}$

$\longrightarrow h=\dfrac{\left(\dfrac{a\sin\alpha}{\cos\alpha}+\dfrac{a\sin\beta}{\cos\beta}\right)}{\left(\dfrac{\sin\beta}{\cos\beta}-\dfrac{\sin\alpha}{\cos\alpha}\right)}$

$\longrightarrow h=\dfrac{\left(\dfrac{a\sin\alpha\cos\beta+a\cos\alpha\sin\beta}{\cos\alpha\cos\beta}\right)}{\left(\dfrac{\sin\beta\cos\alpha-\cos\beta\sin\alpha}{\cos\alpha\cos\beta}\right)}$

$\longrightarrow h=\dfrac{a(\sin\alpha\cos\beta+\cos\alpha\sin\beta)}{\sin\beta\cos\alpha-\cos\beta\sin\alpha}$

$\longrightarrow\underline{\underline{h=\dfrac{a\sin(\alpha+\beta)}{\sin(\beta-\alpha)}}}$

• Hence (2) is the answer

Answer 2

Explanation:

Option b is right dear....