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A tower subtends angles $\alpha, 2 \alpha$ and $3 \alpha$ respectively at points $A, B$ and $C$, all lying on a horizontal line through the foot of the tower, then $\frac{A B}{B C}$ is equal to:
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Verified Answer
The correct answer is:
$1+2 \cos 2 \alpha$
In $\triangle E C D$,
$\tan 3 \alpha=\frac{h}{C D}$
$\Rightarrow \quad C D=h \cot 3 \alpha \quad \ldots(\mathrm{i})$
In $\triangle E B D$,
$\tan 2 \alpha=\frac{h}{B D}$
$\Rightarrow \quad B D=h \cot 2 \alpha \quad \ldots(\mathrm{ii})$
In $\triangle E A D$,
$\tan \alpha=\frac{h}{A D}$
$\Rightarrow \quad A D=h \cot \alpha \quad \ldots(\mathrm{iii})$
From Eqs. (ii) and (iii),
$A D-B D=h \cot \alpha-h \cot 2 \alpha$
$A B=h(\cot \alpha-\cot 2 \alpha) \quad \ldots(\mathrm{iv})$
From Eqs. (i) and (ii),
$B D-C D=h \cot 2 \alpha-h \cot 3 \alpha$
$B C=h(\cot 2 \alpha-\cot 3 \alpha) \quad \ldots(\mathrm{v})$
From Eqs. (iv) and (v),
$\frac{A B}{B C}=\frac{h(\cot \alpha-\cot 2 \alpha)}{2(\cot 2 \alpha-\cot 3 \alpha)}$
$\Rightarrow \frac{A B}{B C}=\frac{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha}-\frac{\cos 3 \alpha}{\sin 3 \alpha}}=\frac{\frac{\sin (2 \alpha-\alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin (3 \alpha-2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}}$
$=\frac{\sin 3 \alpha}{\sin \alpha}=\frac{3 \sin \alpha-4 \sin ^3 \alpha}{\sin \alpha}$
$=3-4 \sin ^2 \alpha$
$=-3-2(1-\cos 2 \alpha)$
$=1+2 \cos 2 \alpha$
$\tan 3 \alpha=\frac{h}{C D}$
$\Rightarrow \quad C D=h \cot 3 \alpha \quad \ldots(\mathrm{i})$
In $\triangle E B D$,
$\tan 2 \alpha=\frac{h}{B D}$
$\Rightarrow \quad B D=h \cot 2 \alpha \quad \ldots(\mathrm{ii})$
In $\triangle E A D$,
$\tan \alpha=\frac{h}{A D}$
$\Rightarrow \quad A D=h \cot \alpha \quad \ldots(\mathrm{iii})$
From Eqs. (ii) and (iii),
$A D-B D=h \cot \alpha-h \cot 2 \alpha$
$A B=h(\cot \alpha-\cot 2 \alpha) \quad \ldots(\mathrm{iv})$
From Eqs. (i) and (ii),
$B D-C D=h \cot 2 \alpha-h \cot 3 \alpha$
$B C=h(\cot 2 \alpha-\cot 3 \alpha) \quad \ldots(\mathrm{v})$
From Eqs. (iv) and (v),
$\frac{A B}{B C}=\frac{h(\cot \alpha-\cot 2 \alpha)}{2(\cot 2 \alpha-\cot 3 \alpha)}$
$\Rightarrow \frac{A B}{B C}=\frac{\frac{\cos \alpha}{\sin \alpha}-\frac{\cos 2 \alpha}{\sin 2 \alpha}}{\frac{\cos 2 \alpha}{\sin 2 \alpha}-\frac{\cos 3 \alpha}{\sin 3 \alpha}}=\frac{\frac{\sin (2 \alpha-\alpha)}{\sin \alpha \sin 2 \alpha}}{\frac{\sin (3 \alpha-2 \alpha)}{\sin 2 \alpha \sin 3 \alpha}}$
$=\frac{\sin 3 \alpha}{\sin \alpha}=\frac{3 \sin \alpha-4 \sin ^3 \alpha}{\sin \alpha}$
$=3-4 \sin ^2 \alpha$
$=-3-2(1-\cos 2 \alpha)$
$=1+2 \cos 2 \alpha$
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