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$A B C$ is a triangle right-angled at $B$. The hypotenuse $(A C)$ is four times the perpendicular $(B D)$ drawn to it from the opposite vertex and $A D < D C$.
What is one of the acute angle of the triangle?
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What is one of the acute angle of the triangle?
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The correct answer is:
$15^{\circ}$
Let $\mathrm{BD}=\mathrm{P}$ and $\mathrm{DE}=x$
$\Rightarrow
\quad A C=4 P$
Let $\mathrm{E}$ be mid-point of $\mathrm{AC}$. Then, $\mathrm{AE}=\mathrm{EC}=\mathrm{BE}=2 \mathrm{P}$
In $\Delta \mathrm{BDE},(\mathrm{BE})^{2}=(\mathrm{BD})^{2}+(\mathrm{ED})^{2}$
$\Rightarrow(2 P)^{2}=(P)^{2}+x^{2}$
$\Rightarrow \quad 4 \mathrm{P}^{2}=\mathrm{P}^{2}+\mathrm{x}^{2}$
$\Rightarrow \quad 3 \mathrm{P}^{2}=\mathrm{x}^{2} \Rightarrow x=\sqrt{3} \mathrm{P}$
Now, $\mathrm{AD}=2 \mathrm{P}-x=2 \mathrm{P}-\sqrt{3} \mathrm{P}=\mathrm{P}(2-\sqrt{3})$
$\mathrm{DC}=2 \mathrm{P}+x=2 \mathrm{P}+\sqrt{3} \mathrm{P}=\mathrm{P}(2+\sqrt{3})$
In $\Delta \mathrm{BAD}, \tan \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{P}}{\mathrm{P}(2-\sqrt{3})}$
$=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2-\sqrt{3}}=2+\sqrt{3}=\tan 75^{\circ}$
$\tan \alpha=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{P}(2-\sqrt{3})}{\mathrm{P}}=2-\sqrt{3}=\tan 15^{\circ}$
$\Rightarrow \quad \alpha=15^{\circ}$
As, $\Delta \mathrm{ABC}$ is right angled at $\mathrm{B}$, from figure $\alpha+\beta=90^{\circ}$ $\Rightarrow \quad 15+\beta=90^{\circ} \Rightarrow \beta=75^{\circ}$
In $\triangle \mathrm{ABC}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad 75^{\circ}+90^{\circ}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad \angle \mathrm{C}=180^{\circ}-165^{\circ}=15^{\circ}$
One of the a cute angle is $15^{\circ}$
$\Rightarrow
\quad A C=4 P$
Let $\mathrm{E}$ be mid-point of $\mathrm{AC}$. Then, $\mathrm{AE}=\mathrm{EC}=\mathrm{BE}=2 \mathrm{P}$
In $\Delta \mathrm{BDE},(\mathrm{BE})^{2}=(\mathrm{BD})^{2}+(\mathrm{ED})^{2}$
$\Rightarrow(2 P)^{2}=(P)^{2}+x^{2}$
$\Rightarrow \quad 4 \mathrm{P}^{2}=\mathrm{P}^{2}+\mathrm{x}^{2}$
$\Rightarrow \quad 3 \mathrm{P}^{2}=\mathrm{x}^{2} \Rightarrow x=\sqrt{3} \mathrm{P}$
Now, $\mathrm{AD}=2 \mathrm{P}-x=2 \mathrm{P}-\sqrt{3} \mathrm{P}=\mathrm{P}(2-\sqrt{3})$
$\mathrm{DC}=2 \mathrm{P}+x=2 \mathrm{P}+\sqrt{3} \mathrm{P}=\mathrm{P}(2+\sqrt{3})$
In $\Delta \mathrm{BAD}, \tan \mathrm{A}=\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{P}}{\mathrm{P}(2-\sqrt{3})}$
$=\frac{1}{2-\sqrt{3}} \times \frac{2+\sqrt{3}}{2-\sqrt{3}}=2+\sqrt{3}=\tan 75^{\circ}$
$\tan \alpha=\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{P}(2-\sqrt{3})}{\mathrm{P}}=2-\sqrt{3}=\tan 15^{\circ}$
$\Rightarrow \quad \alpha=15^{\circ}$
As, $\Delta \mathrm{ABC}$ is right angled at $\mathrm{B}$, from figure $\alpha+\beta=90^{\circ}$ $\Rightarrow \quad 15+\beta=90^{\circ} \Rightarrow \beta=75^{\circ}$
In $\triangle \mathrm{ABC}, \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad 75^{\circ}+90^{\circ}+\angle \mathrm{C}=180^{\circ}$
$\Rightarrow \quad \angle \mathrm{C}=180^{\circ}-165^{\circ}=15^{\circ}$
One of the a cute angle is $15^{\circ}$
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