Question : The base of a triangle is $12\sqrt{3}$ cm and two angles at the base are 30° and 60°, respectively. The altitude of the triangle is:
Option 1: $12$ cm
Option 2: $6$ cm
Option 3: $10\sqrt{3}$ cm
Option 4: $9$ cm
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Correct Answer: $9$ cm
Solution :
Let BD be $x$ cm.
$\therefore$ CD =$ (12\sqrt{3} - x) $cm.
$\angle$ADB = $\angle$ADC = 90°
From $\triangle$ ABD,
$\tan 30°=\frac{AD}{BD}$
$⇒\frac{1}{\sqrt{3}}=\frac{AD}{x}$
So, AD = $\frac{x}{\sqrt{3}}$-----------(1)
From $\triangle$ACD,
$\tan 60°=\frac{AD}{CD}$
$⇒\sqrt{3}=\frac{AD}{ (12\sqrt{3} - x)}$
So, AD = $\sqrt{3}(12\sqrt{3}-x)$----------(2)
From 1 and 2, we get,
$\frac{x}{\sqrt{3}} = \sqrt{3}(12\sqrt{3}-x)$
$⇒x =36\sqrt{3}-3x$
$⇒4x = 36\sqrt{3}$
$⇒x = 9\sqrt{3}$
$\therefore$ AD $=\frac{x}{\sqrt{3}} = \frac{9\sqrt{3}}{\sqrt{3}} = 9$ cm
Hence, the correct answer is $9$ cm.
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