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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