Q.The base of a triangle is \(16\sqrt{3}\) cm, and the two angles adjacent to the base are 30° and 60°. What is the height of the triangle?

From triangle \(∆ADC\), \(\tan 30^\circ = \frac{AD}{DC}\) ⇒ \(\frac{1}{\sqrt{3}} = \frac{AD}{DC}\) ⇒ \(DC = AD\sqrt{3}\) Again, from triangle \(∆ABD\), \(\tan 60^\circ = \frac{AD}{BD}\) ⇒ \(\sqrt{3} = \frac{AD}{BD}\) ⇒ \(BD = \frac{AD}{\sqrt{3}}\) Since \(BD + DC = BC\), \(\frac{AD}{\sqrt{3}} + AD\sqrt{3} = 16\sqrt{3}\) ⇒ \(\frac{AD + 3AD}{\sqrt{3}} = 16\sqrt{3}\) ⇒ \(4AD = 48\) ⇒ \(AD = 12\) ∴ The height of the triangle is 12 cm.
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