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In triangle $ABC,$ the angle bisector of $\angle BAC$ meets $\overline{BC}$ at $D.$ If $\angle BAC = 60^\circ,$ $\angle ABC = 60^\circ,$ and $AD = 24,$ then find the area of triangle $ABC.$

 Mar 25, 2024

Best Answer 

 #1
avatar+394 
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Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

\({x}^{2}+{24}^{2}={(2x)}^{2}\).

We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).

 Mar 25, 2024
 #1
avatar+394 
+2
Best Answer

Because BAC = 60, ABC = 60, ACB should equal 60.

Therefore, ABC is equilateral.

Because AD is an angle bisector, then ABD, is 30, and so ADB = ADC = 90.

Therefore, set BD = x.

\({x}^{2}+{24}^{2}={(2x)}^{2}\).

We get \(x = 8\sqrt{3}\). (These triangle ratios are very special, see if you can find the ratios for a 30-60-90 triangle for yourself)

Therefore the area is \(\frac{8\sqrt{3}*24*2}{2}=192\sqrt{3}\).

hairyberry Mar 25, 2024

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