law of cosines

Geometry Level 3

Given above is △ A B C \triangle ABC with ∠ A = 6 0 ∘ \angle A=60^\circ . Circle O O is tangent at points D , E D,E and F F . Segment B D = 2 BD=2 and segment D C = 5 DC=5 . Find the area of △ A B C \triangle ABC . If your answer is in the form a 3 a\sqrt{3} , submit your answer as a a .


The answer is 10.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Let A E = A F = x AE=AF=x , then by cosine law, we have

49 = ( x + 2 ) 2 + ( x + 5 ) 2 − 2 ( x + 2 ) ( x + 5 ) cos ⁡ 60 49=(x+2)^2+(x+5)^2-2(x+2)(x+5) \cos~60

49 = x 2 + 4 x + 4 + x 2 + 10 x + 25 − 2 ( x 2 + 5 x + 2 x + 10 ) ( 1 2 ) 49=x^2+4x+4+x^2+10x+25-2(x^2+5x+2x+10)\left(\dfrac{1}{2}\right)

49 = 2 x 2 + 14 x + 29 − x 2 − 7 x − 10 49=2x^2 +14x+29-x^2-7x-10

Simplifying further, we get

x 2 + 7 x − 30 = 0 x^2+7x-30=0

By factoring, we get

x = 3 x=3

Therefore, the area of △ A B C \triangle ABC is 1 2 ( A B ) ( A C ) ( sin ⁡ 60 ) = 1 2 ( 5 ) ( 8 ) ( 3 2 ) = \dfrac{1}{2}(AB)(AC)(\sin~60)=\dfrac{1}{2}(5)(8)\left(\dfrac{\sqrt{3}}{2}\right)= 10 3 10\sqrt{3}

Finally, a = 10 \color{plum}\boxed{a=10}

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