Trigonometry! #142

Geometry Level 3

If in a Δ A B C \Delta ABC , 2 cos ⁡ A a + cos ⁡ B b + 2 cos ⁡ C c = a b c + b a c \frac{2\cos A}{a}+\frac{\cos B}{b} + \frac{2\cos C}{c} = \frac{a}{bc}+\frac{b}{ac} then find ∠ A \angle A .

This problem is part of the set Trigonometry .

6 0 ∘ 60^{\circ} 4 5 ∘ 45^{\circ} 9 0 ∘ 90^{\circ} 3 0 ∘ 30^{\circ}

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Mas Mus
May 6, 2015

First, note that 2 b c cos ⁡ A = b 2 + c 2 − a 2 ( 1 ) 2 a c cos ⁡ B = a 2 + c 2 − b 2 ( 2 ) 2 a b cos ⁡ C = a 2 + b 2 − c 2 ( 3 ) ~~~2bc\cos A=b^2+c^2-a^2~~~~(1)\\~~~2ac\cos B=a^2+c^2-b^2~~~~(2)\\~~~2ab\cos C=a^2+b^2-c^2~~~~(3)

Now, multiplying all terms on the question by a b c abc , the equation becomes

2 b c cos ⁡ A + a c cos ⁡ B + 2 a b cos ⁡ C = a 2 + b 2 2bc\cos A+ac\cos B+2ab\cos C=a^2+b^2

Subtitution ( 1 ) (1) and ( 3 ) (3) then ( 2 ) (2) to obtain

( b 2 + c 2 − a 2 ) + a c cos ⁡ B + ( a 2 + b 2 − c 2 ) = a 2 + b 2 2 a c cos ⁡ B = 2 ( a 2 − b 2 ) a 2 + c 2 − b 2 = 2 ( a 2 − b 2 ) a 2 = b 2 + c 2 = b 2 + c 2 − 2 b c cos ⁡ 9 0 ∘ = b 2 + c 2 − 2 b c cos ⁡ A \left(b^2+c^2-a^2\right)+ac\cos B+\left(a^2+b^2-c^2\right)=a^2+b^2\\2ac\cos B=2(a^2-b^2)\\a^2+c^2-b^2=2(a^2-b^2)\\a^2=b^2+c^2=b^2+c^2-2bc\cos90^\circ=b^2+c^2-2bc\cos A

Thus, ∠ A = 9 0 ∘ \boxed{\angle A=90^\circ}

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