Trigonometry! #132

Geometry Level 4

Δ A B C \Delta ABC is a triangle such that sin ⁡ ( 2 A + B ) = sin ⁡ ( C − A ) = − sin ⁡ ( B + 2 C ) = 1 2 \sin (2A+B)=\sin (C-A)=-\sin (B+2C)=\frac{1}{2} If A A , B B and C C are in an arithmetic progression, then determine the values of A A , B B and C C .

Enter the answer as A + 2 B + 3 C A+2B+3C without the degree sign.

This problem is part of the set Trigonometry .


The answer is 390.

Omkar Kulkarni by Omkar Kulkarni , 22, India

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1 solution

Since the three angles are in A.P., we know that B = 6 0 ∘ . B = 60^{\circ}. Without loss of generality assume A < B < C . A \lt B \lt C. Then A = ( 60 − k ) ∘ A = (60 - k)^{\circ} and C = ( 60 + k ) ∘ C = (60 + k)^{\circ} for some real number 0 < k < 60. 0 \lt k \lt 60.

The given equations then become

sin ⁡ ( 18 0 ∘ − 2 k ∘ ) = sin ⁡ ( 2 k ∘ ) = − sin ⁡ ( 18 0 ∘ + 2 k ∘ ) = 1 2 \sin(180^{\circ} - 2k^{\circ}) = \sin(2k^{\circ}) = -\sin(180^{\circ} + 2k^{\circ}) = \dfrac{1}{2}

⟹ sin ⁡ ( 2 k ∘ ) = 1 2 ⟹ 2 k ∘ = 15 0 ∘ \Longrightarrow \sin(2k^{\circ}) = \dfrac{1}{2} \Longrightarrow 2k^{\circ} = 150^{\circ} or 2 k ∘ = 3 0 ∘ . 2k^{\circ} = 30^{\circ}.

Now since 0 < k < 60 0 \lt k \lt 60 we must have 2 k = 30 ⟹ k = 15. 2k = 30 \Longrightarrow k = 15.

Thus A = ( 60 − 15 ) ∘ = 4 5 ∘ , B = 6 0 ∘ A = (60 - 15)^{\circ} = 45^{\circ}, B = 60^{\circ} and C = ( 60 + 15 ) ∘ = 7 5 ∘ C = (60 + 15)^{\circ} = 75^{\circ} , and so without the degrees sign we have that

A + 2 B + 3 C = 45 + 2 ∗ 60 + 3 ∗ 75 = 390 . A + 2B + 3C = 45 + 2*60 + 3*75 = \boxed{390}.

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