Which angle measure should I use?

Geometry Level 3

In a triangle ABC there is a point P in the interior of triangle such that angle APB = 150 and angle PAB = angle PAC = 22 and angle PBC = 30. Find angle APC.

note : all angles are in degrees.


The answer is 142.

Akash Deep by akash deep , 20, India

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1 solution

Aditya Kumar
Jun 24, 2015

A pure trigonometric solution. Use the trigonometric form of Ceva to reach the equation:

sin ⁡ ( 8 ∘ ) sin ⁡ ( 3 0 ∘ ) × sin ⁡ ( 9 8 ∘ − x ∘ ) sin ⁡ ( x ∘ ) = 1 \frac{\sin(8^\circ)}{\sin(30^\circ)} \times \frac{\sin(98^\circ-x^\circ)}{\sin(x^\circ)} = 1

where, x x is ∠ A C P \angle ACP in degrees.

Use the expansion of sin of sum of angles to reach:

sin ⁡ ( 9 8 ∘ ) cot ⁡ ( x ∘ ) − cos ⁡ ( 9 8 ∘ ) = 1 2 sin ⁡ ( 8 ∘ ) \sin(98^\circ)\cot(x^\circ)-\cos(98^\circ)=\frac{1}{2\sin(8^\circ)}

⟹ \implies cos ⁡ ( 8 ∘ ) cot ⁡ ( x ∘ ) + sin ⁡ ( 8 ∘ ) = 1 2 sin ⁡ ( 8 ∘ ) \cos(8^\circ)\cot(x^\circ)+\sin(8^\circ)=\frac{1}{2\sin(8^\circ)}

⟹ cot ⁡ ( x ∘ ) = 1 − 2 sin ⁡ 2 ( 8 ∘ ) 2 sin ⁡ ( 8 ∘ ) cos ⁡ ( 8 ∘ ) = cot ⁡ ( 1 6 ∘ ) \implies \cot(x^\circ) = \frac{1-2\sin^2(8^\circ)}{2\sin(8^\circ)\cos(8^\circ)} = \cot(16^\circ)

So, x = 1 6 ∘ ⟹ ∠ A P C = 14 2 ∘ x=16^\circ \implies \angle APC = 142 ^\circ

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you tell me what is the trignometric form of ceva.? It seems you are quite engaged with olympiad problems. I want to talk to u. Can i have your email. Just for some olympiad related guidance. And this problem is a challenge for pure geometry. Please post a solution with euclidean geometry only.

akash deep - 5 years, 11 months ago

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