Given that triangle \(ABC\) is an isosceles triangle with \(AB=AC\) such that \(P \) is a point inside the triangle and \(\angle BCP = 30^\circ, \angle APB = 150^\circ, \angle CAP=39^\circ \), find the measure of \(\angle BAP \) in degrees.

i draw diagram and trying by applying with trig ceva .
last i come to an equation
2sin(2x-30).sin42=-sin(18-4x)
if i can solve this it will be easy to find that angle

Calling angle ACP as 'z', I got the following equation from Ceva'S Trigonometric Identity:
sin(2z-51)sin(39)=2sin(z)sin(81-z)*sin(81-2z), all angles in degrees. Clearly, z lies somewhere between 26 and 40 degrees. Wolfram returns z=34°. Then angle BAP = 81 -2z =13°.

@Abdullah Ahmed
–
Hi Abdullah,
Since the answer is exact 13°, I suspect that it will be possible to prove this by Euclid's Elements alone w/o using trigonometry. One idea would be to drop a perpendicular from A to BC and work around the symmetry knowing that the perpendicular is also an angle bisector. Perhaps this type of an elegant solution was expected in the first place. If you do find such a solution please write to me at: 1topquark@gmail.com

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## Comments

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TopNewestWe can derive: (if we want to avoid Trigonometry)

Given: PAC=39, PCB=30, APB=150, ABC=ACB;

PAB+ABP=30; BAP+2ABC=141; ABP=2ABC-111; ABP+PBC=30+PCA; ABP-BPC+APC=21; 3APC-2BAC-BPC=114; APC+ABC=171; 2PCA+BAP=81;

I will reply again if I can Solve with above Equations.

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What have you tried? Have you drawn the diagram? What did you notice about it?

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i draw diagram and trying by applying with trig ceva . last i come to an equation 2sin(2x-30).sin42=-sin(18-4x) if i can solve this it will be easy to find that angle

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Calling angle ACP as 'z', I got the following equation from Ceva'S Trigonometric Identity: sin(2z-51)

sin(39)=2sin(z)sin(81-z)*sin(81-2z), all angles in degrees. Clearly, z lies somewhere between 26 and 40 degrees. Wolfram returns z=34°. Then angle BAP = 81 -2z =13°.Log in to reply

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