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.

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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. – Diptangshu Paul · 5 months, 3 weeks ago

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What have you tried? Have you drawn the diagram? What did you notice about it? – Calvin Lin Staff · 6 months, 2 weeks ago

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– Abdullah Ahmed · 6 months, 2 weeks ago

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 angleLog in to reply

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°. – One Top · 6 months, 2 weeks agoLog in to reply

– Abdullah Ahmed · 6 months, 2 weeks ago

thanks very much sir ......very much .Log in to reply

– One Top · 6 months, 2 weeks ago

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.comLog in to reply

– Abdullah Ahmed · 6 months, 2 weeks ago

i must try. if i find any solution i will write youLog in to reply