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Finding angle

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.

Note by Abdullah Ahmed
1 year ago

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We 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 · 11 months, 2 weeks ago

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What have you tried? Have you drawn the diagram? What did you notice about it? Calvin Lin Staff · 1 year ago

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@Calvin Lin 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 Abdullah Ahmed · 1 year ago

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@Abdullah Ahmed 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°. One Top · 1 year ago

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@One Top thanks very much sir ......very much . Abdullah Ahmed · 1 year ago

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@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 One Top · 1 year ago

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@One Top i must try. if i find any solution i will write you Abdullah Ahmed · 12 months ago

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