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.

#Geometry

Easy Math Editor

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Comments

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

Calvin Lin Staff - 5 years 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 angle

Abdullah Ahmed - 5 years ago

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°.

Ajit Athle - 5 years ago

@Ajit Athle – thanks very much sir ......very much .

Abdullah Ahmed - 5 years ago

@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: [email protected]

Ajit Athle - 5 years ago

@Ajit Athle – i must try. if i find any solution i will write you

Abdullah Ahmed - 5 years ago

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 - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$