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Let P be a point inside square ABCD such that PA = 1, PB = 2, and PC = 3. Find angle APB.
Ok so let me get things straight. I know that you guys did this problem before, but I read the post and wondered if there is an easier method of finding angle APB without the cosine method thing.
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Ignore the D and consider only the isosceles right triangle ABC. Draw isosceles right triangle APE (isosceles at P) so that E lies on the other side of AC.
We can see that angle BAP = 45o - angle PAC = angle CAE. Also AB/AC=AP/AE or AB/AP = AC/AE. Thus triangles BAP and CAE are similar, so CE/AE = BP/AP = 2 or CE = 2 * sqrt2. This means PE^2 + CE^2 = CP^2 and so angle PEC = 90. Also angle AEP = 45 because APE is a right isosceles triangle. So angle AEC = 135 and angle APB =135 (remember the similar triangles).
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Ok thanks
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