Posted 25 Apr 2013 Point P is inside ABC, and is such that PAC = 18°, PCA = 57°, PAB = 24°, and PBA = 27°. Show that ABC is isosceles. 0 Share this post Link to post Share on other sites

0 Posted 26 Apr 2013 Find the unknown sides and angles: Let PA = 1 PB = sin(24^{o})/sin(27^{o}) = 0.895914 PC = sin(18^{o})/sin(57^{o}) = 0.368460 Central angles clockwise from CPB: 126^{o} 129^{o } 105^{o} Sides: AB^{2} = PA^{2} + PB^{2} - 2 PA PB cos(129^{o}) = 1.71181106 BC^{2} = PB^{2} + PC^{2} - 2 PB PC cos(126^{o}) = 1.151734499 CA^{2} = PC^{2} + PA^{2} - 2 PC PA cos(105^{o}) = 1.151734499 Angle: sin(PCB) = sin(126^{o}) PB/BC = 0.62932039 PCB = 39^{o} PBC = 15^{o} 0 Share this post Link to post Share on other sites

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Point P is inside ABC, and is such that PAC = 18°, PCA = 57°, PAB = 24°, and PBA = 27°.

Show that ABC is isosceles.

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