Mathematics

$P$ is any point inside the triangle $ABC$. Hence $\angle BPC> \angle BAC$

True

SOLUTION
Let
$\angle PBC = x$ and $\angle PCB = y$
$=> \angle BPC = 180 - (x + y)$

Let
$\angle ABP = z$ and $\angle ACP = w$
$=> \angle BAC = 180 - (x + z) - (y + w )$
$=> \angle BAC = 180 - (x + y) - (w + z )$
$\angle BAC = \angle BPC - ( w+z )$
$=> \angle BPC = \angle BAC + (w + z )$
$=> \angle BPC > \angle BAC$ You're just one step away

Single Correct Medium Published on 09th 09, 2020
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Subjects 10
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