Question

in a isoceles triangles a is the length of two equal sides and A is the equal angle Prove that
it's area is a^2sinA.cosA​

Answer 1

$by \: sin \: \: law \\ area \: = \frac{1}{2} absinc \\ = using \: sine \: law \\ \frac{a}{sin \: a} = \frac{c}{sin \: c} = k \\ a = b \\ c = 180 - 2a\\ area \: = \frac{1}{2} k \: sin(a)ksin \: a \: sin \: \: c \\ = \frac{1}{2} k {}^{2} sin {}^{2} a \: sin(180 - a) \\ = \frac{1}{2} (ksin \: a) {}^{2} sin2a \\ = \frac{1}{2} a {}^{2} \times 2sin \: a \: cos \: a \\ = a {}^{2} sin \: a \: \: cos \: a$