Question

ABCD Is a rhombus find m< ABC

Answer 1

If ABCD is a rhombus, then diagonals are perpendicular.

Therefore $12y^o=90^o\to y^o=\dfrac{90^o}{12^o}=7.5^o$

$m\angle FCB=(4y-2)^o$

Substitute:

$m\angle FCB=(4(7.5)-2)^o=(30-2)^o=28^o$

We know: The sum of the measures of angles in a triangle is equal 180°.

Therefore we have the equation:

$m\angle FBC+28^o+90^o=180^o\\\\m\angle FBC+118^o=180^o\qquad\text{subtract}\ 118^o\ \text{from both sides}\\\\m\angle FBC=62^o$

BD is bisector of ∠ABC. Therefore $m\angle ABF=m\angle FBC=62^o$

$m\angle ABC=m\angle ABF+m\angle FBC\\\\m\angle ABC=62^o+62^o=124^o$