Question

In a quadrilateral ABCD & It; $D={60}^{\circ },AB=\sqrt{3},BC=3,AC=2\sqrt{3}$
Find the magnitude of all angles B and C.

Answer 1

ABCD is a quadrilateral
AB = $\sqrt{3}$, BC = 3, AC =$2\sqrt{3}$
and $\angle A=\angle D={60}^{\circ }$
In $\Delta ABC$,
$A{B}^2+B{C}^2=(\sqrt{3}{)}^2+(3{)}^2=3+9=12$
$A{C}^2=(2\sqrt{3}{)}^2=12$
$\therefore A{B}^2+B{C}^2=A{C}^2$
$\Rightarrow \Delta ABC$ is a right triangle right angled at $\angle B$
$\Rightarrow B={90}^{\circ }$
In Quadrilateral ABCD,
$\angle A+\angle B+\angle C+\angle D={360}^{\circ }$
$\therefore {60}^{\circ }+{90}^{\circ }+\angle C+{60}^{\circ }={360}^{\circ }$
$\Rightarrow {210}^{\circ }+\angle C={360}^{\circ }$
$\Rightarrow \angle C-{360}^{\circ }-{210}^{\circ }-{150}^{\circ }$
Thus, the measure of $\angle Band\angle Cis{90}^{\circ }and{150}^{\circ }$ respectively.