Question

Helppppppp me please

Answer 1

Given:

In $\Delta QRS, \overline{QR}\cong \overline{SQ}$ and $m\angle Q=114^\circ$.

To find:

The $m\angle S$.

Solution:

In $\Delta QRS$,

$\overline{QR}\cong \overline{SQ}$

It means the triangle QRS is an isosceles triangle. We know that the base angles of an isosceles triangle are congruent and their measures are equal.

$\angle R\cong \angle S$      [Base angles of isosceles triangle QRS]

$m\angle R=m\angle S$            ...(i)

In $\Delta QRS$,

$m\angle Q+m\angle R+m\angle S=180^\circ$

$m\angle Q+m\angle S+m\angle S=180^\circ$        [Using (i)]

$114^\circ+2m\angle S=180^\circ$

$2m\angle S=180^\circ-114^\circ$

$2m\angle S=66^\circ$

Divide both sides by 2.

$m\angle S=\dfrac{66^\circ}{2}$

$m\angle S=33^\circ$

Therefore, the $m\angle S$ is 33 degrees.