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Helppppppp me please

Helppppppp me please-example-1
User Raf
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1 Answer

2 votes

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=(66^\circ)/(2)


m\angle S=33^\circ

Therefore, the
m\angle S is 33 degrees.

User Kaykun
by
8.2k points

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