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Solution given:

7.<OYM=15°base angle of isosceles triangle

<OYL=50°base angle of isosceles triangle.

<OYL=<OYM+<MYL

50°=15°+<MYL

<MYL=50°-15°

<MYL=35°

again;

<MOL=35*2=70°central angle is double of a inscribed angle.

18.

Solution given:

<PQR+<PSR=180°sum of opposite angle of a cyclic quadrilateral is supplementary

<PQS+42°+78°=180°

<PQS=180°-120°=60°

<PQS=60°

<SPR=42°inscribed angle on a same arc is equal

:.<QPS=18°+42°=60°

<QSR=18°inscribed angle on a same arc is equal

again.

<PSR=78°

<QSR+<PSQ=78°

18°+<PSQ=78°

<PSQ=78°-18°

<PSQ=60°

In ∆ PQS

<PSQ=60°

<QPS=60°

<PQS=60°

In triangle ∆PQS all the angles are equal.

2

step-by-step explanation: