Question

question 9 in step by step explanation​

Answer 1

Step-by-step explanation:

Jion OC.

In triangle POA and AOC,

PO=AO(radii of same circle)

OA=OA(common)

AP=AC(lenght of tangents are equal)

tranglePOA=AOC(SSS)

LPOA=LAOC(CPCT)(L=angle here)

similarly,

triangle QOB=BOC(SSS)(plz don't confuse it is also proved same as that done above for those 2 triangles)

so, LQOB=BOC(CPCT)

$straight \: line = 180 {}^{0}$

LPOA+LAOC+LQOB+LBOQ=180

LAOC+LAOC+LBOQ+LBOQ=180

2LAOC+2LBOQ=180

2(LAOC+LBOQ)=180

LAOC+LBOQ=180/2

LAOC+LBOQ=90.

Hence proved!

Answer 2

Answer:

answer attached in photo