Question

in the given figure o is the centre of the circle and pr is a diamrter. if angle spq=90 and pa=pq then the value of (a+b+c-d+e-f) is​

Answer 1

Answer:

Given−

PQisadiameterofacirclewithcentreO.

PTisatangenttothecirclefromTatP.

QTintersectsthecircleatR.

∠POR=72

o

.

Tofindout−

∠PTR=?

Solution−

ThediameterPQortheradiusOPmeetsthetangentPT

atP.

∴PQ⊥PTsincetheradiusthroughthepointofcontactof

atangenttoacircleisperpendiculartothetangent.

∴∠QPT=90

o

.

AgaintheminorarcPRsubtends∠PORtothecentreO

and∠PQRtothecicumference.

∴∠PQR=

2

1

×∠POR=

2

1

×72

o

=36

o

.

SoinΔPQT

∠QTP=180

o

−(∠PQR+∠QPT)(byanglesumpropertyoftriangles)

=180

o

−(36

o

+90

o

)=54

o

.

Ans−OptionC.