in the adjoining figure point p is the centre of the circle and line A B is the tangent to the circle at T the radius of the circle is 6cm find PB if TPB=60°
Answers
AB is a tangent to a circle with centre as P and radius PT=6cm.
AB touches the circle at T.
PB intersects the circumference at Q and AB at B such that ∠TPB=60
o
.
To find out- PB
Solution-
PT=PQ=6cm ...(given)
TQ is joined.
Then, ΔPQT is isosceles.
∴∠PTQ=∠PQT
i.e ∠PTQ+∠PQT=2(∠PTQor∠PQT)
So, by the angle sum property of triangles, we have
∠TPQ+2∠PTQ=180
o
⟹60
o
+2∠PTQ=180
o
⟹∠PTQ=60
o
=∠PQT
So, ΔPQT is equilateral.
∴TP=PQ=TQ=6cm. .......(i)
Now, ∠QTB=∠PTB−∠PTQ.
But ∠PTB=90
o
, since the tangent at a point of a circle is perpendicular to its radius which meets the tangent at the point of contact.
∴∠QTB=90
o
−60
o
=30
o
Now, in ΔTQB we have,
∠TQB=180
o
−60
o
=120
o
...linear pair
∴ By the angle sum property of triangles, we have
∠QTB+∠TQB+∠QBT=180
o
⟹∠QBT+120
o
+30
o
=180
o
∴∠QBT=30
o
.
So, ΔTQB is isosceles with QT=QB=6cm. ....(by i).
∴PB=PQ+QB=(6+6)cm=12cm