Question

In Fig. 10.87 , BOA is a diameter of a circle and the tangent at a point P meets BA produced at T . If $\angle$PBO = 30⁰ , then find $\angle$PTA .

figure

Answer 1

In the given figure, PT is the tangent to the circle with centre O.
$\angle$PBO = 30$°$
OP = OB (Radii)
So, $\angle$PBO = $\angle$BPO = 30$°$
$\angle$BPA = 90$°$ (Angle made by the diameter on the arc of the circle)
In ∆APB,
$\angle$BPA + $\angle$PBO + $\angle$PAB = 180$°$
⇒ 90$°$ + 30$°$ + $\angle$PAB = 180$°$
⇒ $\angle$PAB = 60$°$
In ∆OPA,
OA = OP (Radii)
So, $\angle$OPA = OAP = 60$°$
Hence, $\angle$AOP = 180$°$ − 60$°$ − 60$°$ = 60$°$
In ∆OPT,
$\angle$OPT + $\angle$PTO + $\angle$POT = 180$°$
⇒ 90$°$ + $\angle$PTO + 60$°$ = 180$°$
⇒ $\angle$PTO = 30$°$