Question

from a point p,two tangent AP and PB are drawn to a circle C(O,r) if OP=2r,then find angleAPB.​

Answer 1

$\sf\bf \: \large{\underbrace{\pink{Required \: answer:}}}$

$\rm \dag \: \large{\underline{\underline{Question:}}}$

from a point p,two tangent PA and PB are drawn to a circle C(O,r) if OP=2r,then find < APB.

$\rm\dag\large{\underline{\underline{Given:}}}$

from a point p,two tangent PA and PB are drawn to a circle C(O,r) if OP=2r.

$\rm\dag \large{\underline{\underline{To \: Prove:}}}$

We have to show that if OP = 2r than ∆ APB is an equilateral triangle.

$\rm \dag \: \large{\underline{\underline{Soluation:}}}$

PA and PB tangent to a circle

$\sf < OAP = 90°$

$\sf in \: ∆ \: OPA$

$\sf: \longrightarrow \: Sin < OPA = \frac{ \:OA }{ \: OP} = \frac{r}{2r}$

[ OP is a the diameter = 2 × radius ]

$\sf:\longrightarrow Sin < OPA = \frac{1}{2} = Sin 30°$

$\sf < OPA =30°$

$\sf Similarly < OPB = 30°$

$\rm < APB = \: < OPA + < OPB = 30° + 30°=60°$

Now, PA = PB (tangent from an external point to circle)

< PAB = <PBA —— (1) [ angle opposite to equal sides are equal ]

Now, from ∆ APB , Sum of angle = 180°

$\sf:\longrightarrow \: < PAB + < PBA + < APB = 180°$

$\sf:\longrightarrow \: < PAB + < PAB + 60° = 180°[by (1)]$

$\sf:\longrightarrow \: 2< PAB = 120°$

$\sf:\longrightarrow \: < PAB = 60° \: \: \: \: \: \: \longrightarrow(2)$

From ( 1 ) and ( 2 )

$\rm < PAB + < PAB +APB = 60°$

All angles are equal in an equilateral triangle. ( ie 60°)

Hence, ∆ PAB is an equilateral triangle .

Answer 2

Answer:

\sf\bf \: \large{\underbrace{\pink{Required \: answer:}}}

Requiredanswer:

\rm \dag \: \large{\underline{\underline{Question:}}}†

Question:

from a point p,two tangent PA and PB are drawn to a circle C(O,r) if OP=2r,then find < APB.

\rm\dag\large{\underline{\underline{Given:}}}†

Given:

from a point p,two tangent PA and PB are drawn to a circle C(O,r) if OP=2r.

\rm\dag \large{\underline{\underline{To \: Prove:}}}†

ToProve:

We have to show that if OP = 2r than ∆ APB is an equilateral triangle.

\rm \dag \: \large{\underline{\underline{Soluation:}}}†

Soluation:

PA and PB tangent to a circle

\sf < OAP = 90°<OAP=90°

\sf in \: ∆ \: OPAin∆OPA

\sf: \longrightarrow \: Sin < OPA = \frac{ \:OA }{ \: OP} = \frac{r}{2r}:⟶Sin<OPA=

OP

OA

=

2r

r

[ OP is a the diameter = 2 × radius ]

\sf:\longrightarrow Sin < OPA = \frac{1}{2} = Sin 30°:⟶Sin<OPA=

2

1

=Sin30°

\sf < OPA =30°<OPA=30°

\sf Similarly < OPB = 30°Similarly<OPB=30°

\rm < APB = \: < OPA + < OPB = 30° + 30°=60°<APB=<OPA+<OPB=30°+30°=60°

Now, PA = PB (tangent from an external point to circle)

< PAB = <PBA —— (1) [ angle opposite to equal sides are equal ]

Now, from ∆ APB , Sum of angle = 180°

\sf:\longrightarrow \: < PAB + < PBA + < APB = 180°:⟶<PAB+<PBA+<APB=180°

\sf:\longrightarrow \: < PAB + < PAB + 60° = 180°[by (1)]:⟶<PAB+<PAB+60°=180°[by(1)]

\sf:\longrightarrow \: 2 < PAB = 120°:⟶2<PAB=120°

\sf:\longrightarrow \: < PAB = 60° \: \: \: \: \: \: \longrightarrow(2):⟶<PAB=60°⟶(2)

From ( 1 ) and ( 2 )

\rm < PAB + < PAB +APB = 60°<PAB+<PAB+APB=60°

All angles are equal in an equilateral triangle. ( ie 60°)

Hence, ∆ PAB is an equilateral triangle .