Two tangents pa and pb are drawn to a circle with centre o such that < apb = 1200. prove that op=2ap

Given: PA and PB are tangents and ∠APB = 120°

And, OP is joined.

Required to prove: OP = 2 AP

Construction: Taken M as the mid point of OP and joined AM, join also OA and OB.

Proof:

In right $\vartriangle OAP$

$\angle OPA=\frac{1}{2}\angle APB=\frac{1}{2}({{120}^{\circ }})={{60}^{\circ }}$

$\angle AOP={{90}^{\circ }}-{{60}^{\circ }}={{30}^{\circ }}$ Hypotenuse $OP$  of$\vartriangle OAP$  whose mid point is $M$ [from construction]

So, $MO=MA=MP$

$\angle OAM=\angle AOM={{30}^{\circ }}$  and

$\angle PAM={{90}^{\circ }}-{{30}^{\circ }}={{60}^{\circ }}$

Thus, $\vartriangle AMP$  is an equilateral triangle

$MA=MP=AP$

But,$M$ is mid-point of $OP$

So,

$OP=2MP=2AP$

– Hence proved.

Given : From a point P. Out side the circle with centre O, PA and PB are tangerts drawn and ∠APB = 120°

OP is joined To prove : OP = 2 AP

Const: Take mid point M of OP and join AM, join also OA and OB.

Proof : In right ∆OAP,

∠OPA = \( \dfrac{1}{2}\) ∠APB = \( \dfrac{1}{2}\) x 120° = 60°

∠AOP = 90° – 60° = 30°

M is mid point of hypotenuse OP of ∆OAP

MO = MA = MP

∠OAM = ∠AOM = 30° and ∠PAM = 90° – 30° = 60°

∆AMP is an equilateral triangle

MA = MP = AP

But M is mid point of OP

OP = 2 MP = 2 AP

Hence proved.

Two tangent segments PA and PB are drawn to a circle with center O such that ∠APB =120°. Prove that OP = 2AP

A + P

OP bisects ∠APB

∠APO = ∠OPB =`1/2`∠𝐴𝑃𝐵 =`1/2`× 120° = 60°

At point A

OA ⊥ AP, ∠OAP = 90°

In ΔPDA, cos 60° =`(AP)/(DP)`

`1/2=(AP)/(DP)`⇒ 𝐷𝑃 = 2𝐴𝑃

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

  Is there an error in this question or solution?

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Two tangent segments PA and PB are drawn to a circle with centre O such that ∠ APB =120∘. Prove that OP = 2AP.

Solution

Given: O is the centre of the circle. PA and PB are tangents drawn to a circle and ∠APB = 120°.

To prove: OP = 2AP

Proof:

In ΔOAP and ΔOBP, OP = OP (Common) ∠OAP = ∠OBP (90°) (Radius is perpendicular to the tangent at the point of contact) OA = OB (Radius of the circle) ∴ ΔOAP is congruent to ΔOBP (RHS criterion)

∠OPA = ∠OPB = 120°2 = 60° (CPCT)

In ΔOAP,
cos∠OPA = cos 60° = APOP
∴ 12=APOP
Thus, OP = 2AP

Mathematics

RD Sharma

Standard X

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Two tangent segment PA and PB are drawn to a circle center O such that angle APB =120 . Prove that OP =2 AP.

Solution

Triangles PAO and PBO can be proved congruent using RHS criterion.

Thus, ∠APO=∠BPO (CPCT)

Given that ∠APB=120∘
APB=APO+BPO=2APO=120
APO=60∘ In triangle APO

cos60=12=APOP

Thus, OP = 2AP Hence Proved