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.
Answered by Sakshi | 2 months agoTwo 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
2
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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
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 ProvedSuggest Corrections
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