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  1. Problems
  2. /Two tangents TP and TQ are drawn from a point T to a circle with centre O…
MediumMathematicsHigh School (9–10)GeneralGeometry › CirclesNumerical

Two tangents TP and TQ are drawn from a point T to a circle with centre O…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Two tangents TP and TQ are drawn from a point T to a circle with centre O. They meet at an angle of 70∘70^\circ70∘. Find ∠POQ\angle POQ∠POQ and ∠PTO\angle PTO∠PTO.

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Work it out on paper, then type your number. The answer itself stays hidden.

Need a reminder? The key ideaShowHide

The two tangents drawn from an outside point T to a circle are equal: TP=TQTP = TQTP=TQ. In the right triangle OPT, PT2=OT2−OP2PT^2 = OT^2 - OP^2PT2=OT2−OP2. The line OT bisects the angle PTQPTQPTQ and also the angle POQPOQPOQ. In the figure OPTQ, ∠POQ+∠PTQ=180∘\angle POQ + \angle PTQ = 180^\circ∠POQ+∠PTQ=180∘. A tangent from an outside point has the same length on both sides of OT.

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. The angle at the centre. ∠POQ+∠PTQ=180∘\angle POQ + \angle PTQ = 180^\circ∠POQ+∠PTQ=180∘, so ∠POQ=180∘−70∘=110∘\angle POQ = 180^\circ - 70^\circ = 110^\circ∠POQ=180∘−70∘=110∘.

Step 2. OT bisects the angle at T. ∠PTO=70∘÷2=35∘\angle PTO = 70^\circ \div 2 = 35^\circ∠PTO=70∘÷2=35∘.

Step 3. Check. ∠POT=110∘÷2=55∘\angle POT = 110^\circ \div 2 = 55^\circ∠POT=110∘÷2=55∘, and 35∘+55∘+90∘=180∘35^\circ + 55^\circ + 90^\circ = 180^\circ35∘+55∘+90∘=180∘.

Answer: ∠POQ=110∘\angle POQ = 110^\circ∠POQ=110∘ and ∠PTO=35∘\angle PTO = 35^\circ∠PTO=35∘.

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  1. Tangents from an outside pointTangents to a Circle

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