Question

ln the figure the line PQ is a tangent to the circle. If angle PAC=65 degree and angle QAB=50 degree. find the measure of the following angles. angle BAC,angle ACB,angle D​

Answer 1

Answer:

∠BAQ=30∘

Since AB is the bisector of ∠CAQ

∠CAB=∠BAQ=30∘

∠CAQ=∠CAB+∠BAQ=60∘

We can also see that: ∠PAC+∠CAQ=180∘

So,∠PAC=120∘

Since AD bisects ∠PAC

So,∠PAD=∠DAC=2∠PAC=60∘

So,∠DAB=90∘

And we know that only diameter subtends an angle of 90∘on the circle

Hence part (I) is proved

Now we see that ∠CAB & ∠ACBare equal because they share a same side in front on them

Finally we can see two angles same in ΔABCSo it is an Isosceles triangle, Hence (II)

Answer 2

Step-by-step explanation:

AO and OB are radii of the circle.

 side AO=BO  so, ∠OAB=∠OBA    [Isosceles triangle AOB]

Angle subtended by chord at the centre of a circle is double of the angle subtended at it's circumference.

Therefore ∠AOB=2∠ACB

∠AOB=100∘

In triangle AOB 

 The sum of all three angle will be 180∘

So, ∠AOB+∠OBA+∠OAB=180∘

   100∘+∠OBA+∠OAB=180∘

  100∘+2∠OAB=180∘       [∠OAB=∠OBA]

∠OAB=280∘

∠OAB=40∘

Therefore option B is the answer.