Question

OP is equal to the diameter of the circle. prove that ABP is an equilateral triangle.

Answer 1

Heya...❤❤❤

here is ur answer...

Join A to B.
We have
OP= Diameter
Or, OQ + OP= diameter
Or, Radius + PQ = diameter (that is OQ=radius)
Or, PQ = diameter – radius
Or, PQ = radius
Or, OQ = PQ = radius
Thus OP is the hypotenuse of the right angle triangle AOP.
So, in ∆AOP sin of angle P = OP /AO
= 1/2
So, P = 30°
Hence, APB = 60°
Now, As in ∆AOP, AP=AB
So, PAB = PBA =60°
Hence,
∆ABP is equilateral triangle.

hope it helps...

plzz mark me as brainliest my dear !!!

❤❤❤

Answer 2

$\huge{hey-mate}$
→→→→→→→→→→→→→→→→→→→→→

$SOLUTION$ => Join A to B.
We have
OP= Diameter
Or, OQ + OP= diameter
Or, Radius + PQ = diameter (that is OQ=radius)
Or, PQ = diameter – radius
Or, PQ = radius
Or, OQ = PQ = radius
Thus OP is the hypotenuse of the right angle triangle AOP.
So, in ∆AOP sin of angle P = OP /AO
= 1/2
So, P = 30°
Hence, APB = 60°
Now, As in ∆AOP, AP=AB
So, PAB = PBA =60°
Hence,
∆ABP is equilateral triangle.

######Hope its hlp uh!!!####.
.
$<marquee > Sumit Tiwari from Azamgarh up$