Vikas TU
Last Activity: 5 Years ago
AB = AC (given)
=> ABC = ACB = x deg ( angles opposite to equal sides of a triangle)
BC = BP ( given)
=> BCP=BPC = 'x' deg
& PQ = AQ (given)
=> QPA = QAP = 'a' deg
=> PQB = 2a deg ( exterior angle = the sum of interior opposite angles)
=> PBQ = 2a deg ( as PQ = PB , given)
PBC = (x-2a) deg
Now, in triangle ABC
2x + a = 180 ( angle sum property of a triangle) ……………………… (1)
In triangle BCP
x + x + x-2a = 180 ( angle sum property of a tri)
=> 3x - 2a = 180 ……………… (2)
By solving eq (1), & (2)
Eq(1): 4x + 2a = 360
Eq(2): 3x - 2a -= 180
=> 7x = 540
=> x = 540/7 deg
=>a=180/7(by substituting the value of x)
=> AQP = 180 - 2a = 180- 2*180/7
=> AQP = 180–360/7 = 900/7 deg
=> AQP = (Pi * 900)/(7*180) = 5pi/7
AQP = 5pi/7 radian
