Question

ABC is an isosceles triangle with angle B = angle C. Point P is the midpoint of AB. If angle A is 30 degrees, what is angle PCB?

Answer 1

Let `PCB = x` so that
`CPB = 105 - x`

Now #(((BC)/2)/(AB)) = cos75 = (BC)/(4*BP)=> (BC) / (BP) = 4 cos75 #

Applying the sine rule to triangle PCB,
`(BC)/sin(105-x) = (BP)/sin(x)`
so #(BC)/(BP) = sin(105-x) / sin(x) => 4cos75*sinx=sin(105-x) => sin105 cosx - cos105 sinx = 4 cos75 sinx=> sin 75 cosx + cos75 sinx = 4 cos75 sinx=> sin75*cosx=3cos75sinx=> tanx=1/3cos75#

Finally `x = 51.206 degrees`