in triangle ABC with BP bisecting AC; Ac=2; the measure of <BAC is 45 degrees; and the measure of <BCA is 60. Find BP

Question

Problem of the day 

Answer 1

Challenging....

 

AP = PC = 1

 

Angle ABC = 75

 

Let Angle APB = t

 

Then ANgle PBC = 75-t

 

 Angle BPA = 135 - t and angle BPC = 45+t

 

Let BP = h even though it technically is not the height

 

 

From the law of sines:

sin 60/h = sin (t+45)/y = sin(75-t)

                      AND
sin45/h = sin(135-t)/x = sint

So using only the first and last equations

sin60/h = sin(75-t) and sin 45 / h = sin t

Then
sin 60/ sin(75-t) = h and sin 45 / sin t = h

 

They are both equal to h, so

sin 60/ sin(75-t) = sin 45 / sin t

Cross multiplying:
sin 60 * sin t = sin 45 * sin (75 - t)

 

Finally
sin 60 * sin t - sin 45 * sin(75-t) = 0

 

The left side is a function of t, which can be solved numerically.

Per desmos.com, since t must be in the first quadrant as it is less than 90 degrees,
t = sqrt(3)/3 which is about 33 degrees

Again by the law of sines
sin 33/1 = sin 45/h

h = sin 45 / sin 33 = 1.2983