Angle BAC and the exterior angle CBD of triangle ABC are each bisected. If the angle formed by these bisectors when they meet is 50 degrees, find the number of degrees in angle C of triangle ABC.

The answer I got was 100, after creating a system of equations and solving. I divided the variable x and y by 2 to represent the angle bisection.

I would like to see if I am correct, can someone check it for me? Step by step!

Anyways, I had problem uploading images the past few days. It would always say it is an unknown error. I had to drag my image into a new tab to create a url for it in order for me to upload.

CalculatorUser Aug 25, 2019

#1**+2 **

Here's what I come up with :

See the following image :

Let the intersection point of the bisectors = E

Let angle CAB =x angle ABC = y angle ACB = z angle DBC = (180-y)/2

angle EAB = x/2 angle ABE = y+ (180-y)/2

So we have two triangles ABC and AEB

So in ABC .....x + y + z = 180 (1)

And in AEB .....x/2 + 50 + y + (180-y)/2=180 (2)

We can simplify (2) as follows :

x/2 + (180-y)/2 + y = 130 multiply through by 2

x + 180-y + 2y =260

x + y = 80 (3)

Subbing (3) into (1) we have that

(80) + z =180

z = 100 = angle ACB = angle "C"

Which matches your result, CU !!!!

CPhill Aug 25, 2019