SOLUTION: The point P on the side BC of triangle ABC divides BC in the ratio 1:2 i.e. BP:PC=1:2 , angle ABC =45° , angle APC=60° , calculate angle ACB.

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Question 1172758: The point P on the side BC of triangle ABC divides BC in the ratio 1:2 i.e. BP:PC=1:2 , angle ABC =45° , angle APC=60° ,
calculate angle ACB.
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There is probably a relatively simple solution method, because the answer turns out to be "nice". But I'm not seeing an easy solution. So....

We can arbitrarily let BP=1 and PC=2.

With angle ABC 45 degrees and angle APC 60 degrees, angle APB is 120 degrees; that makes angle BAP 15 degrees.

Let x be the measure of angle ACB that we are looking for; then the measure of angle CAP is 120-x.

Let y be the length of AP.

Then in triangle BAP the law of sines gives us

1%2Fsin%2815%29+=+y%2Fsin%2845%29

which gives us

y+=+sin%2845%29%2Fsin%2815%29

And in triangle APC the law of sines gives us

y%2Fsin%28x%29+=+2%2Fsin%28120-x%29

which gives us

y+=+2sin%28x%29%2Fsin%28120-x%29

Now we have two expressions for y -- one a constant and the other an expression in x. Set them equal to each other:

2sin%28x%29%2Fsin%28120-x%29+=+sin%2845%29%2Fsin%2815%29

Graphing the two expressions on a graphing calculator shows that x, the measure of angle ACB we are looking for, is 75 degrees.

ANSWER: angle ACB is 75 degrees.

I will be watching with curiosity to see if another tutor comes up with a simpler solution method....