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.

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) (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

which gives us

And in triangle APC the law of sines gives us

which gives us

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

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....

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