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I'm new to this forum, so I apologize if this Q is in the wrong thread -it's a mixture of trigonometry and algebra. I've tried to solve it but i don't get the correct answer. I am given a triangle, whose sides i had to calculate, which i have done, and the results are correct.

so the triangle OPA has sides: OP= sqr(x^2 + 100) and PA = sqr(x^2 -16x +80), and AO=10

Thereafter i have to show that cos OPA = cos(60°)=((x^2)-8x+40) / sqr(((x^2)-16x+80)((x^2)+100)), which i have also done.

After a few other Q I am asked to

**find the positive value of x such that the angle OPA = 60degrees**, which is the problem i cannot solve. I have tried to solve for x with the above equation but it just becomes too complicated.

I have also tried to just solve for x by saying sin(60degrees)= 10 / sqr(x^2 +100) , where i get x=5.77 but the result should be 5.63.

Is there any other way to find the positive value than having to solve for x in the complicated equation above?