Hi EP!

Note, NTS, that PQR can't be a triangle because

PQ + PR  = QR

But the triangle inequality says that

PQ + PR  must be > QR

Here's a similar prob that I answered earlier :

https://web2.0calc.com/questions/geometry_2526

See if you can solve it now....

Ok, let me give this a shot....

The Law of Cosines state that $cos PMQ = - cos PMR$

$PQ^2 = QM^2 + PM^2 - 2(QM * PM) * (-cos PMR) \\ PR^2 = RM^2 + PM^2 - 2 (RM * PM) * ( cos PMR)$

Subsituting in the values we already know from the problem. we get

$8^2 = 7.5^2 + PM^2 + 2(7.5 * PM)cos(PMR) \\ 9^2 = 7.5^2 + PM^2 - 2(7.5 * PM) cos (PMR)$

Now, adding these two, we get

$8^2 + 9^2  = 2 * 7.5^2 + 2PM^2$
$145= 112.5+ 2PM^2 \\ 32.5/ 2 = PM^2 \\ 16.25 = PM^2 \\ PM=\sqrt{16.25}\approx 4.03112887415$

Is this correct, CPhill? I'm not sure. Can u verify?

Thanks! :)

Hi EP! :)