Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule? I wonder if there is another way to do this.
I did this question by using Cosine Rule.
I tried to use similar triangle, but it seems that there is not enough information to support the property. The best I could is Side and Angle + unknown..
Edited
COSINE RULE
$$102^{2} = 126^{2} + 60^{2} - [ 2*{60}*{126}*Cos{A}] $$
Where A = angle (RPQ)
After finding the angle, I then use trigonometry to find
$$Cos{A} = \frac{PS}{60}$$
Writing $h$ for $RS$ and $a$ for $PS$, we have $a^2 + h^2 = 60^2$ (angle $PSR$ is right because angle $QSR$ is) and $(126-a)^2 + h^2 = 102^2$.  If I subtract the second equation from the first I learn that $a^2 - (126-a)^2 = 60^2 - 102^2$.  If I do some algebra I see that the $a^2$ terms on the left hand side of this equation drop out, and I can solve directly for $a$. Knowing $a$ makes the rest easy.
Use Heron's formula to find the area, where the semi-perimeter $s$ is $\frac{60+102+126}{2} = 144$.
Then $A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{144(144-60)(144-102)(144-126)} = 3024$, and $A = \frac{1}{2}bh \Rightarrow h = \frac{A}{(1/2)b} = \frac{3024}{1/2 \cdot 126} = 48$, hence $RS = 48$.
$PS$ and $QS$ can be found using Pythagoras: they are $\sqrt{60^2-48^2} = 36$ and $\sqrt{102^2-48^2} = 90$ respectively.

As a sanity check, $PS+QS = 36+90=126$ which is the length of $PQ$.