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
$$Cos{A} = \frac{PS}{60}$$
 A: 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.
A: Yep. Here's the clever way:

*

*Simplify the lengths:

*

*

*PR=10



*

*

*RQ=17



*

*

*PQ=21



Consider the largest side as the base (PQ in this case)
Think of two right angles with hypotenuses 10 and 17.

 8-15-17 and 6-8-10. Note that 8 is common in both and 6+15=21. Hence the scaled-down RS = 8

Now, fill the rest in the diagram, and then, scale up.
A: Hint:  Write the Pythagorean formula for each of the right triangles.  You have two equations in two unknowns.
A: Hint:
$$\begin{cases}PS^2+SR^2&=60^2,\\QS^2+SR^2&=102^2,\\QS+PS&=126.\end{cases}$$
From this,
$$QS-PS=\frac{QS^2-PS^2}{QS+PS}=54$$ and the rest easily follows.

$$PS=36,QS=90,SR=48.$$


Alternatively, express that the heights of the two right triangles are the same:
$$60^2-PS^2=102^2-(126-PS)^2,$$
or after simplification
$$252\,PS=9072.$$
A: 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 $PQ$.
