Finding lengths related to an altitude of $\triangle PQR$ without Cosine Rule?

Question

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

$$Cos{A} = \frac{PS}{60}$$

Answer 1

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.

Answer 2

Hint: Write the Pythagorean formula for each of the right triangles. You have two equations in two unknowns.

Answer 3

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.

Answer 4

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.

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

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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

$$252\,PS=9072.$$

Answer 5

Three equations three unknowns. (and one assumption that all terms are positive)

1. $PS^2 + RS^2 = 60^2$
2. $RS^2 + QS^2 = 102^2$
3. $PS+QS = 126$

They aren't linear so they may not satisfy the simple "$k$ (linear and independant) equations; $k$ unknowns = solvable" mantra, but they probably do.

Substitute for one variable and combine to one equation with one unknown.

$x = PS$ and so $x^2 + (102^2 - (126-x)^2)=60^2$

or in other words $102^2 -126^2 +2\cdot 126x = 60^2$ so

$PS = x = \frac {60^2 + 126^2 -102^2}{2\cdot 126}$

$QS = 126 - \frac {60^2 + 126^2 -102^2}{2\cdot 126}$

$RS =\sqrt{60^2 -(\frac {60^2 + 126^2 -102^2}{2\cdot 126})^2 }$

.... of course you can make things easier by dividing out common factors and noting the difference of square rule that $126^2 - 102^2 = (126-102)(126+102)$ which may be things easier to calculate (especially if you divided out the common factor of $6$).....

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Alternatively

$PS^2 + RS^2 =60^2$ and $RS^2 + QS^2 = 102^2$ so

$QS^2 - PS^2 = 102^2 - 60^2$ and

$(QS + PS)(QS-PS) = (102+60)(102-60)$

$126(QS-PS) = 162*42$

$(QS-PS) = \frac {162*42}{126}\frac {6*27*6*7}{6*21} = 9*6 =54$.

$OS-PS = 54$ and $QS+PS = 126$ so $QS=\frac {54+126}2=\frac 62(9+21)=3*30=90$

And $PS = \frac {126-54}2 = \frac 62{21-9} =3*12 = 36$

And $RS = \sqrt{60^2 - 36^2} = \sqrt{102^2 - 90^2}=$

$6\sqrt{10^2 -6^2} = 6\sqrt{17^2-15^2}=$

$6\sqrt{(10-6)(10+6)} = 6\sqrt{(17-15)(17+15)}=$

$6\sqrt{4*16} = 6\sqrt {2*32}=$

$6*2*4 = 6*8 = 48$.

Answer 6

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.

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