# Length of a side of a double angle triangle [closed]

I need some help on this problem: In the triangle $$PQR$$, $$\angle R = 2\angle P, PR=5, QR=4$$. What is the length of $$PQ$$? I constructed the triangle and have no idea what to do next. The only idea I had was that maybe the $$4$$ and $$5$$ represent a $$3-4-5$$ Pythagorean triple, but after closer examination, the angles would be $$30$$ and $$60$$ degrees if it were are right triangle. Please help

• Note that, if it were a right triangle with side lengths $3,4,5$ the angles wouldn't be $30^{\circ}$ and $60^{\circ}$. In fact, the $90-60-30$ triangle has the side lengths of $a,\sqrt3 a, 2a$ Commented Jul 31 at 23:37
• 1/ Can you draw the figure and include it? $\quad$ 2/ Are you familiar with sine rule? If so, can you apply it to this situation? Commented Jul 31 at 23:41
• Not a 3-4-5 triangle. But the triangle it does represent has a remarkable dissection, or two. Commented Aug 1 at 0:18

Here is diagram of the problem,

Applying law of sines gives, $$\dfrac x{\sin 2\alpha}=\dfrac 4{\sin \alpha}\Rightarrow x\sin\alpha=8\sin\alpha\cos\alpha\Rightarrow x=8\cos\alpha$$ Now by applying law of cosines we have, $$QR^2=PQ^2+PR^2-2PR.PQ\cos(\angle QPR)$$ $$x^2+25-10x\cos\alpha=16$$ Substituting $$x=8\cos\alpha$$ results in,

$$64\cos^2\alpha-80\cos^2\alpha=-9\Rightarrow \cos\alpha= \pm \dfrac34$$ As $$x$$ cannot be negative, we have $$x=8\times\dfrac34=6$$

• Or substitute $\cos\alpha = \frac x8$ into $x^2 +25-10x\cos\alpha = 16$, in order to find $x$. Commented Aug 1 at 2:54

First render $$\angle Q=180°-3\angle P$$ (why?). Then the Law of Sines gives

$$\dfrac{\sin(3\angle P)}{5}=\dfrac{\sin(\angle P)}{4}\tag{1}$$

Now instead of memorizing the triple angle identities we may render

$$\sin(3\angle P)-\sin(\angle P)=2\cos(2\angle P)\sin(\angle P)$$

$$=2\sin(\angle P)-4\sin^3(\angle P)\tag{2}$$

where we use only the double angle formula for cosine and the trigonometric sum-product relations.

Then from (1) we have $$\sin(3\angle P)=(5/4)\sin(\angle P)$$; substituting this into (2) and solving (since $$\angle P$$ is not the largest angle it must be acute, so the sine is positive) yields

$$\sin(\angle P)=\sqrt7/4$$

Then from the double angle formula for sine we have

$$\sin(\angle R)=2\sin(\angle P)\cos(\angle P)=(2)[\sin(\angle P)](3/4)=(3/2)\sin(\angle P)$$

So the Law of Sines implies $$PQ=(3/2)QR=6$$.

Let $$\alpha$$ be the value of $$\angle P$$ or $$\frac12 \angle R$$. Construct the angle bisector of $$\angle R$$ to meet $$PQ$$ at $$S$$:

$$\triangle PSR$$ has equal base angles $$\alpha$$, so it is isosceles:

$$PS=RS$$

Since $$\angle RPQ = \angle SRQ = \alpha$$ and $$\angle PQR = \angle RQS$$,

$$\triangle PQR \sim \triangle RQS\quad\text{(AAA)}$$

Then these corresponding lengths are in ratio:

\begin{align*} \frac{PQ}{RQ} &= \frac{PR+RQ}{RS+SQ}\\ \frac{PQ}{4} &= \frac{5+4}{PS+SQ}\\ PQ^2 &= 4(5+4)\\ PQ &= 6 \end{align*}