# Prove that a triangle is right angled …

... if one side is double the other and the angle opposite these sides differ by $\frac \pi 3$

It would be better if this was proven using only algebraic trigonometric identities(sine/cosine rules, etc.)

Let A,B,C be the angles and a,b,c be the respective sides opposite to them.

\begin{align} b &=2a \\ |B-A|&=\frac \pi 3 \\ \sin(B-A) &= \frac {\sqrt{3}} 2\\ \sin B \cos A-\cos B \sin A &=\frac {\sqrt{3}} 2\\ \frac b k \frac {b^2+c^2-a^2}{2bc}-\frac a k \frac {a^2+c^2-b^2}{2ac}&=\frac{\sqrt 3} 2 \\ \frac 1 {kc} (b^2-a^2)&=\frac {\sqrt 3} 2 \\ \frac {3a^2} {kc} &= \frac {\sqrt 3} 2 \end{align}

And I'm stuck at that.

Another approach \begin{align} \frac a {\sin A } &= \frac b {\sin B} \\ \frac {a}{\sin A} &= \frac {2a} {\sin (\frac \pi 3+A)}\\ \frac {\sqrt 3} 2\cos A-\frac 1 2 \sin A&= 2\sin A\\ \tan A&= \frac {\sqrt 3} 5\\ \end{align}

That doesn't lead anywhere too

• If $B$ is the bigger angle, $b$ should be $2a$, rather than $a = 2b$. – Zvi Rosen Sep 24 '13 at 17:02
• Oh right, corrected – user80551 Sep 24 '13 at 17:07
• What is the angle "opposite to two sides" in a triangle?! – DonAntonio Sep 24 '13 at 17:21

## 3 Answers

You have a mistake computing $\sin \left(\frac{\pi}{3}+A\right)$, the correct formula is

$$\sin (x+y) = \sin x\cos y + \cos x\sin y,$$

which leads here to

$$\frac12\sin A + \frac{\sqrt{3}}{2}\cos A = 2\sin A,$$

and further to the well-known

$$\tan A = \frac{1}{\sqrt{3}},$$

whence $A = \frac{\pi}{6}$.

Let $\;\Delta ABC\;$ be our triangle, with sides $\;|AB|=x\;,\;\;|BC|=2x\;$ , and

$$\angle A=\beta\;,\;\;\angle C=\alpha\;\;,\;\;\beta-\alpha=\frac\pi3$$

Apply now the Law of Sines:

$$\frac{|AB|}{\cos\alpha}=\frac{|BC|}{\sin\beta}\iff \frac x{\sin\alpha}=\frac{2x}{\sin\left(\alpha+\frac\pi3\right)}=\frac{2x}{\frac12\sin\alpha+\frac{\sqrt3}2\cos\alpha}\implies$$

$$\implies \frac x{\sin\alpha}=\frac{4x}{\sin\alpha+\sqrt3\cos\alpha}\implies\sin\alpha+\sqrt3\cos\alpha=4\sin\alpha\implies$$

$$\implies\tan\alpha=\frac1{\sqrt3}\implies \alpha=\frac\pi6$$

and thus

$$\beta=\alpha+\frac\pi3=\frac\pi6+\frac\pi3=\frac\pi2$$

Visual aid for in case you're interested in developing a synthetic proof: 