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Consider the triangle ABC with angle A being 70 degrees, and the side lengths satisfying:

$BC^2=AC(AB+AC)$

Is there any intuitive way of finding the measure of angle B? I noticed that the given equality is similar to the Pythagorean Theorem, but not quite.

Thanks for any suggestions!

nak17
  • 377

2 Answers2

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From the given $BC^2=AC(AB+AC)$, we have from the sine rule

$$\sin^2A=\sin B (\sin C+ \sin B)$$

Note

$$\begin{align} & \sin^2A-\sin B (\sin C+ \sin B)\\ =& \sin^2A-\sin^2 B- \sin B \sin C\\ =& \frac12(\cos 2B -\cos 2A) - \sin B \sin (A+B)\\ =& \sin (A-B)\sin (A+B) -\sin B \sin (A+B)\\ =& \sin (A+B)[\sin (A-B) -\sin B]=0\\ \end{align}$$ which leads to $\sin (A-B) =\sin B$ or $ B=\frac12A=35^\circ$.

Quanto
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Let $a = BC, b = AC$, and $c = AB$. Using the cosine rule we have:

$$a^2 = b^2+c^2 - 2bc \cos 70º$$

and since $a^2 = c \cdot b + b^2$, then:

$$c \cdot b + b^2 = b^2+c^2 - 2bc \cos 70º$$ $$bc = c^2 - 2bc \cos 70º$$

and this information allows you to find $b$ in terms of $c$.

Then substitute back into the cosine rule which allows you to relate $a$ with $b$. A straightforward application of the sine rule will allow you to find angle $B$.

I get that $B = 35º$.

Edit: there is a proof on AoPS (and another proof here) that if $\angle A = 2 \angle B$, then $BC^2 = AC(AB+AC)$. This result gives $B = 35º$.

Toby Mak
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