If the sides of the Triangle are $a$, $b$ and $\sqrt{(a^2+ab+b^2)}$, then find the value of the greatest Angle.

Question

If the sides of the Triangle are $a$, $b$ and $\sqrt{(a^2+ab+b^2)}$, then find the value of the greatest Angle.

Answer Given in my notebook - $\frac{2\pi}{3}$ radians.

The Question is from the Chapter Solutions of Triangles and I'm aware of Sine and Cosine Rule of the triangles.

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My try.

In any case, $a$ or $b$, no matter which side is greater, the 3rd side (Let's denote it by c) $ \sqrt{(a^2+ab+b^2)} $ will have the greatest angle (i.e. greatest angle will be opposite to this side).

Now, let's take Opposite Angles of sides a, b and c as $A$, $B$, and $C$ respectively. Then the only way to find the angle formed at angle C of the triangle can be easily calculated using Cosine Rule (if all the angles are given, which in this case is true), so putting the formula I get -

$$\cos C = \frac{a^2+b^2-c^2}{2ab} = \sqrt{(a^2+ab+b^2)}$$

Now, in the following case, I've to get the Value of Cosine function as $120^o or \frac{2\pi}{3}$ which I'm not sure how to do.

Can anyone help me with that?

Thanks :)

Answer 1

$$\cos C=\frac{a^2+b^2-c^2}{2ab}\tag{1}$$ as you nearly say, but $c$ is $\sqrt{a^2+ab+b^2}$. Substituting that into $(1)$ should give you the $C$ that you want.

Answer 2

In Triangle ABC,

the side $c=\sqrt{a^2+ab+b^2}$ ----------------------(1)

and by cosine Rule, -

$\cos C = \frac{a^2+b^2-c^2}{2ab}$ ------------------------------(2)

So, substituting (1) in (2) gives,

$$\cos C = \frac{a^2+b^2-(a^2+ab+b^2)}{2ab}$$ $$\Rightarrow \cos C = \frac{-ab}{2ab}$$ $$\iff C = 2n\pi \pm \frac{2\pi}{3}$$ (for $n$ as any integer)

The rule of the Sum angle Property only stands correct only when $n=1$, thus $120^o$ or $\frac{2\pi}{3}$ is the reqd. answer