How to find the type of triangle when given the ratio of it's sides? 
Q.The sides of a triangle are in ratio 4 : 6 : 7, then the triangle is: 
(A) acute angled 
(B) obtuse angled
   (C) right angled
   (D) impossible


It's definitely not (C) right-angled since $7^2 ≠ 6^2+4^2$
Is it possible to use trigonometry here even though the triangle is not right-angled?
 A: Let $ABC$ be a triangle with sides $a,b,c$.
If $c^2=a^2+b^2$, then the angle at $C$ is a right angle.
If $c^2\lt a^2+b^2$, then the angle at $C$ is acute.
If $c^2\gt a^2+b^2$, then the angle at $C$ is obtuse.
We can think of these facts as coming from the Cosine Law
$$c^2=a^2+b^2-2ab\cos(\angle C).$$
Remark: But we don't need trigonometry to see the answer. Take two sticks of length $a$ and $b$, and join them by a hinge. When the hinge puts the two sticks at right angles, the square of the distance between their two ends is $a^2+b^2$. 
If we spread the ends further apart, then the angle at the hinge becomes greater than $90^\circ$, and in that case if $c$ is the distance between the two ends, we have $c^2\gt a^2+b^2$. 
If on the other hand we bring the two ends closer together, the angle at the hinge becomes less than $90^\circ$, and $c^2\lt a^2+b^2$.
A: The law of cosines for $\theta$ the angle opposite $7$ (the only one that can be greater than $\frac \pi 2$) is $7^2=4^2+6^2-2\cdot 4 \cdot 6 \cos \theta$.  If $\theta \gt \frac \pi 2, \cos \theta \lt 0$, so we would have $7^2=49  \gt 4^2 + 6^2=16+36=52$, so the triangle is acute.
