Given that two of the sides in a triangle are $2$ and $4$ and that the angle between them is $60^{\text{o}},$ the triangle is
a) An isoceles
b) Right angeled
c) can't be decided
d) None of the above
Is this as simple as applying the cosine theorem to get the third side $$s^2=4^2+2^2-2\cdot4\cdot2\cdot{60}=16+4-8=12\Leftrightarrow s=2\sqrt{3}.$$
Now I have to figure out the two other angles by using the same theorem. $$2^2=4^2+(2\sqrt{3})^2-2\cdot4\cdot2\sqrt{3}\cos{\theta}\Leftrightarrow \cos{\theta}=\frac{\sqrt{3}}{2}$$
thus $\theta =30^{\text{o}}$, meaning that the third angle has to be $180-60-30=90.$ b) is the correct answer.
Is there a way to come to this conclusion faster than having to compute all the sides and angles in the triangle?
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$\begingroup$If I tell you that a triangle has the sides $4$ and $4$ and that the angle between them is $60^o$ you will say that the triangle is equilateral.
Now if I take an half of one side and joint it with the other vertex I'll obtain an "half" of it: a right angled triangle.
We can apply the Law of Cosines to find that the third side is $s=2\sqrt{3}$, as you have done.
Is there a way to come to this conclusion faster than having to compute all the sides and angles in the triangle?
You do not need to explicitly calculate the right angle in order to know that the triangle is a right triangle.
Let $x=2$ be the given side and $y=4$ be the second. Observe that
$$x^2 + s^2 \,\,\,= \,\,\,2^2 + \left(2\sqrt{3}\right)^2 \,\,\,= \,\,\,4 + 12\,\,\, =\,\,\, 16 \,\,\,= \,\,\,4^2 \,\,\,= \,\,\,y^2$$
This means that $(x,s,y)$ is a Pythagorean triple, so the triangle must be a right triangle.
Alternatively, you could recognize the $2:1$ ratio as a feature of $30^\circ - 60^\circ-90^\circ$ triangles (it's common in contest math or even SAT prep to just memorize the ratios in $30^\circ - 60^\circ-90^\circ$ triangles).
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