Nature of the $\triangle$

Question

In $\triangle$ ABC, the $\angle BAC$ is a root of the equation $3^{1\over2} \cos x + \sin x = {1\over2}.$ Then what kind of triangle is the $\triangle$ ABC.

$\sin(\pi/3)\cos x + \cos(\pi/3)\sin x = sin(\frac{\pi}{3}+x) = 1/4$ — r9m, May 08 '14 at 05:09
@r9m Yes. I got. It means that $x\in (0,{2\pi\over3})$. But, then how to ascertain the nature of the $\triangle$? — Ruddie, May 08 '14 at 05:13

score 2 · Accepted Answer · answered May 08 '14 at 05:12

2

Multiply both sides of the equation by $1/2$ and use the expansion formula for $\sin(A+B)$ to get $$\sin(x+60^\circ)=\frac{1}{4}\Rightarrow x\approx 105.522^\circ$$ So the triangle is an obtuse triangle.

answered May 08 '14 at 05:12

Samrat Mukhopadhyay

16,277

score 0 · Answer 2 · answered May 08 '14 at 05:27

0

You can't know, because as it was said in comments, $$\sin(x+60°)=\frac{1}{4}$$

means $x\approx 105.522°$ or $x\approx -45.522°$. So either the triangle is obtuse or not.

answered May 08 '14 at 05:27

Xoff

10,310

Nature of the $\triangle$

2 Answers2