# Find the type of triangle from equation.

In triangle $ABC$, the angle($BAC$) is a root of the equation

$$\sqrt{3}\cos x + \sin x = \frac{1}{2}.$$

Then the triangle $ABC$ is

a) obtuse angled

b) right angled

c) acute angled but not equilateral

d) equilateral.

• No. It is (sqrt of 3 * cosx) + sinx = 1/2 – lk42392 Apr 22 '15 at 11:37

We have $$\dfrac{\sqrt3}2 \cos(x) + \dfrac12 \sin(x) = \dfrac14 \implies \sin(x+\pi/3) = \dfrac14$$ We hence obtain $x+\pi/3 = \arcsin(1/4)$ or $x+\pi/3 = \pi - \arcsin(1/4)$ Note that $\arcsin(1/4) < \arcsin(1/2) = \dfrac{\pi}6 < \dfrac{\pi}3$. Since $x \in (0,\pi)$, we have $$x+\pi/3 = \pi - \arcsin(1/4) \implies x = \pi - \left(\arcsin(1/4)+\pi/3\right) > \pi/2$$ Hence, the triangle is obtuse.

$\frac{\sqrt{3}}{2}\cdot \cos x + \frac{1}{2}\sin x = \frac{1}{4}$

$\sin ({x + \frac{\pi}{3}}) = \frac{1}{4}$

• It may be helpful to the OP if you say (slightly more explicitly) what trig identity you are using (sum of angles formula). – TravisJ Apr 22 '15 at 11:59

given that $$\sqrt{3}cosx+sinx=\frac{1}{2}$$ Now, diving the above equation by $\sqrt{(\sqrt{3})^2+(1)^2}=2$ we get \begin{align} \frac{\sqrt{3}}{2}cosx+\frac{1}{2}sinx=\frac{1}{4}\\ cosxcos\frac{\pi}{6}+sinxsin\frac{\pi}{6}=\frac{1}{4}\\ cos\left(x-\frac{\pi}{6}\right)=\frac{1}{4}\\ x-\frac{\pi}{6}=cos^{-1}\frac{1}{4}\\ x=\frac{\pi}{6}+cos^{-1}\frac{1}{4}\\ \end{align} $$x\approx1.841714847... rad\approx105.52^{o}>90^{o}$$ Thus, the triangle is an obtuse angled triangle. Hence, the option (a) is correct.

$\sqrt3 cosx \ + \ sinx = \frac{1}{2}$ $\rightarrow$ $\ 2cos(30)cos(x) \ + 2sin(30)sin(x)\ = \frac{1}{2}$ $\rightarrow$ $2cos(x-30)$ = $\frac{1}{2}$ $\rightarrow$ $x-30$ = $arccos(\frac{1}{4})$ and from here you can deduce the angle and so the type of traingle.