3
$\begingroup$

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.

Thanks in advance.

$\endgroup$
  • $\begingroup$ No. It is (sqrt of 3 * cosx) + sinx = 1/2 $\endgroup$ – lk42392 Apr 22 '15 at 11:37
5
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ It may be helpful to the OP if you say (slightly more explicitly) what trig identity you are using (sum of angles formula). $\endgroup$ – TravisJ Apr 22 '15 at 11:59
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.