There was someone on Facebook asking this question, I wanted to help, but I'm stuck right know. By the way, here's the question:

$ABC$ is a triangle in which

$$2\sin(A) = 3\sin(B) = 4\sin(C)$$

Find the measure of the smallest angle

My attempt:

Suppose I set the triangle like this screenshot below (sorry, it's a bit messy):

enter image description here

Using the sine rule, and dividing three sides with $12$, we have:

$$\frac{\sin(A)}{6} = \frac{\sin(B)}{4} = \frac{\sin(C)}{3}$$

From that, I assume:

$$x=6; \quad y=3; \quad z=4$$

Then, using cosine rule, I have:

$$\cos(A) = \frac{4^2 + 3^3 - 6^2}{2\cdot 4 \cdot 3} = \frac{-11}{24}$$

This is where I'm stuck. Why negative? Did I do something wrong with the assumption or the calculations?

  • 2
    $\begingroup$ Not -11/4, -11/24. And minus value of cosine means the angle is bigger than 90deg. $\endgroup$
    – MH.Lee
    Oct 31, 2021 at 0:43
  • $\begingroup$ @Nightflight Thank you. I did input -11/4 on my calculator, got the answer in complex number and turned out it was a typo. $\endgroup$
    – user516076
    Oct 31, 2021 at 1:18
  • 1
    $\begingroup$ $C$ is the smallest angle. Now apply cosine rule. $\endgroup$
    – ACB
    Oct 31, 2021 at 1:38
  • $\begingroup$ The smallest side of the triangle will be opposite the smallest of the angles. So the smallest angle will be opposite the side of length $3$, i.e., angle $C$ as ACB has indicated. $\endgroup$ Oct 31, 2021 at 13:25

1 Answer 1


$ABC$ is a triangle in which $\bbox[lightgreen]{2\sin(A)=3\sin(B)=4\sin(C)}$.

We will use the fact that $\bbox[lightblue]{\angle{A}+\angle{B}+\angle{C}=180^{0}}$.

Therefore $$ \left\{ \begin{array}{l} 2\sin(B+C)=3\sin(B),\\ 3\sin(B)=4\sin(C).\\ \end{array} \right. \iff \left\{ \begin{array}{l} 2\sin(B)\cos(C)+2\sin(C)\cos(B)=3\sin(B),\\ 3\sin(B)=4\sin(C).\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} 2\cos(C)+2\sin(C)\cot(B)=3,\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \left\{ \begin{array}{l} 2\cos(C)+2\sin(C)\cot(B)=3,\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} 2\left(1-\frac{9\sin^{2}(B)}{16}\right)^{0.5}+\frac{6\cos(B)}{4}=3,\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \left\{ \begin{array}{l} 2\left(\frac{7+9\cos^{2}(B)}{16}\right)^{0.5}+\frac{6\cos(B)}{4}=3,\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} {(7+9\cos^{2}(B))^{0.5}}+{3\cos(B)}=6,\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \left\{ \begin{array}{l} ({(7+9\cos^{2}(B))^{0.5}})^{2}=(6-{3\cos(B)})^{2},\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} {(7+9\cos^{2}(B))}=(36+{9\cos^{2}(B)}-36\cos(B)),\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \left\{ \begin{array}{l} \cos(B)=\frac{29}{36},\\ \sin(C)=\frac{3\sin(B)}{4}.\\ \end{array} \right. \iff \\ \iff \left\{ \begin{array}{l} \cos(B)=\frac{29}{36},\\ \sin(C)=\frac{3(1-\cos^{2}(B))^{0.5}}{4}.\\ \end{array} \right. \iff \left\{ \begin{array}{l} B=\arccos\left(\frac{29}{36}\right)=36^{0},\\ C=\arcsin\left(\frac{\sqrt{455}}{48}\right)=26^{0}.\\ \end{array} \right. \implies \left\{ \begin{array}{l} \angle{B}=36^{0},\\ \angle{C}=26^{0},\\ \angle{A}=118^{0}. \end{array} \right. $$ Therefore, $\bbox[lightgreen]{\angle{C}\text{ is the smallest angle}}$.

Good luck!


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