Question

With usual notations, if the lengths of the sides of a triangle are 7 cm, $4\sqrt{3}$ cm and $\sqrt{13}$ cm, then the measures of the smallest angle is

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{6}$
• C. $\frac{\pi}{3}$
• D. $\frac{\pi}{4}$

Answer 1

Answer: (B)

$\frac{\pi}{6}$

Answer 2

The smallest side is $c = \sqrt{13}$ (since $\sqrt{13} \approx 3.61$ cm, which is less than $4\sqrt{3} \approx 6.93$ cm and $7$ cm). Hence, the smallest angle is opposite side $\sqrt{13}$.

Using the cosine rule for the angle $\theta$ opposite side $c$:

$$\cos\theta = \frac{a^2 + b^2 - c^2}{2ab}$$

Substitute the values:

$$\cos\theta = \frac{7^2 + (4\sqrt{3})^2 - (\sqrt{13})^2}{2 \times 7 \times 4\sqrt{3}} = \frac{49 + 48 - 13}{56\sqrt{3}} = \frac{84}{56\sqrt{3}} = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$$

Thus,

$$\theta = \frac{\pi}{6}$$