What is the smallest angle of given triangle?

Question

Calculate the smallest angle in a triangle with side lengths 7 cm, 8 cm and 9 cm.

A. 35∘ 20′

B. 58∘41′

C. 60∘

D. 48∘19′

do i need Trigonometric ratios table?

Answer 1

The side lengths are roughly the same, so we expect all angles to be somewhat near $\frac \pi3$. We know that the sines are proportional to the sidelengths (sine theorem, $\frac{\sin\alpha}a=\frac{\sin\beta}b=\frac{\sin\gamma}c$), hence the fact that the shortest side is $12.5\%$ smaller (and the longest is $12.5\%$ longer) than the middle side, suggests that the sine of the smallest angle is about $12.5\%$ smaller than the sine of $\frac \pi3$ (similarly, the sine of the largest is about as much larger, and themiddle angle is still about sixty degrees). But at that angle, the slope of sine is $\frac 12$, so we raise to $25\%$ and thus expect an angle about $\frac\pi3-\frac\pi{12}=\frac\pi4=45^\circ$. Honestly, we probably deducted too much, so by simply applying these rule-of-thumb calculations (no calculator, no slide-rule, no trigonometric table), we pick D from the suggested options.

Remark: Using a calculator to numerically determine the three angles, we find that they are $$ 48^\circ11'22.9''\qquad 58^\circ24'42.7''\qquad 73^\circ23'54.4''$$ whichjustifies our approximations, but shows an error in the problem statement. Note however that the smallest angles is approximately $48.19^\circ$, which is not the same as the stated $48^\circ19'$.

Answer 2

Note that by the laws of sines we can deduce that the smaller angle is the opposite to $c = 7 cm$ then by the law of cosines

$$c^2 = a^2 + b^2 - 2ab\cos\gamma\implies\cos \gamma=\frac{a^2+b^2-c^2}{2ab}=\frac23\implies \gamma=48.19°$$