Minimize the Angle

Geometrical solution:

Let the triangle be $ABC$ , where $a=8$ and $b=13$ . Then the area of the triangle:

$\begin{aligned} [ABC] & = \frac 12 ab \sin \angle C \\ 40 & = \frac 12 \times 8 \times 13 \sin \angle C \\ \sin \angle C & = \frac {10}{13} & \small \color{#3D99F6} \implies \cos \angle C = \pm \frac {\sqrt{69}}{13} \\ \implies \angle C & \approx 50.28^\circ , \ 129.72^\circ \end{aligned}$

We note that the smallest angle is $\angle A$ since $a=8$ is the shortest side length. And the larger $\angle C \approx 129.72^\circ$ gives a smaller $\angle A$ . Let us now find the side length $c$ by the cosine rule so that we can use the sine rule to find $\angle A$ .

$\begin{aligned} c^2 & = a^2+b^2 - 2ab \color{#3D99F6} \cos \angle C & \small \color{#3D99F6} \text{For }C > 90^\circ, \ \cos \angle C = - \frac {\sqrt{69}}{13} \\ & = 8^2+13^2+2(8)(13)\frac {\sqrt{69}}{13} \\ \implies c & = \sqrt{233+16\sqrt{69}} \end{aligned}$

By the sine rule:

$\begin{aligned} \frac {\sin \angle A}a & = \frac {\sin \angle C}c \\ \frac {\sin \angle A}8 & = \frac {\frac {10}{13}}{\sqrt{233+16\sqrt{69}}} \\ \sin \angle A & = \frac {80}{13\sqrt{233+16\sqrt{69}}} \\ \implies \angle A & \approx \boxed{18.766}^\circ \end{aligned}$

The smallest angle will be opposite the smallest side, which is either the side with length $8,$ or the other missing side (call it side $c$ ).

Thus, we can consider the case where $c \leq 8$ . If this is the case, then $\dfrac{1}{2}(13)(8)\sin{(\theta)} = 40 \implies \theta = \arcsin{\left( \dfrac{10}{13} \right)} \approx 50.285^{\circ}$ .

In the case that $c > 8$ , we set the semi-perimeter equal to $s$ to get that $\sqrt{s(s-8)(s-13)(s-c)} = 40 \implies s(s-8)(s-13)(s-c) = 1600.$

Since $s = \dfrac{8+13+c}{2} = \dfrac{21+c}{2} \implies \dfrac{21+c}{2}\left( \dfrac{5+c}{2} \right)\left( \dfrac{c-5}{2} \right)\left( \dfrac{21-c}{2} \right) = 1600 \\ \therefore \left( 21+c \right)\left( 5+c \right)\left( c-5 \right)\left( 21-c \right) = 25600 \implies \left( 441-c^2 \right)\left( c^2-25 \right) = 25600 \\ \text{So } c^2 \left( 466 - c^2 \right) = 36625$

• Finding the roots of the quartic equation:

Since $c^2 \left( 466 - c^2 \right)$ is a shifted even function (a shifted quartic that opens downward), we take the following steps to find the roots of the equation:

Guess: $c = \sqrt{\dfrac{466}{2}} = \sqrt{233}$ . Plugging this into the equation gives $233^2 = 36625$ but $233^2 - 36625 = 17664$ .

Take $\triangle c$ to be our error in $c$ and this gives that $\pm \sqrt{233^2-36625} = \triangle c = \pm \sqrt{17664} = \pm 16\sqrt{69}$ .

Thus, $c = \sqrt{233 \pm 16\sqrt{69}}$ . We neglect the two negative roots because the side length of a triangle cannot be negative. Of these two remaining solutions, we will only consider $c = \sqrt{233 + 16\sqrt{69}}$ because by maximizing the length of side $c$ , we minimize the angle opposite the side with length $8$ .

Thus, we can rewrite the area of the triangle in terms of the angle opposite the side with length $8$ (call it angle $B$ ), as:

$\dfrac{1}{2}(13)\left( \sqrt{233 + 16\sqrt{69}} \right)\sin{B} = 40 \implies B = \arcsin{\left( \dfrac{80}{13\sqrt{233 + 16\sqrt{69}}} \right)}$ .

Hence, $B \approx 18.766^{\circ}$ .

Angle $B$ is clearly the smaller of the two cases tested here so the smallest possible angle in the triangle is an angle of $18.766^{\circ}$ .