Question

In a triangle, two sides that measure 8 centimeters and 11 centimeters form an angle that measures 82°. to the nearest tenth of a degree, determine the measure of the smallest angle in the triangle.

Answer 1

We can use the Сosine formula to solve this problem.
First we must find the third side (АС) of the triangle:

$AC^2=AB^2+BC^2-2*AB*BC*cos82^o \\ AC^2=8^2+11^2-2*8*11*0.1392 \\ AC^2=64+121-24.4992= 160.508 \\ AC= \sqrt{160.508} \approx 12.67 \ cm$

The smallest angle of the triangle lies opposite the smallest side, so we need to find m∠C.

$cosC= \frac{AC^2+BC^2-AB^2}{2*AC*BC} \\ \\ cosC= \frac{12.67^2+11^2-8^2}{2*12.67*11}= \frac{160.53+121-64}{278.74} = \frac{217.53}{278.74} \approx 0.7804$

Now we can use Bradis's Table (I don't know the name in English, maybe Trigonometric Table?) to find m∠C:

m∠С = 38°42' = 38.7°

Answer: 38.7°


I hope this helps