Chapter 8.3, Problem 17AYU

In Problems 17-32, solve each triangle.

$a\text{ = 3}$, $b\text{ = 4}$, $c{\text{ = 40}}^{\circ }$

To determine

To find: The remaining angle(s) and side(s) of the given triangle.

Answer to Problem 17AYU

Solution:

$\text{c} \approx 2.57, \text{B} \approx 91.4°$ and $\text{C} \approx 48.6°$

Explanation of Solution

Given:

$a = 3, b = 4, C = 40°$

Formula used:

$c² = a² + b² - 2ab\text{cos}C$

$b² = a² + c² - 2ac\text{cos}B$

Sum of the interior angles in a triangle is $180°$

Calculation:

$a = 3, b = 4 and C = 40°$

$c² = a² + b² - 2ab\text{cos}C$

$c² = 3² + 4² - 2 * \left(3\right)\left(4\right)\text{cos}40°$

$c = \sqrt{25 - 24\text{cos}40}$

$c \approx 2.57$

$b² = a² + c² - 2ac\text{cos}B$

$\text{cos}B = \frac{a^2 + c^2 - b²}{2ac}$

$\text{cos}B = \frac{3^2 + 2.57² - 4²}{2\left(2.57\right)\left(3\right)}$

$B = {\text{cos}}^{ - 1}\left(\frac{ - .3951}{15.42}\right) \approx 91.4°$

$A + B + C = 180°$

$A = 180° - 40° - 91.4°$

$A = 48.6°$