Chapter 8.3, Problem 41AYU

In Problems 33-42, solve each triangle using either the Law of Sines or the Law of Cosines.

$b\text{ = 5}$, $c\text{ = 12}$, $A{\text{ = 60}}^{\circ }$

To determine

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

Answer to Problem 41AYU

Solution:

$a \approx 10.4, B \approx 24.5 \text{and} C = 95.5°$

Explanation of Solution

Given:

$b = 5, c = 12, A = 60°$

Formula used:

$a² = b² + c² - 2bc\text{cos}A$

Calculation:

$b = 5, c = 12, A = 60°$

$a² = b² + c² - 2bc\text{cos}A$

$a² = 5² + 12² - 2 * 5 * 12 * \text{cos}60$

$a² = 169 - 60$

$a = \surd 109 \approx 10.4$

$\frac{\text{sin}60}{\surd 13} = \frac{\text{sin}B}{5}$

$B = {\text{sin}}^{ -1}\frac{5\text{sin}60}{10.4} = 24.5°$

$C = 180 - 60 - 24.5 = 95.5°$