Chapter 8.4, Problem 28AYU

In Problems 29-34, use the results of Problem 27 or 28 to find the area of each triangle. Round answers to two decimal places.

$A={50}^{\circ }$, $C={20}^{\circ }$, $a=3$

To determine

To find: The area of the given triangle. (Round answers to two decimal places)

Answer to Problem 28AYU

Solution:

$K \approx 1.89$

Explanation of Solution

Given:

$\text{A} = 50°, \text{C} = 20°, \text{a} = 3$.

Use the formula $= \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}a}$, $K = \frac{b^2\text{sin}A\text{sin}C}{2\text{sin}B}$ or $K = \frac{c^2\text{sin}A\text{sin}B}{2\text{sin}C}$

Formula used:

$A + B + C = 180°$

$K = \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}A}$, $K = \frac{b^2\text{sin}A\text{sin}C}{2\text{sin}B}$ or $K = \frac{c^2\text{sin}A\text{sin}B}{2\text{sin}C}$

Calculation:

$\text{A} = 50°, \text{C} = 20°, \text{a} = 3$.

$A + B + C = 180°$

$C = 180 - 20 - 50$

$C = 110°$

$Area = K = \frac{a^2\text{sin}B\text{sin}C}{2\text{sin}A}$

$K = \frac{3^2\text{sin}20\text{sin}110}{2\text{sin}50}$

$K = \frac{3^2\text{sin}20\text{sin}110}{2\text{sin}50}$

$K \approx 1.89$