Chapter 8.4, Problem 29AYU

To determine

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

Answer to Problem 29AYU

Solution:

$K \approx 2.27$

Explanation of Solution

Given:

$\text{A} = 70°, \text{C} = 10°, \text{a} = 5$.

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:

$\text{A} = 70°, \text{C} = 10°, \text{a} = 5$.

$A + B + C = 180°$

$B = 180 - 70 - 10$

$B = 100°$

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

$K = \frac{5^2\text{sin}100\text{sin}10}{2\text{sin}70}$

$K \approx 2.27$