Chapter 8.4, Problem 5AYU

To determine

To find: The area of the given triangle.

Answer to Problem 5AYU

Solution:

$area \approx 2.83$

Explanation of Solution

Given:

Formula used:

$K = \frac{1}{2} ac\text{sin}B$

Calculation:

$a = 2, B = 45°, c = 4$

Area of the triangle is given by

$K = \frac{1}{2} ac\text{sin}B$

$K = \frac{1}{2} * 2 * 4 \text{sin}45°$

$K = 4 * \frac{1}{\sqrt{2}} = 2\surd 2 \approx 2.83$

$Area \approx 2.83$