Mathematics: What is the perimeter of the largest triangle

What is the perimeter of the largest triangle

Nathaliasmiles

01.04.2022

$x \approx 51.058\°$

or in radians

$x \approx 0.8911$

Step-by-step explanation:

We have the opposite side of angle x and hypotenuse of the right triangle.

$\text{sin}(\theta)= \frac{\text{opposite}}{\text{hypotenuse}}$

$\text{sin}(x)= \frac{14}{18}$

$\text{sin}(x)= \frac{7}{9}$

$x \approx 51.058\°$

Let's find the other side.

$18^2=14^2+b^2\\324=196+b^2\\b=\sqrt{128}$

$b=8\sqrt{2}$

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