triangle abc is a right triangle and sin(53°) = 4/x. solve for x and round to the nearest whole number…

Explanation:

Step1: Solve for x using sine definition

Given $\sin(53^{\circ})=\frac{4}{x}$, and $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$. We can rewrite the equation as $x = \frac{4}{\sin(53^{\circ})}$. Since $\sin(53^{\circ})\approx0.7986$, then $x=\frac{4}{0.7986}\approx5$.

Step2: Recall co - function identity

In a right - triangle, if one acute angle is $\theta$, the other is $90^{\circ}-\theta$. Also, $\sin\theta=\cos(90^{\circ}-\theta)$. Here, if one angle is $53^{\circ}$, the other non - right angle is $37^{\circ}$, and $\sin(53^{\circ})=\cos(37^{\circ})$. In right - triangle $ABC$, $\sin(53^{\circ})=\frac{4}{x}$, and $\cos(53^{\circ})=\frac{y}{x}$. Since we found $x\approx5$ and $\sin(53^{\circ})=\frac{4}{x}$, by the Pythagorean theorem $y=\sqrt{x^{2}-4^{2}}=\sqrt{25 - 16}=3$. So $\cos(53^{\circ})=\frac{y}{x}$.

Answer:

$\cos(53^{\circ})=\frac{y}{x}$ (second option from the right - hand side list)