Question

9)
a) $x=\frac{7\sqrt{6}}{6}, y=7$
b) $x=7, y=\frac{7\sqrt{2}}{2}$
c) $x=\frac{14\sqrt{3}}{3}, y=\frac{7\sqrt{2}}{2}$
d) $x=7, y=\frac{7\sqrt{6}}{6}$

Explanation:

Step1: Identify triangle type

This is a 45-45-90 right triangle, so the two legs are equal, and the hypotenuse is leg $\times \sqrt{2}$.

Step2: Solve for $y$

Use $\tan(45^\circ) = \frac{\text{opposite}}{\text{adjacent}}$. Since $\tan(45^\circ)=1$,
$\tan(45^\circ) = \frac{\frac{7\sqrt{2}}{2}}{y} \implies 1 = \frac{\frac{7\sqrt{2}}{2}}{y} \implies y = \frac{7\sqrt{2}}{2}$

Step3: Solve for $x$

Use $\cos(45^\circ) = \frac{\text{adjacent}}{\text{hypotenuse}}$. $\cos(45^\circ)=\frac{\sqrt{2}}{2}$,
$\cos(45^\circ) = \frac{y}{x} \implies \frac{\sqrt{2}}{2} = \frac{\frac{7\sqrt{2}}{2}}{x}$
Multiply both sides by $x$: $\frac{\sqrt{2}}{2}x = \frac{7\sqrt{2}}{2}$
Divide both sides by $\frac{\sqrt{2}}{2}$: $x = 7$

Answer:

B) $x = 7$, $y = \frac{7\sqrt{2}}{2}$