Question

choose the correct values for x and y in the right triangle. x = 5√6 / 2 x = 5 y = 10√3 y = 10

Explanation:

Step1: Use tangent function

In a right - triangle, $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$. Given $\theta = 60^{\circ}$ and the adjacent side to the $60^{\circ}$ angle is $x\sqrt{3}$, and the opposite side is $x$. We know that $\tan60^{\circ}=\sqrt{3}=\frac{x}{x\sqrt{3}}$. Cross - multiplying gives $x\sqrt{3}\times\sqrt{3}=x$. Also, we can use the cosine function. $\cos60^{\circ}=\frac{1}{2}=\frac{x\sqrt{3}}{y}$.

Step2: Find the value of x

We know that in a 30 - 60 - 90 right - triangle, if the side opposite the $30^{\circ}$ angle is $a$, the side opposite the $60^{\circ}$ angle is $a\sqrt{3}$, and the hypotenuse is $2a$. Let the side opposite the $30^{\circ}$ angle be $x$. The side opposite the $60^{\circ}$ angle is $x\sqrt{3}$.
If we consider the relationship between the sides of a 30 - 60 - 90 triangle, and assume the side opposite the $30^{\circ}$ angle is $x$. We know that $\tan60^{\circ}=\frac{\text{opposite}}{\text{adjacent}}=\frac{x\sqrt{3}}{x}=\sqrt{3}$. Also, from $\cos60^{\circ}=\frac{x\sqrt{3}}{y}=\frac{1}{2}$, we can rewrite it as $y = 2x\sqrt{3}$.
Let's use the Pythagorean theorem $x^{2}+(x\sqrt{3})^{2}=y^{2}$. Substituting $y = 2x$ (from the 30 - 60 - 90 triangle ratio).
If we assume the side opposite the $30^{\circ}$ angle is $x$, and the side opposite the $60^{\circ}$ angle is $x\sqrt{3}$. Given the side opposite the $60^{\circ}$ angle is $5\sqrt{3}$, then $x\sqrt{3}=5\sqrt{3}$, so $x = 5$.

Step3: Find the value of y

Since in a 30 - 60 - 90 triangle, the hypotenuse $y$ and the side opposite the $30^{\circ}$ angle $x$ have the relationship $y = 2x$. Since $x = 5$, then $y=10$.

Answer:

$x = 5$, $y = 10$