Question

calculate the exact value of x. * 1 point $5\sqrt{2}$ $45^{circ}$ $60^{circ}$ $\frac{5\sqrt{3}}{2}$ $10\sqrt{2}$ $\frac{5\sqrt{6}}{6}$ $\frac{5\sqrt{6}}{4}$

Explanation:

Step1: Find the length of the common side in the left - hand right - triangle

In a right - triangle with hypotenuse \(c = 5\sqrt{2}\) and one angle \(\theta=60^{\circ}\), we use the sine function. Let the common side (opposite to the \(60^{\circ}\) angle) be \(h\).
We know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). So, \(h = 5\sqrt{2}\sin60^{\circ}\).
Since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\), then \(h = 5\sqrt{2}\times\frac{\sqrt{3}}{2}=\frac{5\sqrt{6}}{2}\).

Step2: Find the value of \(x\) in the right - hand right - triangle

In the right - hand right - triangle with one angle \(\alpha = 45^{\circ}\), and the side we just found (\(h\)) as the hypotenuse of this \(45 - 45-90\) triangle.
In a \(45 - 45-90\) triangle, if the hypotenuse is \(h\) and the legs are \(x\) (since the two legs are equal in a \(45 - 45-90\) triangle), and we know that \(\sin45^{\circ}=\frac{x}{h}\). Also, \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\) and \(h=\frac{5\sqrt{6}}{2}\).
We can also use the property of a \(45 - 45-90\) triangle: \(x=\frac{h}{\sqrt{2}}\) (because in a \(45 - 45-90\) triangle, \(h = x\sqrt{2}\)).
Substitute \(h=\frac{5\sqrt{6}}{2}\) into \(x=\frac{h}{\sqrt{2}}\), we get \(x=\frac{\frac{5\sqrt{6}}{2}}{\sqrt{2}}=\frac{5\sqrt{6}}{2\sqrt{2}}\).
Rationalize the denominator: \(\frac{5\sqrt{6}}{2\sqrt{2}}=\frac{5\sqrt{6}\times\sqrt{2}}{2\sqrt{2}\times\sqrt{2}}=\frac{5\sqrt{12}}{4}=\frac{5\times2\sqrt{3}}{4}=\frac{5\sqrt{3}}{2}\) (another way: \(\frac{5\sqrt{6}}{2\sqrt{2}}=\frac{5\sqrt{3\times2}}{2\sqrt{2}}=\frac{5\sqrt{3}\times\sqrt{2}}{2\sqrt{2}}=\frac{5\sqrt{3}}{2}\)).

Answer:

\(\frac{5\sqrt{3}}{2}\)