Question

solve for x
(right triangle with one angle 60°, one leg √5, hypotenuse x)

Explanation:

Step1: Identify Trig Ratio

The triangle is right - angled, with an angle of \(60^{\circ}\) and the side opposite to \(60^{\circ}\) is \(\sqrt{5}\), and \(x\) is the hypotenuse. We use the sine ratio, where \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). So, \(\sin60^{\circ}=\frac{\sqrt{5}}{x}\).

Step2: Recall \(\sin60^{\circ}\) Value

We know that \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\). Substituting this into the equation from Step 1, we get \(\frac{\sqrt{3}}{2}=\frac{\sqrt{5}}{x}\).

Step3: Solve for \(x\)

Cross - multiply to solve for \(x\). \(x\times\sqrt{3}=2\times\sqrt{5}\), then \(x = \frac{2\sqrt{5}}{\sqrt{3}}\). Rationalize the denominator by multiplying the numerator and denominator by \(\sqrt{3}\): \(x=\frac{2\sqrt{5}\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}=\frac{2\sqrt{15}}{3}\)

Answer:

\(\boldsymbol{\frac{2\sqrt{15}}{3}}\)