Question

17)
triangle: right-angled, angle 60° at the left vertex, hypotenuse labeled ( 2sqrt{2} ), vertical leg labeled ( a ), horizontal leg labeled ( b )

Explanation:

Step1: Identify trigonometric ratios

We have a right - triangle with hypotenuse \(c = 2\sqrt{2}\), angle \(\theta=60^{\circ}\), opposite side to \(\theta\) is \(a\) and adjacent side is \(b\). We can use the cosine and sine functions. The cosine of an angle in a right - triangle is defined as \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\) and sine of an angle is defined as \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)

Step2: Calculate \(b\) (adjacent side)

Using \(\cos\theta=\frac{b}{c}\), where \(\theta = 60^{\circ}\) and \(c = 2\sqrt{2}\). We know that \(\cos60^{\circ}=\frac{1}{2}\)
So, \(\frac{1}{2}=\frac{b}{2\sqrt{2}}\)
Multiply both sides by \(2\sqrt{2}\): \(b = 2\sqrt{2}\times\frac{1}{2}=\sqrt{2}\)

Step3: Calculate \(a\) (opposite side)

Using \(\sin\theta=\frac{a}{c}\), where \(\theta = 60^{\circ}\) and \(c = 2\sqrt{2}\). We know that \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\)
So, \(\frac{\sqrt{3}}{2}=\frac{a}{2\sqrt{2}}\)
Multiply both sides by \(2\sqrt{2}\): \(a=2\sqrt{2}\times\frac{\sqrt{3}}{2}=\sqrt{6}\)

Answer:

If we want to find \(a\), \(a = \sqrt{6}\); if we want to find \(b\), \(b=\sqrt{2}\)