Question

4 - 2 and 4 - 3 practice online
geometry (s1 p01 rm h218) sharp, jeffrey / unit 4 - right triangle trigonometry / lesson 4 - 3 half an e
find the length of leg ab.
ab =
units
find the length of leg bc.
bc =
units

Explanation:

Step1: Use the sine function for \(AB\)

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle C = 30^{\circ}\), \(\sin30^{\circ}=\frac{AB}{AC}\). Given \(AC = 10\), and \(\sin30^{\circ}=\frac{1}{2}\), then \(AB=\sin30^{\circ}\times AC\).

$$AB=\frac{1}{2}\times10 = 5$$

But if we use the cosine function for \(\angle A\) (\(\cos60^{\circ}=\frac{1}{2}\)), \(\cos60^{\circ}=\frac{AB}{AC}\), \(AB=\cos60^{\circ}\times AC=\frac{1}{2}\times10 = 5\)

Step2: Use the cosine function for \(BC\)

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle C=30^{\circ}\), \(\cos30^{\circ}=\frac{BC}{AC}\). Given \(AC = 10\) and \(\cos30^{\circ}=\frac{\sqrt{3}}{2}\), then \(BC=\cos30^{\circ}\times AC\)

$$BC=\frac{\sqrt{3}}{2}\times10 = 5\sqrt{3}$$

Or using the sine function for \(\angle A\) (\(\sin60^{\circ}=\frac{\sqrt{3}}{2}\)), \(\sin60^{\circ}=\frac{BC}{AC}\), \(BC=\sin60^{\circ}\times AC=\frac{\sqrt{3}}{2}\times10 = 5\sqrt{3}\)

Answer:

\(AB = 5\) units, \(BC=5\sqrt{3}\) units