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which equation can be solved to find one of the missing side lengths in…

Question

which equation can be solved to find one of the missing side lengths in the triangle?

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, for an angle \(\theta\), \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, the hypotenuse \(AB = 12\) units, angle at \(B\) is \(60^{\circ}\), side \(a\) (adjacent to \(60^{\circ}\)) and side \(b\) (opposite to \(60^{\circ}\)).

Step2: Apply \(\cos\theta\) ratio for side \(a\)

Using \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), with \(\theta = 60^{\circ}\), adjacent side \(=a\) and hypotenuse \(=12\). So, \(\cos60^{\circ}=\frac{a}{12}\).

Step3: Apply \(\sin\theta\) ratio for side \(b\)

Using \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), with \(\theta = 60^{\circ}\), opposite side \(=b\) and hypotenuse \(=12\). So, \(\sin60^{\circ}=\frac{b}{12}\).

Answer:

Equations \(\cos60^{\circ}=\frac{a}{12}\) or \(\sin60^{\circ}=\frac{b}{12}\) can be used to find the missing side lengths.