Question

cos m∠b = 2√5 / 5
sin m∠b = √5
cos m∠b = √5 / 2
sin m∠b = 2√5 / 5

Explanation:

Step1: Recall the definition of sine in a right - triangle

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle B\), the opposite side to \(\angle B\) is \(AC = 3\) and the hypotenuse is \(BC=3\sqrt{5}\).

Step2: Calculate \(\sin m\angle B\)

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Step3: Recall the definition of cosine in a right - triangle

In a right - triangle, \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle B\), the adjacent side to \(\angle B\) is \(AB = 6\) and the hypotenuse is \(BC = 3\sqrt{5}\).

Step4: Calculate \(\cos m\angle B\)

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Answer:

\(\cos m\angle B=\frac{2\sqrt{5}}{5}\)