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find the indicated side of the triangle. b = \\frac{?}{\\sqrt{}}

Question

find the indicated side of the triangle.
b = \frac{?}{\sqrt{}}

Explanation:

Step1: Use trigonometric ratio

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\). Here \(\theta = 45^{\circ}\), the hypotenuse \(c = 10\), and the side \(b\) is opposite to the \(45^{\circ}\) angle. So \(\sin45^{\circ}=\frac{b}{10}\).

Step2: Substitute the value of \(\sin45^{\circ}\)

Since \(\sin45^{\circ}=\frac{\sqrt{2}}{2}\), we have \(\frac{\sqrt{2}}{2}=\frac{b}{10}\).

Step3: Solve for \(b\)

Cross - multiply: \(b = 10\times\frac{\sqrt{2}}{2}\). Simplify \(b=\frac{10}{\sqrt{2}}\) (by rationalizing the denominator \(10\times\frac{\sqrt{2}}{2}=\frac{10\sqrt{2}}{2} = \frac{10}{\sqrt{2}}\) as \(2=\sqrt{2}\times\sqrt{2}\)).

Answer:

\(b=\frac{10}{\sqrt{2}}\)