Question

using trig to find a side
question
solve for ( x ). round to the nearest tenth, if
necessary.
answer
attempt 2 out of 3
( x = )

Explanation:

Step1: Identify the trigonometric ratio

In a right - triangle, we know that \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\). Here, \(\angle Q = 40^{\circ}\), the side \(SR = 35\) is the opposite side with respect to \(\angle Q\), and \(x\) (side \(SQ\)) is the hypotenuse. So, we use the sine ratio: \(\sin Q=\frac{SR}{SQ}\).

Step2: Substitute the values into the sine formula

We have \(\sin40^{\circ}=\frac{35}{x}\).

Step3: Solve for \(x\)

Cross - multiply to get \(x\times\sin40^{\circ}=35\). Then \(x = \frac{35}{\sin40^{\circ}}\).
Since \(\sin40^{\circ}\approx0.6428\), then \(x=\frac{35}{0.6428}\approx54.45\).

Answer:

\(x\approx54.5\)