Question

question 33 (1 point)
for the given triangle, use one of the reciprocal ratios to determine θ to the nearest degree.
15
9
12
a) 39
b) 37
c) 40
d) 38
page 3

Explanation:

Step1: Identify the reciprocal ratio

We know that \(\sin\theta=\frac{opposite}{hypotenuse}\), \(\cos\theta=\frac{adjacent}{hypotenuse}\), \(\tan\theta=\frac{opposite}{adjacent}\). The reciprocal ratios are \(\csc\theta=\frac{hypotenuse}{opposite}\), \(\sec\theta=\frac{hypotenuse}{adjacent}\), \(\cot\theta=\frac{adjacent}{opposite}\). Here, for angle \(\theta\), the opposite side \(= 9\), adjacent side \(=12\), hypotenuse \(=15\). Let's use \(\sin\theta=\frac{9}{15}=\frac{3}{5}\), then \(\csc\theta=\frac{15}{9}=\frac{5}{3}\). Or use \(\cos\theta=\frac{12}{15}=\frac{4}{5}\), then \(\sec\theta=\frac{15}{12}=\frac{5}{4}\). Or use \(\tan\theta=\frac{9}{12}=\frac{3}{4}\), then \(\cot\theta=\frac{12}{9}=\frac{4}{3}\). Let's use \(\sin\theta=\frac{9}{15}\) (since \(\sin^{-1}\) is a common inverse - trigonometric function).

Step2: Calculate the angle

We know that \(\theta=\sin^{-1}(\frac{9}{15})\). Since \(\frac{9}{15} = 0.6\), then \(\theta=\sin^{-1}(0.6)\). Using a calculator, \(\sin^{-1}(0.6)\approx37^{\circ}\)

Answer:

B. 37