Question

which shows the correct trig equation? tanθ = 15/8 cosθ = 8/15 sinθ = 8/15 sinθ = 15/8

Explanation:

Step1: Recall trigonometric ratios

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\).
For the given right - triangle with respect to angle \(\theta\), the opposite side \(= 8\), the hypotenuse \(= 15\), and the adjacent side can be found using the Pythagorean theorem \(a=\sqrt{15^{2}-8^{2}}=\sqrt{225 - 64}=\sqrt{161}\) (but we don't need it for checking the given options).

Step2: Check each option

• For \(\tan\theta=\frac{15}{8}\): Since \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), and opposite \( = 8\) (not \(15\)), this is wrong.
• For \(\cos\theta=\frac{8}{15}\): Since \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), and adjacent\(

eq8\) (adjacent\(=\sqrt{161}\)), this is wrong.

• For \(\sin\theta=\frac{8}{15}\): Since \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), with opposite \( = 8\) and hypotenuse \(=15\), this is correct.
• For \(\sin\theta=\frac{15}{8}\): Since \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\text{opposite}<\text{hypotenuse}\) (\(8 < 15\)), \(\sin\theta=\frac{15}{8}>1\) which is not possible as \(- 1\leqslant\sin\theta\leqslant1\).

Answer:

\(\sin\theta=\frac{8}{15}\) (the green option)