Question

if $cos\theta=\frac{6}{9}$, find the value of $sin\theta$ to 2 decimal points.

Explanation:

Step1: Use Pythagorean theorem

By Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = 9\), \(a = 6\), and \(b=x\). So \(x=\sqrt{9^{2}-6^{2}}=\sqrt{81 - 36}=\sqrt{45}\).

Step2: Calculate \(\sin\theta\)

Since \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), and opposite side is \(x = \sqrt{45}\), hypotenuse is \(9\). Then \(\sin\theta=\frac{\sqrt{45}}{9}\approx\frac{6.708}{9}\approx0.75\).

Answer:

\(\frac{\sqrt{45}}{9}\) or \(0.75\)