given that $sin\theta = \frac{21}{29}$, what is the value of $cos\theta$, for $0^{circ}<\theta<90^{circ}$?\n…

Answer:

D. $\sqrt{\frac{20}{29}}$

Explanation:

Step1: Recall the Pythagorean identity

$\sin^{2}\theta+\cos^{2}\theta = 1$

Step2: Substitute the given value of $\sin\theta$

$(\frac{21}{29})^{2}+\cos^{2}\theta=1$

Step3: Solve for $\cos^{2}\theta$

$\cos^{2}\theta=1 - (\frac{21}{29})^{2}=\frac{841 - 441}{841}=\frac{400}{841}$

Step4: Take the square - root

$\cos\theta=\pm\sqrt{\frac{400}{841}}=\pm\frac{20}{29}$

Step5: Determine the sign of $\cos\theta$

Since $0^{\circ}<\theta<90^{\circ}$, $\cos\theta>0$. So $\cos\theta=\frac{20}{29}=\sqrt{\frac{20}{29}}$ (after rationalizing the denominator in the original multiple - choice form).