if $cos\thetaapprox0.3090$, which of the following represents approximate values of $sin\theta$ and…

if $cos\thetaapprox0.3090$, which of the following represents approximate values of $sin\theta$ and $\tan\theta$, for $0^{circ}<\theta<90^{circ}$?\n$sin\thetaapprox0.9511$; $\tan\thetaapprox0.3249$\n$sin\thetaapprox0.9511$; $\tan\thetaapprox3.0780$\n$sin\thetaapprox3.2362$; $\tan\thetaapprox0.0955$\n$sin\thetaapprox3.2362$; $\tan\thetaapprox10.4731$

if $cos\thetaapprox0.3090$, which of the following represents approximate values of $sin\theta$ and $\tan\theta$, for $0^{circ}<\theta<90^{circ}$?\n$sin\thetaapprox0.9511$; $\tan\thetaapprox0.3249$\n$sin\thetaapprox0.9511$; $\tan\thetaapprox3.0780$\n$sin\thetaapprox3.2362$; $\tan\thetaapprox0.0955$\n$sin\thetaapprox3.2362$; $\tan\thetaapprox10.4731$

Answer

Answer:

sin θ≈0.9511; tan θ≈3.0780

Explanation:

Step1: Use the Pythagorean identity

We know that $\sin^{2}\theta+\cos^{2}\theta = 1$. Given $\cos\theta\approx0.3090$, then $\sin^{2}\theta=1 - \cos^{2}\theta$. So $\sin^{2}\theta=1-(0.3090)^{2}=1 - 0.095481 = 0.904519$. Then $\sin\theta=\sqrt{0.904519}\approx0.9511$ (since $0^{\circ}<\theta<90^{\circ}$, $\sin\theta>0$).

Step2: Calculate the tangent value

We know that $\tan\theta=\frac{\sin\theta}{\cos\theta}$. Substituting $\sin\theta\approx0.9511$ and $\cos\theta\approx0.3090$, we get $\tan\theta=\frac{0.9511}{0.3090}\approx3.0780$.