Question

if $\sin y^\circ = \frac{s}{8}$ and $\tan y^\circ = \frac{s}{t}$, what is the value of $\cos y^\circ$? options: $\cos y^\circ = 8s$, $\cos y^\circ = 8t$, $\cos y^\circ = \frac{t}{8}$, $\cos y^\circ = \frac{8}{t}$ (with a right triangle image showing angle $y^\circ$)

Explanation:

Step1: Recall the tangent identity

We know that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\). For \(\theta = y^{\circ}\), we have \(\tan y^{\circ}=\frac{\sin y^{\circ}}{\cos y^{\circ}}\).

Step2: Substitute the given values

We are given that \(\sin y^{\circ}=\frac{s}{8}\) and \(\tan y^{\circ}=\frac{s}{t}\). Substituting these into the tangent identity:
\(\frac{s}{t}=\frac{\frac{s}{8}}{\cos y^{\circ}}\)

Step3: Solve for \(\cos y^{\circ}\)

First, rewrite the right - hand side as \(\frac{s}{8\cos y^{\circ}}\). So we have the equation \(\frac{s}{t}=\frac{s}{8\cos y^{\circ}}\).
Since \(s
eq0\) (if \(s = 0\), \(\sin y^{\circ}=0\) and \(\tan y^{\circ}=0\), but we can still solve the equation algebraically), we can cross - multiply:
\(s\times8\cos y^{\circ}=s\times t\)
Divide both sides by \(s\) (assuming \(s
eq0\)):
\(8\cos y^{\circ}=t\)
Then, \(\cos y^{\circ}=\frac{t}{8}\)

Answer:

\(\cos y^{\circ}=\frac{t}{8}\) (the option corresponding to \(\cos y^{\circ}=\frac{t}{8}\))