Question

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summary
a) what is θ? round to two decimal places, if necessary. enter the degree symbol by using the toolbar or typing deg.
θ = 32°
correct - good job!
b) what is b? round to two decimal places, if necessary.
b = 5.62

Explanation:

Part b)

Step1: Identify triangle type

It's a right - triangle, so we can use trigonometric ratios. We know angle \(A = 58^{\circ}\), side \(a = 9\) (adjacent to angle \(A\)) and we want to find side \(b\) (opposite to angle \(A\)). The tangent ratio is \(\tan(A)=\frac{\text{opposite}}{\text{adjacent}}=\frac{b}{a}\).

Step2: Solve for \(b\)

We know \(A = 58^{\circ}\) and \(a = 9\). From \(\tan(A)=\frac{b}{a}\), we can re - arrange to get \(b=a\times\tan(A)\).
Substitute \(a = 9\) and \(A = 58^{\circ}\) into the formula. \(\tan(58^{\circ})\approx1.6003\). Then \(b = 9\times\tan(58^{\circ})\approx9\times1.6003 = 14.4027\)? Wait, no, wait. Wait, maybe I mixed up opposite and adjacent. Wait, in the right - triangle, angle \(A\) has adjacent side \(a\) and opposite side \(b\)? Wait, no, maybe angle \(A\) is at the base, so the right angle is at the bottom - right. So side \(a\) is adjacent to angle \(A\), side \(b\) is opposite to angle \(A\)? Wait, no, maybe I made a mistake. Wait, the other angle \(\theta=32^{\circ}\), so angle \(A = 58^{\circ}\), angle \(\theta = 32^{\circ}\), right angle \(90^{\circ}\). Let's use \(\tan(\theta)=\frac{a}{b}\), since \(\theta = 32^{\circ}\), \(a = 9\). So \(\tan(32^{\circ})=\frac{9}{b}\), then \(b=\frac{9}{\tan(32^{\circ})}\). \(\tan(32^{\circ})\approx0.6249\), so \(b=\frac{9}{0.6249}\approx14.40\)? Wait, but the given answer in the image is \(5.62\). Wait, maybe I mixed up \(a\) and \(b\). Wait, maybe \(a\) is the opposite side to angle \(\theta\) and \(b\) is the adjacent side. Let's re - examine. The right - triangle: angle \(A = 58^{\circ}\), right angle, so the third angle \(\theta=90 - 58=32^{\circ}\). If we consider angle \(\theta = 32^{\circ}\), then the side opposite to \(\theta\) is \(a = 9\)? No, that can't be. Wait, maybe the triangle has angle \(A = 58^{\circ}\), side \(a\) is adjacent to \(A\), side \(b\) is opposite to \(A\), and hypotenuse \(c\). Wait, maybe we should use \(\tan(58^{\circ})=\frac{b}{a}\), so \(b=a\times\tan(58^{\circ})\). \(a = 9\), \(\tan(58^{\circ})\approx1.6003\), \(b = 9\times1.6003\approx14.40\). But the image shows a value of \(5.62\). Wait, maybe it's a typo or I misread the side. Wait, maybe \(a\) is the hypotenuse? No, the right angle is marked. Wait, maybe the side \(a = 9\) is the opposite side to angle \(\theta\) (which is \(32^{\circ}\)). So \(\sin(\theta)=\frac{a}{c}\), \(\cos(\theta)=\frac{b}{c}\), or \(\tan(\theta)=\frac{a}{b}\). So \(\tan(32^{\circ})=\frac{9}{b}\), then \(b=\frac{9}{\tan(32^{\circ})}\approx\frac{9}{0.6249}\approx14.40\). But the image has \(b = 5.62\). Wait, maybe the side \(a = 9\) is the hypotenuse? No, the right angle is there. Wait, maybe I made a mistake in the angle. Wait, the angle \(A = 58^{\circ}\), so the other non - right angle is \(32^{\circ}\). Let's use \(\tan(58^{\circ})=\frac{b}{a}\), \(a = 9\), so \(b = 9\times\tan(58^{\circ})\approx9\times1.6003 = 14.40\). But the given answer in the image is \(5.62\). Wait, maybe the side \(a\) is the opposite side to \(58^{\circ}\), and we want to find the adjacent side. So \(\cot(58^{\circ})=\frac{b}{a}\), \(\cot(58^{\circ})=\frac{1}{\tan(58^{\circ})}\approx0.6249\), so \(b=a\times\cot(58^{\circ})=9\times0.6249\approx5.62\). Ah! That's it. So if \(a\) is the opposite side to angle \(A = 58^{\circ}\), then to find the adjacent side \(b\) (relative to angle \(A\)), we use \(\cot(A)=\frac{b}{a}\) or \(\tan(\theta)=\frac{a}{b}\) where \(\theta = 32^{\circ}\). So \(b=a\times\cot(58^{\circ})=9\times\frac{1}{\tan(58^{\circ})}\approx9\times0.6249 = 5.62\).

Answer:

\(b\approx5.62\)