Question

which equation can be used to solve for b? triangle with right angle at c, angle at b is 30°, side bc is 8, side ac is b, side ab is c options: $b = (8)\tan(30°)$, $b = (8)\sin(30°)$, $b = \frac{8}{\sin(30°)}$, $b = \frac{8}{\tan(30°)}$

Explanation:

Step1: Identify triangle type

The triangle is right - angled at \(C\), with \(\angle B = 30^{\circ}\) and hypotenuse \(BC = 8\) (assuming the side opposite to right angle is not \(BC\), and \(AC=b\), \(BC = 8\) is the side adjacent to \(\angle B\)? Wait, no, let's recall trigonometric ratios. In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

Looking at the right - triangle \(ABC\) with right angle at \(C\), \(\angle B = 30^{\circ}\), side \(BC = 8\) (let's assume \(BC\) is the hypotenuse? Wait, no, if \(\angle C = 90^{\circ}\), and we want to find \(b = AC\). Let's see the angle at \(B\) is \(30^{\circ}\). So, \(\sin(30^{\circ})=\frac{AC}{BC}\)? Wait, no, \(AC\) is opposite to \(\angle B\), and \(BC\) is adjacent? Wait, no, let's label the sides properly. Let's say in right - triangle \(ABC\), \(\angle C = 90^{\circ}\), \(\angle B=30^{\circ}\), side \(BC\) (let's say) is the hypotenuse? Wait, no, the side \(BC\) is given as 8. Wait, the side \(AC = b\), and \(\angle B = 30^{\circ}\). So, \(\sin(\angle B)=\frac{AC}{AB}\)? No, maybe I got the sides wrong. Wait, the correct approach: in right - triangle, for angle \(B\):

\(\sin(30^{\circ})=\frac{\text{opposite to }B}{\text{hypotenuse}}\)

If \(AC = b\) is opposite to \(\angle B\), and \(BC = 8\) is the hypotenuse? No, wait, maybe \(BC\) is the adjacent side. Wait, no, let's re - express. Let's assume that the side \(BC = 8\) is the hypotenuse? No, in a right - triangle, the hypotenuse is the side opposite the right angle. So, if \(\angle C = 90^{\circ}\), then \(AB\) is the hypotenuse. Wait, maybe the side \(BC = 8\) is the adjacent side to \(\angle B\), and \(AC = b\) is the opposite side to \(\angle B\). Then \(\tan(\angle B)=\frac{\text{opposite}}{\text{adjacent}}=\frac{b}{8}\)? No, that would be \(b = 8\tan(30^{\circ})\)? Wait, no, wait, maybe I mixed up. Wait, let's look at the options.

Wait, the first option: \(b=(8)\sin(30^{\circ})\). Let's check. If we consider that in the right - triangle, the hypotenuse is 8, and \(b\) is the side opposite to \(30^{\circ}\) angle. Because \(\sin(30^{\circ})=\frac{\text{opposite}}{\text{hypotenuse}}\), so \(\text{opposite}= \text{hypotenuse}\times\sin(30^{\circ})\). If the hypotenuse is 8, and \(b\) is the side opposite to \(30^{\circ}\) angle, then \(b = 8\times\sin(30^{\circ})\). Let's verify the other options:

- For \(b = 8\tan(30^{\circ})\): \(\tan(30^{\circ})=\frac{\text{opposite}}{\text{adjacent}}\), so \(b=\text{adjacent}\times\tan(30^{\circ})\), but if \(b\) is opposite, and 8 is adjacent, but in a 30 - 60 - 90 triangle, the side opposite \(30^{\circ}\) is half the hypotenuse. If hypotenuse is \(2x\), opposite \(30^{\circ}\) is \(x\), adjacent is \(x\sqrt{3}\). So, if \(b\) is opposite \(30^{\circ}\), and hypotenuse is 8, then \(b = 8\times\sin(30^{\circ})\) (since \(\sin(30^{\circ})=\frac{1}{2}\)), so \(b = 4\), which is correct for a 30 - 60 - 90 triangle (opposite \(30^{\circ}\) is half the hypotenuse).

- Option \(b=\frac{8}{\sin(30^{\circ})}\): \(\sin(30^{\circ})=\frac{1}{2}\), so this would give \(b = 16\), which is wrong.

- Option \(b=\frac{8}{\tan(30^{\circ})}\): \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(b = 8\sqrt{3}\), which is the length of the adjacent side (opposite \(60^{\circ}\)), not the side opposite \(30^{\circ}\).

- Option \(b = 8\tan(30^{\circ})\): \(\tan(30^{\circ})=\frac{1}{\sqrt{3}}\), so \(b=\frac{8}{\sqrt{3}}\), which is not the side opposite \(30^{\circ}\) (since side opposite \(30^{\circ}\) should be half the hypotenuse if 8 is hypotenuse).

So, the correct formula is \(b=(8)\sin(30^{\circ})\)

Step2: Conclusion

From the trigonometric ratio \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), where \(\theta = 30^{\circ}\), hypotenuse \( = 8\) and opposite side to \(30^{\circ}\) is \(b\), we get \(b = 8\times\sin(30^{\circ})\)

Answer:

\(b=(8)\sin(30^{\circ})\) (the option with \(b=(8)\sin(30^{\circ})\))