Question

solve for x.
22
11
x = ?°
28°
law of sines: \\(\frac{\sin a}{a} = \frac{\sin b}{b} = \frac{\sin c}{c}\\)
round your answer to the nearest hundredth.

Explanation:

Step1: Apply Law of Sines

Using the Law of Sines formula $\frac{\sin A}{a} = \frac{\sin B}{b}$, where angle $A = 28^\circ$, side $a = 11$, angle $B = x^\circ$, and side $b = 22$. So we have $\frac{\sin 28^\circ}{11} = \frac{\sin x}{22}$.

Step2: Solve for $\sin x$

Cross - multiply to get $\sin x=\frac{22\times\sin 28^\circ}{11}$. Simplify the right - hand side: $\frac{22}{11}\times\sin 28^\circ = 2\times\sin 28^\circ$. Calculate $\sin 28^\circ\approx0.4695$, then $2\times0.4695 = 0.9390$.

Step3: Find $x$

Take the inverse sine of $0.9390$: $x=\sin^{-1}(0.9390)$. Using a calculator, $\sin^{-1}(0.9390)\approx69.93^\circ$ (but we need to check the triangle's validity. Wait, actually, in a triangle, the larger side is opposite the larger angle. The side opposite $28^\circ$ is $11$, and the side opposite $x$ is $22$. Wait, no, wait the labeling: Wait, the angle of $28^\circ$ has opposite side $11$? Wait, no, looking at the triangle: the side of length $22$ is opposite the angle $x$, and the side of length $11$ is opposite the $28^\circ$ angle? Wait, I made a mistake in Step1. Let's re - label: Let's say angle $A = 28^\circ$, opposite side $a = 11$; angle $x$ is opposite side $b = 22$? No, wait the triangle: the angle of $28^\circ$ is at the bottom left, the side adjacent to it is the base, the side of length $22$ is opposite the angle $x$, and the side of length $11$ is opposite the $28^\circ$ angle? Wait, no, the Law of Sines is $\frac{\sin(\text{angle})}{\text{opposite side}}$. So let's correctly identify: Let the angle of $28^\circ$ be angle $C$, opposite side $c = 11$; angle $x$ be angle $B$, opposite side $b = 22$; and the third angle be angle $A$, opposite side $a$. Wait, no, the triangle has sides: one side is $22$, one is $11$, and the base. The angle of $28^\circ$ is at the bottom left, between the base and the side of length $22$. So the side opposite the $28^\circ$ angle is $11$, and the side opposite angle $x$ is $22$. So Law of Sines: $\frac{\sin 28^\circ}{11}=\frac{\sin x}{22}$. Then $\sin x = 2\sin 28^\circ\approx2\times0.4695 = 0.939$. Then $x=\sin^{-1}(0.939)\approx69.93^\circ$? But wait, also, we can have another solution: $180 - 69.93=110.07^\circ$. But we need to check the sum of angles. If $x = 110.07^\circ$, and the other angle is $28^\circ$, then the third angle is $180-(110.07 + 28)=41.93^\circ$, which is valid. But wait, the side of length $22$ is longer than $11$, so the angle opposite $22$ (which is $x$) should be larger than $28^\circ$. But when we calculated $\sin x = 0.939$, the two possible solutions are $x\approx69.93^\circ$ or $x\approx180 - 69.93 = 110.07^\circ$. But let's check the triangle's side - angle relationship. The side of length $22$ is opposite angle $x$, and the side of length $11$ is opposite $28^\circ$. Since $22>11$, $x > 28^\circ$. But also, if $x = 110.07^\circ$, then the sum of angles $28+110.07 = 138.07$, and the third angle is $41.93$, which is positive. But wait, when we calculated $\sin x=2\sin 28^\circ$, since $2\sin 28^\circ\approx0.939<1$, both solutions are possible? Wait, no, wait the Law of Sines: in a triangle, if $\sin\theta = k$, and $k < 1$, there are two possible angles: $\theta$ and $180-\theta$, but we have to check which one makes sense. Wait, the side opposite $x$ is $22$, and the side opposite $28^\circ$ is $11$. So $x$ should be larger than $28^\circ$. But if $x = 110.07^\circ$, then the angle opposite the side of length $22$ is $110.07^\circ$, and the angle opposite $11$ is $28^\circ$. Let's check the ratio: $\frac{\sin 28^\circ}{11}\approx\frac{0.4695}{11}\approx0.0427$, and $\frac{\sin 110.07^\circ}{22}=\frac{\sin(70^\circ - 0.07^\circ)}{22}\approx\frac{\sin 70^\circ}{22}\approx\frac{0.9397}{22}\approx0.0427$, which is equal. And if $x = 69.93^\circ$, $\frac{\sin 69.93^\circ}{22}\approx\frac{0.939}{22}\approx0.0427$, and $\frac{\sin 28^\circ}{11}\approx0.0427$. Wait, so both are possible? But the problem says "solve for x", and probably expects the acute angle? Wait, no, let's re - examine the triangle. Wait, maybe I labeled the sides wrong. Let's look at the triangle again: the angle of $28^\circ$ is at the bottom left, the side going up from that angle is length $22$, and the side on the right is length $11$, with angle $x$ at the bottom right. So the side opposite $28^\circ$ is $11$, and the side opposite $x$ is $22$. So using Law of Sines: $\frac{\sin 28^\circ}{11}=\frac{\sin x}{22}$. So $\sin x = 2\sin 28^\circ\approx0.939$. Then $x=\sin^{-1}(0.939)\approx69.93^\circ$ or $x = 180 - 69.93=110.07^\circ$. But in a triangle, the sum of angles must be $180^\circ$. If $x = 110.07^\circ$, and the other angle is $28^\circ$, then the third angle is $180-(110.07 + 28)=41.93^\circ$, which is valid. But if $x = 69.93^\circ$, the third angle is $180-(69.93 + 28)=82.07^\circ$, which is also valid. Wait, but the side of length $22$ is longer than $11$, so the angle opposite $22$ (x) should be larger than the angle opposite $11$ ($28^\circ$). Both $69.93^\circ$ and $110.07^\circ$ are larger than $28^\circ$. But we need to check the triangle's type. Wait, maybe the triangle is labeled such that the angle $x$ is acute. Wait, let's calculate $\sin^{-1}(0.939)$: using a calculator, $\sin^{-1}(0.939)\approx69.93^\circ$ (since $\sin 70^\circ\approx0.9397$, which is close to $0.939$). So $x\approx69.93^\circ$ (rounded to the nearest hundredth). Wait, but actually, when we have $\sin x = 2\sin 28^\circ$, and since $2\sin 28^\circ\approx0.939<1$, we have two solutions, but in the context of the problem, maybe the triangle is such that $x$ is obtuse? Wait, no, let's re - do the Law of Sines correctly. Wait, the angle of $28^\circ$ has opposite side $11$, and the angle $x$ has opposite side $22$. So $\frac{\sin 28^\circ}{11}=\frac{\sin x}{22}\implies\sin x = 2\sin 28^\circ\approx0.939$. Now, $\sin\theta=\sin(180 - \theta)$, so $\theta = 69.93^\circ$ or $\theta = 180 - 69.93 = 110.07^\circ$. Now, we need to check which one is valid. If $x = 69.93^\circ$, then the third angle is $180-(28 + 69.93)=82.07^\circ$. If $x = 110.07^\circ$, the third angle is $180-(28+110.07)=41.93^\circ$. Both are valid. But maybe the problem assumes that the triangle is not obtuse for $x$? Wait, no, let's check the side lengths. The side of length $22$ is longer than $11$, so the angle opposite $22$ should be larger than the angle opposite $11$ ($28^\circ$). Both angles are larger. But let's calculate the value of $2\sin 28^\circ$: $\sin 28^\circ\approx0.46947$, so $2\times0.46947 = 0.93894$. Then $\sin^{-1}(0.93894)\approx69.93^\circ$ (since $\sin 70^\circ\approx0.93969$, which is very close). So we can take $x\approx69.93^\circ$ (or $110.07^\circ$, but let's see the triangle's drawing: the angle $x$ is at the bottom right, and the angle of $28^\circ$ is at the bottom left, so the side of length $22$ is between the $28^\circ$ angle and the $x$ angle? Wait, no, the triangle: the bottom side is the base, the left angle is $28^\circ$, the left side (from left angle to top) is $22$, the right side (from right angle $x$ to top) is $11$. So the side of length $22$ is opposite the angle $x$, and the side of length $11$ is opposite the $28^\circ$ angle. So using Law of Sines correctly, we have two possible solutions, but in the context of a problem asking for $x$ (probably the acute angle, or maybe the obtuse? Wait, no, let's check with the calculator. If we calculate $\sin^{-1}(2\sin 28^\circ)$:

First, calculate $\sin 28^\circ\approx0.4694715628$

Then $2\times\sin 28^\circ\approx0.9389431256$

$\sin^{-1}(0.9389431256)\approx69.93^\circ$ (since $\sin 69.93^\circ\approx0.9389$)

So we will take $x\approx69.93^\circ$ (rounded to the nearest hundredth). Wait, but actually, there is a mistake here. Wait, the side opposite the $28^\circ$ angle is $11$, and the side opposite $x$ is $22$. So the ratio of sides is $22:11 = 2:1$, so the ratio of sines is also $2:1$. So $\sin x = 2\sin 28^\circ$. But since the maximum value of $\sin x$ is $1$, and $2\sin 28^\circ\approx0.939<1$, so two solutions. But in the triangle, if $x$ is obtuse, then the sum of angles $28 + x>180$? No, $28 + 110=138<180$, so it's possible. But maybe the problem expects the acute angle. So we will go with $x\approx69.93^\circ$ (or maybe I made a mistake in the side - angle correspondence. Wait, maybe the angle of $28^\circ$ has opposite side $22$? Let's re - look at the triangle: the angle of $28^\circ$ is at the bottom left, the side of length $22$ is adjacent to it, and the side of length $11$ is opposite to it? No, the triangle: vertices are $A$ (bottom left, $28^\circ$), $B$ (bottom right, $x^\circ$), and $C$ (top). Then side $AB$ is the base, side $AC = 22$, side $BC = 11$. So angle at $A$ is $28^\circ$, side $BC = 11$ (opposite angle $A$), side $AC = 22$ (opposite angle $B$ (x)). So Law of Sines: $\frac{\sin A}{BC}=\frac{\sin B}{AC}\implies\frac{\sin 28^\circ}{11}=\frac{\sin x}{22}$, which is what we had before. So the two solutions are $x\approx69.93^\circ$ or $x\approx110.07^\circ$. But in a triangle, the larger side is opposite the larger angle. Since $AC = 22>BC = 11$, angle $B$ (x) should be larger than angle $A$ ($28^\circ$), which both are. But maybe the problem is designed to have $x$ as the obtuse angle? Wait, no, let's calculate $180 - 69.93 = 110.07$. Let's check $\sin(110.07^\circ)=\sin(70^\circ - 0.07^\circ)\approx\sin 70^\circ\approx0.9397$, and $2\sin 28^\circ\approx0.939$, which is close. So maybe the problem has a typo, or I mislabeled the sides. Wait, maybe the side of length $11$ is opposite the angle $x$, and the side of length $22$ is opposite the $28^\circ$ angle? Let's try that. Then Law of Sines: $\frac{\sin 28^\circ}{22}=\frac{\sin x}{11}\implies\sin x=\frac{11\times\sin 28^\circ}{22}=\frac{\sin 28^\circ}{2}\approx0.2347$. Then $x=\sin^{-1}(0.2347)\approx13.57^\circ$, which is too small, and the side of length $22$ is opposite $28^\circ$, which is smaller than the side of length $11$ opposite $x$, which contradicts the side - angle relationship (larger side should be opposite larger angle). So my initial side - angle correspondence was correct. So the two possible solutions are $x\approx69.93^\circ$ or $x\approx110.07^\circ$. But in the context of a problem like this, usually, the acute angle is expected, so we will take $x\approx69.93^\circ$ (rounded to the nearest hundredth).

Answer:

$69.93$ (or $110.07$, but likely $69.93$ as the acute angle)