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in δnop, o = 48 inches, n = 47 inches and ∠n = 72°. find all possible values of ∠o, to the nearest degree.
answer attempt 2 out of 3
⊕ additional angle ⊖ no possible angles
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# Explanation: ## Step1: Apply the Law of Sines In $\triangle NOP$, by the Law of Sines, we have $\frac{\sin\angle O}{o}=\frac{\sin\angle N}{n}$. Substituting the given values $o = 48$, $n = 47$, and $\angle N=72^{\circ}$, we get $\frac{\sin\angle O}{48}=\frac{\sin72^{\circ}}{47}$. ## Step2: Solve for $\sin\angle O$ First, calculate $\sin72^{\circ}\approx0.9511$. Then, $\sin\angle O=\frac{48\times\sin72^{\circ}}{47}\approx\frac{48\times0.9511}{47}\approx\frac{45.6528}{47}\approx0.9713$. ## Step3: Find the possible angles for $\angle O$ We know that if $\sin\theta = x$, then $\theta=\arcsin(x)$ or $\theta = 180^{\circ}-\arcsin(x)$ (in the range $0^{\circ}<\theta<180^{\circ}$ for a triangle angle). So, $\angle O=\arcsin(0.9713)\approx76^{\circ}$ or $\angle O = 180^{\circ}-76^{\circ}=104^{\circ}$. Now we need to check if these angles are valid. For $\angle O = 76^{\circ}$, the third angle $\angle P=180^{\circ}-72^{\circ}-76^{\circ}=32^{\circ}$, which is positive. For $\angle O = 104^{\circ}$, the third angle $\angle P=180^{\circ}-72^{\circ}-104^{\circ}=4^{\circ}$, which is also positive. But we also need to check the side lengths. Since $o = 48$ and $n = 47$, $o>n$, so $\angle O>\angle N$ (since in a triangle, larger side is opposite larger angle). $\angle N = 72^{\circ}$, so $\angle O$ should be greater than $72^{\circ}$. Both $76^{\circ}$ and $104^{\circ}$ are greater than $72^{\circ}$, but wait, let's re - check the Law of Sines application. Wait, $o = 48$, $n = 47$, so side $o$ is opposite $\angle N$? No, wait, in triangle notation, in $\triangle NOP$, side $n$ is opposite $\angle N$, side $o$ is opposite $\angle O$, side $p$ is opposite $\angle P$. So $\angle N$ is opposite side $n = 47$, $\angle O$ is opposite side $o = 48$. So since $o>n$, $\angle O>\angle N$ (72°). Now, $\sin\angle O\approx0.9713$, so $\angle O\approx76^{\circ}$ or $104^{\circ}$. But let's check the sum of angles. For $\angle O = 76^{\circ}$, sum is $72 + 76+32 = 180$, good. For $\angle O=104^{\circ}$, sum is $72 + 104 + 4=180$, good. But wait, is there a mistake in the side - angle correspondence? Wait, no, the Law of Sines is $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$, where $a$ is opposite $\angle A$, $b$ opposite $\angle B$, etc. So in $\triangle NOP$, $\angle N$ is at vertex $N$, so side opposite $\angle N$ is $OP$, which is $o$? Wait, no, standard notation: in $\triangle ABC$, side $a$ is opposite $\angle A$, side $b$ opposite $\angle B$, side $c$ opposite $\angle C$. So in $\triangle NOP$, vertex $N$, $O$, $P$. So side opposite $N$ is $OP$ (let's call it $o$), side opposite $O$ is $NP$ (let's call it $n$), side opposite $P$ is $NO$ (let's call it $p$). Oh! I made a mistake in the side - angle correspondence. So $\angle N$ is opposite side $o$ (length 48), $\angle O$ is opposite side $n$ (length 47). So that means $\frac{\sin\angle N}{o}=\frac{\sin\angle O}{n}$. So correct formula: $\frac{\sin\angle N}{o}=\frac{\sin\angle O}{n}$, so $\frac{\sin72^{\circ}}{48}=\frac{\sin\angle O}{47}$. Then $\sin\angle O=\frac{47\times\sin72^{\circ}}{48}\approx\frac{47\times0.9511}{48}\approx\frac{44.7017}{48}\approx0.9313$. Now, $\arcsin(0.9313)\approx68^{\circ}$, and the other angle is $180 - 68=112^{\circ}$. Now, check the side - angle relationship: side $n = 47$ is opposite $\angle O$, side $o = 48$ is opposite $\angle N$. So since $o>n$, $\angle N>\angle O$? Wait, no, if side $o$ (opposite $\angle N$) is longer than side $n$ (opposite $\angle O$), then $\angle N>\angle O$. $\angle N = 72^{\circ}$, so $\angle O$ should be less than $72^{\circ}$? Wait, this is the key mistake. So let's start over. Correct Step1: Law of Sines in $\triangle NOP$ Let's define: $\angle N = 72^{\circ}$, side opposite $\angle N$ is $OP = o=48$ inches. $\angle O$ is an angle, side opposite $\angle O$ is $NP=n = 47$ inches. $\angle P$ is the third angle, side opposite $\angle P$ is $NO = p$ inches. By Law of Sines: $\frac{\sin\angle N}{o}=\frac{\sin\angle O}{n}$ So $\frac{\sin72^{\circ}}{48}=\frac{\sin\angle O}{47}$ Correct Step2: Solve for $\sin\angle O$ $\sin\angle O=\frac{47\times\sin72^{\circ}}{48}$ $\sin72^{\circ}\approx0.9511$ $\sin\angle O=\frac{47\times0.9511}{48}\approx\frac{44.7017}{48}\approx0.9313$ Correct Step3: Find possible $\angle O$ $\angle O=\arcsin(0.9313)\approx68^{\circ}$ (since $\arcsin(0.9313)\approx68^{\circ}$) and the other possible angle is $180^{\circ}-68^{\circ}=112^{\circ}$. But now, since side $o = 48$ (opposite $\angle N = 72^{\circ}$) is longer than side $n = 47$ (opposite $\angle O$), we must have $\angle N>\angle O$ (because in a triangle, longer side is opposite larger angle). So $\angle O<\angle N = 72^{\circ}$. Now, $\angle O = 68^{\circ}<72^{\circ}$, which is valid. $\angle O = 112^{\circ}>72^{\circ}$, which would mean that side $n$ (opposite $\angle O$) should be longer than side $o$ (opposite $\angle N$), but $n = 47NP$ (48 > 47), then $\angle N>\angle O$ (because side opposite $\angle N$ is longer than side opposite $\angle O$). So $\angle O$ must be less than $\angle N = 72^{\circ}$. Now, $\sin\angle O=\frac{47\times\sin72^{\circ}}{48}\approx0.9313$, so $\angle O=\arcsin(0.9313)\approx68^{\circ}$ (since $\arcsin(0.9313)$ is in the first quadrant, and the second - quadrant angle $180 - 68 = 112^{\circ}$ would make $\angle O>72^{\circ}$, which contradicts the side - angle relationship, and also makes the third angle negative). So the only valid angle for $\angle O$ is approximately $68^{\circ}$? Wait, no, let's calculate $\arcsin(0.9313)$ more accurately. $\sin68^{\circ}\approx0.9272$, $\sin69^{\circ}\approx0.9336$. So $0.9313$ is between $\sin68^{\circ}$ and $\sin69^{\circ}$. Let's use a calculator: $\arcsin(0.9313)\approx68.2^{\circ}\approx68^{\circ}$ (to the nearest degree). Wait, but let's check the initial problem again. The problem says "Find all possible values of $\angle O$". Maybe my side - angle correspondence was wrong. Let's use the standard notation: in $\triangle ABC$, $a$ is opposite $\angle A$, $b$ opposite $\angle B$, $c$ opposite $\angle C$. So in $\triangle NOP$, let's let $\angle N$ be $\angle A$, side $NP$ be $a$, $\angle O$ be $\angle B$, side $NO$ be $b$, $\angle P$ be $\angle C$, side $OP$ be $c$. Then Law of Sines: $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$. So if $\angle N = A = 72^{\circ}$, side $NP=a = 47$ (opposite $\angle N$), side $OP = c=48$ (opposite $\angle P$), side $NO = b$ (opposite $\angle O = B$). Then $\frac{\sin A}{a}=\frac{\sin B}{b}$ is not the right formula. I think the confusion is in the notation. Let's use the given notation: in $\triangle NOP$, the sides are labeled as $o = 48$ (opposite $\angle N$), $n = 47$ (opposite $\angle O$), and we need to find $\angle O$. So with $\angle N = 72^{\circ}$, $o = 48$, $n = 47$. Then by Law of Sines: $\frac{\sin\angle N}{o}=\frac{\sin\angle O}{n}$ => $\sin\angle O=\frac{n\sin\angle N}{o}=\frac{47\times\sin72^{\circ}}{48}\approx\frac{47\times0.9511}{48}\approx0.931$. Now, $\angle O=\arcsin(0.931)\approx68^{\circ}$ or $\angle O = 180 - 68=112^{\circ}$. Now, check if these angles are valid. For $\angle O = 68^{\circ}$, $\angle P=180 - 72 - 68 = 40^{\circ}$, all angles positive. For $\angle O = 112^{\circ}$, $\angle P=180 - 72 - 112=-4^{\circ}$, which is invalid. So only $\angle O\approx68^{\circ}$ is valid? But wait, the problem has an option "Additional Angle" and "No Possible Angles". Wait, maybe I made a mistake in the side lengths. Wait, $o = 48$, $n = 47$, so $o>n$, so the angle opposite $o$ ($\angle N$) should be greater than the angle opposite $n$ ($\angle O$). So $\angle N = 72^{\circ}>\angle O$, so $\angle O$ must be less than $72^{\circ}$. $\arcsin(0.931)\approx68^{\circ}$, which is less than $72^{\circ}$, and $180 - 68 = 112^{\circ}$ which is greater than $72^{\circ}$, so $112^{\circ}$ is invalid. So the only possible angle is approximately $68^{\circ}$. Wait, but let's calculate $\frac{47\times\sin72^{\circ}}{48}$ more accurately. $\sin72^{\circ}\approx0.9510565163$, $47\times0.9510565163 = 44.69965626$, $44.69965626\div48\approx0.9312428387$. $\arcsin(0.9312428387)\approx68.2^{\circ}\approx68^{\circ}$ (to the nearest degree). # Answer: $68$ (and we can check that there is no other valid angle, so the possible value of $\angle O$ to the nearest degree is $68^{\circ}$)

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Question

Explanation:

Step1: Apply the Law of Sines

Step2: Solve for \(\sin\angle O\)

Step3: Find the possible angles for \(\angle O\)

Answer: