Question

question
in δnop, o = 48 inches, n = 47 inches and ∠n = 72°. find all possible values of ∠o, to the nearest degree.
answer attempt 3 out of 3
additional angle no possible angles
68
°
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Explanation:

Step1: Apply Law of Sines

In $\triangle NOP$, by the Law of Sines, we have $\frac{\sin O}{o}=\frac{\sin N}{n}$. Substituting the given values $o = 48$, $n = 47$, and $\angle N=72^\circ$, we get $\frac{\sin O}{48}=\frac{\sin 72^\circ}{47}$.

Step2: Solve for $\sin O$

First, calculate $\sin 72^\circ\approx0.9511$. Then, $\sin O=\frac{48\times\sin 72^\circ}{47}\approx\frac{48\times0.9511}{47}\approx\frac{45.6528}{47}\approx0.9713$.

Step3: Find possible angles for $O$

Since $\sin O\approx0.9713$, we find $O=\sin^{-1}(0.9713)\approx76^\circ$ or $O = 180^\circ - 76^\circ=104^\circ$. Now we need to check if these angles are valid. The sum of angles in a triangle is $180^\circ$. For $O = 76^\circ$, the third angle $P=180 - 72 - 76 = 32^\circ$ (valid). For $O = 104^\circ$, the third angle $P=180 - 72 - 104 = 4^\circ$ (also valid as all angles are positive and sum to $180^\circ$). But we need to check the side lengths. Since $o = 48$ and $n = 47$, $o>n$, so $\angle O>\angle N$ (since larger side 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 we check the sine value. Wait, actually, when we calculated $\sin O\approx0.9713$, the reference angle is $\approx76^\circ$, and the other angle is $180 - 76 = 104^\circ$. But we need to see if $o>n$ implies $\angle O>\angle N$. Since $o = 48>n = 47$, $\angle O>\angle N = 72^\circ$. Both $76^\circ$ and $104^\circ$ are greater than $72^\circ$, but let's check the sum. Wait, maybe I made a mistake earlier. Wait, the Law of Sines: $\frac{\sin O}{o}=\frac{\sin N}{n}$, so if $o>n$, then $\sin O>\sin N$. $\sin N=\sin 72^\circ\approx0.9511$, and $\sin O\approx0.9713>0.9511$, so $O$ can be in the first or second quadrant. But we need to check if $O + N<180^\circ$. For $O = 76^\circ$, $76 + 72 = 148<180$, good. For $O = 104^\circ$, $104 + 72 = 176<180$, also good. But wait, the problem says "find all possible values". But let's check the side lengths again. Wait, $o = 48$, $n = 47$, so $o$ is slightly longer than $n$, so $\angle O$ is slightly larger than $\angle N$. So $76^\circ$ is close, but $104^\circ$: let's see, $\sin 104^\circ=\sin(76^\circ)\approx0.9713$, so both are valid. But maybe the problem expects the acute angle first? Wait, no, we need to check the triangle. Wait, maybe I miscalculated. Wait, $\frac{\sin O}{48}=\frac{\sin 72^\circ}{47}$, so $\sin O=\frac{48}{47}\sin 72^\circ\approx1.0213\times0.9511\approx0.9713$. Wait, $\frac{48}{47}\approx1.0213$, so $\sin O\approx1.0213\times0.9511\approx0.9713$, which is less than $1$, so two solutions. But let's check the angle sums. For $O = 76^\circ$, $P = 180 - 72 - 76 = 32^\circ$. For $O = 104^\circ$, $P = 180 - 72 - 104 = 4^\circ$. Both are valid. But the problem says "to the nearest degree" and maybe we need to check which one is correct. Wait, maybe I made a mistake in the side labels. Wait, in triangle $NOP$, the sides are labeled as $o$ (opposite $\angle O$), $n$ (opposite $\angle N$), so side $o$ is opposite $\angle O$, side $n$ opposite $\angle N$. So if $o = 48$, $n = 47$, then $o>n$, so $\angle O>\angle N = 72^\circ$. So both $76^\circ$ and $104^\circ$ are greater than $72^\circ$, but let's check the sine value. Wait, $\sin 76^\circ\approx0.9703$, $\sin 104^\circ=\sin(76^\circ)\approx0.9703$? Wait, no, $\sin 104^\circ=\sin(180 - 76)=\sin 76^\circ\approx0.9703$, but my calculation earlier was $\approx0.9713$, maybe due to rounding. Let's recalculate: $\sin 72^\circ\approx0.9510565163$, $48\times0.9510565163 = 45.65071278$, $45.65071278\div47\approx0.9712917613$, so $\sin O\approx0.97129$, so $O=\sin^{-1}(0.97129)\approx76.3^\circ\approx76^\circ$, and the other angle is $180 - 76.3 = 103.7^\circ\approx104^\circ$. Now, we need to check if both triangles are possible. The ambiguous case (SSA) occurs when $a < b\sin A$ (no triangle), $a = b\sin A$ (one right triangle), $b\sin A < a < b$ (two triangles), and $a\geq b$ (one triangle). Wait, here, in standard SSA notation, let's let $a = o = 48$, $b = n = 47$, $A=\angle O$, $B=\angle N = 72^\circ$. Wait, no, standard SSA is side $a$ opposite $\angle A$, side $b$ opposite $\angle B$. So here, side $n$ (47) is opposite $\angle N$ (72°), side $o$ (48) opposite $\angle O$. So it's SSA with $b = 47$, $B = 72°$, $a = 48$. So we check $a\sin B$: $48\sin 72°\approx48\times0.9511\approx45.65$, and $b = 47$. So $a\sin B = 45.65 < b = 47 < a = 48$? Wait, no, $b = 47$, $a = 48$, so $b < a$, so it's the case where $b > a\sin B$ (since $47 > 45.65$) and $b < a$ (47 < 48), so two triangles? Wait, no, the ambiguous case is when $a < b$ and $a > b\sin A$. Wait, maybe I mixed up the notation. Let's use the standard notation: in $\triangle ABC$, $a$ opposite $\angle A$, $b$ opposite $\angle B$, $c$ opposite $\angle C$. So here, let $\angle N = B = 72°$, side $n = b = 47$, side $o = a = 48$, opposite $\angle O = A$. So we have $a = 48$, $b = 47$, $B = 72°$. Then, by Law of Sines, $\frac{\sin A}{a}=\frac{\sin B}{b}$, so $\sin A=\frac{a\sin B}{b}=\frac{48\sin 72°}{47}\approx\frac{48\times0.9511}{47}\approx0.9713$. So $\angle A=\sin^{-1}(0.9713)\approx76°$ or $104°$. Now, check if $\angle A + \angle B < 180°$ for both. For $\angle A = 76°$, $76 + 72 = 148 < 180$, good. For $\angle A = 104°$, $104 + 72 = 176 < 180$, good. Also, since $a > b$ (48 > 47), $\angle A > \angle B$ (72°), which is true for both 76° and 104°. But wait, the problem says "Find all possible values of $\angle O$". But maybe in the problem, the triangle is labeled such that it's a valid triangle, and maybe only one solution? Wait, maybe I made a mistake in the side labels. Wait, the problem says "In $\triangle NOP$", so vertices $N$, $O$, $P$. So angle $N$ is at vertex $N$, between sides $O$ and $P$? Wait, no, in triangle notation, $\triangle NOP$ has vertices $N$, $O$, $P$, so side opposite $N$ is $OP$ (length $o$), side opposite $O$ is $NP$ (length $n$)? Wait, no, standard notation: in $\triangle ABC$, side $a$ is opposite $\angle A$, so in $\triangle NOP$, side opposite $\angle N$ is $OP$ (length $o$), side opposite $\angle O$ is $NP$ (length $n$)? Wait, that's probably my mistake! Oh no, I mixed up the side labels. So in $\triangle NOP$, vertex $N$, $O$, $P$. So side opposite $\angle N$ is $OP$ (let's call it $o$), side opposite $\angle O$ is $NP$ (let's call it $n$). Wait, the problem says "o = 48 inches, n = 47 inches and $\angle N = 72^\circ$". So side $o$ is opposite $\angle N$? No, that can't be. Wait, the problem says "In $\triangle NOP$, o = 48 inches, n = 47 inches and $\angle N = 72^\circ$". So probably, in the triangle, side $o$ is opposite $\angle O$, side $n$ is opposite $\angle N$, and side $p$ is opposite $\angle P$. So vertex $N$: angle $\angle N$, between sides $O$ and $P$; side opposite $\angle N$ is $OP$ (length $o$)? No, that's confusing. Wait, the standard notation is that in $\triangle ABC$, side $a$ is opposite $\angle A$, side $b$ opposite $\angle B$, side $c$ opposite $\angle C$. So in $\triangle NOP$, side opposite $\angle N$ is $OP$ (let's denote as $o$), side opposite $\angle O$ is $NP$ (denote as $n$), side opposite $\angle P$ is $NO$ (denote as $p$). So the problem says $o = 48$ (opposite $\angle N$), $n = 47$ (opposite $\angle O$), and $\angle N = 72^\circ$. Oh! That's my mistake earlier. So side $o$ is opposite $\angle N$, side $n$ opposite $\angle O$. So now, Law of Sines: $\frac{\sin O}{n}=\frac{\sin N}{o}$. So that's $\frac{\sin O}{47}=\frac{\sin 72^\circ}{48}$. Now, let's recalculate.

Step1 (corrected): Apply Law of Sines correctly

$\frac{\sin O}{47}=\frac{\sin 72^\circ}{48}$. So $\sin O=\frac{47\times\sin 72^\circ}{48}$.

Step2 (corrected): Solve for $\sin O$

$\sin 72^\circ\approx0.9511$, so $\sin O=\frac{47\times0.9511}{48}\approx\frac{44.7017}{48}\approx0.9313$.

Step3 (corrected): Find possible angles for $O$

Now, $\sin O\approx0.9313$, so $O=\sin^{-1}(0.9313)\approx68^\circ$ or $O = 180^\circ - 68^\circ = 112^\circ$. Now check validity. Sum of angles: for $O = 68^\circ$, $P = 180 - 72 - 68 = 40^\circ$ (valid). For $O = 112^\circ$, $P = 180 - 72 - 112 = -4^\circ$ (invalid, since angle can't be negative). Wait, that's the mistake! I had the side labels wrong. So side $o$ is opposite $\angle N$, so $o = 48$ (opposite $\angle N = 72^\circ$), side $n = 47$ (opposite $\angle O$). So since $o > n$ (48 > 47), then $\angle N > \angle O$ (because larger side opposite larger angle). Wait, $\angle N = 72^\circ$, so $\angle O$ must be less than $72^\circ$. Ah! That's the key. So my initial side labeling was wrong. So side opposite $\angle N$ is $o = 48$, side opposite $\angle O$ is $n = 47$. So $o > n$ implies $\angle N > \angle O$, so $\angle O < 72^\circ$. So now, $\sin O=\frac{47\times\sin 72^\circ}{48}\approx0.9313$, so $O=\sin^{-1}(0.9313)\approx68^\circ$ (since $68^\circ < 72^\circ$), and the other angle $180 - 68 = 112^\circ$ is greater than $72^\circ$, which would imply $\angle O > \angle N$, but $n < o$ implies $\angle O < \angle N$, so $112^\circ$ is invalid. So only $68^\circ$ is valid. Let's verify: $\frac{\sin 68^\circ}{47}\approx\frac{0.9272}{47}\approx0.0197$, $\frac{\sin 72^\circ}{48}\approx\frac{0.9511}{48}\approx0.0198$, which are approximately equal, so that's correct.

Answer:

$68^\circ$ (and we checked the other angle is invalid, so the only possible value is $68^\circ$)