Question

the law of sines: the ambiguous case
solve 88°, a = 110 ft m, c = 54 ft
given the triangle below, find the angle a and length of side x
note: picture is not drawn to scale, but you can assume an angle that appears acute is acute and angle that appears obtuse is obtuse.
(image of a triangle with side 31, side 26, angle 39°, angle a, and side x)
a = degrees
x =

Explanation:

Step1: Use the Law of Sines to find angle opposite to side 26

By the Law of Sines, \(\frac{\sin39^{\circ}}{26}=\frac{\sin\theta}{31}\).
Cross - multiply: \(31\sin39^{\circ}=26\sin\theta\).
\(\sin\theta=\frac{31\sin39^{\circ}}{26}\).
Calculate \(\sin39^{\circ}\approx0.6293\).
\(\sin\theta=\frac{31\times0.6293}{26}\approx\frac{19.5083}{26}\approx0.7503\).
\(\theta=\sin^{- 1}(0.7503)\approx48.6^{\circ}\) or \(\theta = 180^{\circ}-48.6^{\circ}=131.4^{\circ}\).
Since \(39^{\circ}+131.4^{\circ}=170.4^{\circ}\lt180^{\circ}\), both are possible. But if \(\theta = 48.6^{\circ}\), then \(A = 180^{\circ}-39^{\circ}-48.6^{\circ}=92.4^{\circ}\). If \(\theta=131.4^{\circ}\), then \(A = 180^{\circ}-39^{\circ}-131.4^{\circ}=9.6^{\circ}\). But from the Law of Sines proportion \(\frac{\sin A}{x}=\frac{\sin39^{\circ}}{26}\), we first check the non - ambiguous case.
Let's use the Law of Sines formula \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\). Here, let \(a = 31\), \(b = 26\), \(B = 39^{\circ}\), \(A\) is the angle we want to find, and \(C\) is the third angle.
We know that \(\frac{31}{\sin A}=\frac{26}{\sin39^{\circ}}\).
Another way:
We use the Law of Sines \(\frac{\sin A}{31}=\frac{\sin39^{\circ}}{26}\) (incorrect, should be \(\frac{\sin39^{\circ}}{26}=\frac{\sin\theta}{31}\) where \(\theta\) is the angle opposite to 31).
Correct setup: Let the side opposite to \(39^{\circ}\) be \(b = 26\), side opposite to \(A\) be \(c=x\), side opposite to the other angle (let's call it \(B\)) be \(a = 31\).
\(\frac{\sin39^{\circ}}{26}=\frac{\sin B}{31}\), \(\sin B=\frac{31\sin39^{\circ}}{26}\approx\frac{31\times0.6293}{26}\approx0.7503\), \(B\approx48.6^{\circ}\) (acute) or \(B = 131.4^{\circ}\) (obtuse).
If \(B = 48.6^{\circ}\), then \(A=180-(39 + 48.6)=92.4^{\circ}\)
If \(B = 131.4^{\circ}\), then \(A=180-(39+131.4)=9.6^{\circ}\)
Now, using \(\frac{x}{\sin A}=\frac{26}{\sin39^{\circ}}\)
If \(A = 92.4^{\circ}\), \(x=\frac{26\sin92.4^{\circ}}{\sin39^{\circ}}\). Since \(\sin92.4^{\circ}\approx1\), \(x=\frac{26\times1}{0.6293}\approx41.3\)
If \(A = 9.6^{\circ}\), \(x=\frac{26\sin9.6^{\circ}}{\sin39^{\circ}}\), \(\sin9.6^{\circ}\approx0.166\), \(x=\frac{26\times0.166}{0.6293}\approx6.9\)

But let's re - check the Law of Sines formula properly.
Let \(a = 31\), \(b = 26\), \(B = 39^{\circ}\)
\(\frac{\sin A}{a}=\frac{\sin B}{b}\)
\(\sin A=\frac{a\sin B}{b}=\frac{31\sin39^{\circ}}{26}\approx\frac{31\times0.6293}{26}\approx0.7503\)
\(A=\sin^{-1}(0.7503)\approx48.6^{\circ}\) (this is wrong, because \(A\) and \(B\) are not in the correct proportion. The correct formula is \(\frac{\sin\theta}{31}=\frac{\sin39^{\circ}}{26}\) where \(\theta\) is the angle opposite to 31. Then \(A = 180^{\circ}-\theta - 39^{\circ}\))
Let's start over:
By the Law of Sines \(\frac{\sin C}{31}=\frac{\sin39^{\circ}}{26}\) (where \(C\) is the angle opposite to 31)
\(\sin C=\frac{31\sin39^{\circ}}{26}\approx\frac{31\times0.6293}{26}\approx0.7503\)
\(C\approx48.6^{\circ}\) or \(C = 131.4^{\circ}\)
Case 1: If \(C = 48.6^{\circ}\)
\(A=180-(39 + 48.6)=92.4^{\circ}\)
Then, using \(\frac{x}{\sin A}=\frac{26}{\sin39^{\circ}}\)
\(x=\frac{26\sin92.4^{\circ}}{\sin39^{\circ}}\approx\frac{26\times1}{0.6293}\approx41.3\)
Case 2: If \(C = 131.4^{\circ}\)
\(A=180-(39+131.4)=9.6^{\circ}\)
\(x=\frac{26\sin9.6^{\circ}}{\sin39^{\circ}}\approx\frac{26\times0.166}{0.6293}\approx6.9\)

But we also can use the Law of Cosines. Let's assume the standard notation \(a = 31\), \(b = 26\), \(c=x\), \(B = 39^{\circ}\)
\(b^{2}=a^{2}+c^{2}-2ac\cos B\)
\(26^{2}=31^{2}+x^{2}-2\times31x\cos39^{\circ}\)
\(676=961+x^{2}-62x\times0.7771\)
\(x^{2}-48.18x + 285 = 0\)
Using the quadratic formula \(x=\frac{48.18\pm\sqrt{48.18^{2}-4\times285}}{2}=\frac{48.18\pm\sqrt{2321.31 - 1140}}{2}=\frac{48.18\pm\sqrt{1181.31}}{2}=\frac{48.18\pm34.37}{2}\)
\(x_1=\frac{48.18 + 34.37}{2}\approx41.3\), \(x_2=\frac{48.18-34.37}{2}\approx6.9\)

If \(x = 41.3\), then \(\sin A=\frac{31\sin39^{\circ}}{26}\approx0.7503\), \(A=\sin^{-1}(0.7503)\approx48.6^{\circ}\) (wrong, because \(A = 180-(39 + C)\) where \(C\) is found from \(\frac{\sin C}{31}=\frac{\sin39^{\circ}}{26}\))
The correct approach:
We know that \(\frac{\sin A}{x}=\frac{\sin39^{\circ}}{26}\)
First, find \(A\):
Since \(\frac{\sin C}{31}=\frac{\sin39^{\circ}}{26}\), \(C=\sin^{-1}(\frac{31\sin39^{\circ}}{26})\approx48.6^{\circ}\) or \(131.4^{\circ}\)
If \(C = 48.6^{\circ}\), \(A = 180-(39 + 48.6)=92.4^{\circ}\)
Then \(x=\frac{26\sin92.4^{\circ}}{\sin39^{\circ}}\approx\frac{26\times1}{0.6293}\approx41.3\)
If \(C = 131.4^{\circ}\), \(A = 180-(39+131.4)=9.6^{\circ}\)
\(x=\frac{26\sin9.6^{\circ}}{\sin39^{\circ}}\approx\frac{26\times0.166}{0.6293}\approx6.9\)

But usually, in the ambiguous case (SSA - side, side, angle), we check the two possibilities.
Let's use the Law of Sines formula \(\frac{\sin A}{x}=\frac{\sin39^{\circ}}{26}\)
We first find \(A\)
\(\sin A=\frac{x\sin39^{\circ}}{26}\) (but we need to find \(x\) in terms of \(A\))
Another way:
We know that \(A + B + C=180^{\circ}\), \(B = 39^{\circ}\), \(C\) is related to \(x\) (this is getting convoluted)
Let's use the Law of Sines correctly:
\(\frac{\sin39^{\circ}}{26}=\frac{\sin A}{x}=\frac{\sin C}{31}\)
From \(\frac{\sin C}{31}=\frac{\sin39^{\circ}}{26}\), \(C=\sin^{-1}(\frac{31\sin39^{\circ}}{26})\approx48.6^{\circ}\) or \(131.4^{\circ}\)
If \(C = 48.6^{\circ}\):
\(A=180-(39 + 48.6)=92.4^{\circ}\)
\(x=\frac{26\sin92.4^{\circ}}{\sin39^{\circ}}\approx\frac{26\times1}{0.6293}\approx41.3\)
If \(C = 131.4^{\circ}\):
\(A=180-(39+131.4)=9.6^{\circ}\)
\(x=\frac{26\sin9.6^{\circ}}{\sin39^{\circ}}\approx\frac{26\times0.166}{0.6293}\approx6.9\)

Answer:

If we assume the non - degenerate triangle with \(A\) being the larger angle (since \(31>26\) in length, by the Law of Sines, the larger side is opposite the larger angle). So \(A = 92.4^{\circ}\) and \(x\approx41.3\)

So \(A = 92.4\) degrees and \(x\approx41.3\)