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How many solutions does a triangle with side lengths [tex]a = 22[/tex], [tex]A = 117^\circ[/tex], and [tex]b = 25[/tex] have?

Write your answer in numeric form.

Answer here: __________

Sagot :

First, we need to determine if the given triangle with side lengths [tex]\(a = 22\)[/tex], [tex]\(A = 117^\circ\)[/tex], and [tex]\(b = 25\)[/tex] can form a valid triangle, and if so, how many possible solutions there are.

We can use the Law of Sines to find the possible values for angle [tex]\(B\)[/tex]:

[tex]\[ \sin(A) = \sin(117^\circ) \][/tex]

Next, according to Law of Sines:

[tex]\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} \][/tex]

Rewriting this equation, we have:

[tex]\[ \sin(B) = \frac{b \cdot \sin(A)}{a} \][/tex]

We must ensure that the value calculated for [tex]\(\sin(B)\)[/tex] falls within the range of possible sine values, which is [tex]\(-1\)[/tex] to [tex]\(1\)[/tex]. If it falls outside this range, it means no solution is possible.

Assuming [tex]\(\sin(B)\)[/tex] is valid, we solve for [tex]\(B\)[/tex]:

1. First, compute the principal value [tex]\(B_1\)[/tex], which is given by the arcsine function.
2. Additionally, consider the supplementary angle [tex]\(B_2\)[/tex], because the sine function is positive in both the first and second quadrants for one complete cycle [tex]\(0^\circ\)[/tex] to [tex]\(180^\circ\)[/tex]. Thus:

[tex]\[ B_2 = 180^\circ - B_1 \][/tex]

Once we have the two possible values for [tex]\(B\)[/tex], we need to compute the corresponding values for angle [tex]\(C\)[/tex]:

[tex]\[ C_1 = 180^\circ - A - B_1 \\ C_2 = 180^\circ - A - B_2 \][/tex]

For the triangle to be a valid one, [tex]\(C\)[/tex] must be positive. We then check the validity of [tex]\(C_1\)[/tex] and [tex]\(C_2\)[/tex]:

- If [tex]\(C_1 > 0\)[/tex], then it's a valid angle combination.
- If [tex]\(C_2 > 0\)[/tex], then it's another valid angle combination.

Counting the valid combinations will give us the number of possible solutions.

The detailed checking process yields:

[tex]\[ \text{Number of solutions} = 1 \][/tex]