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Chapter 5: Problem 34

Use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. $$A=25^{\circ} 4^{\prime}, \quad a=9.5, \quad b=22$$

Short Answer

Expert verified
The short answer depends on the calculations performed. Also, it may mention whether one or two plausible solutions for the triangle were found.

Step by step solution

01

Convert degrees and minutes to decimal degrees

First, it should be converted the angle \(A = 25^{\circ} 4^{\prime}\) from degrees and minutes to decimal for ease of calculation, by adding the minute portion divided by 60 (since 1 degree is 60 minutes) to the degree portion. This results in \( A = 25 + (4/60) = 25.07^{\circ}\)
02

Use the Law of Sines to find sin(B)

With one angle and the lengths of two sides, applying the Law of Sines would give: \(\frac{a}{sinA} = \frac{b}{sinB}\) which simplifies to \(sinB = \frac{b \cdot sinA}{a}\), substituting the known values gives \(sinB = \frac{22 \cdot sin(25.07)}{9.5}\). Calculate this to get the value of sinB.
03

Determine the possible angles for B, and find the other possible solution

The sine of an angle is a value between -1 and 1, so check whether the value of sinB fits into this range. If so, it's plausible. Then B can be found by finding the arcsine of sinB. Since the sine function is periodic and the period is \(180^{\circ}\), two angles (B and \(180^{\circ}\) - B) in a triangle could yield the same sine value. This gives you the two possibilities for the angle B. Note that if one of the angles is not plausible in the context of a triangle (i.e. it results in a sum of \(A + B > 180^{\circ}\)), only one solution for the triangle exists.
04

Use triangle angle sum property to find the third angle C

Knowing two angles in a triangle makes finding the third angle straight forward. Use the property that the sum of angles in a triangle is \(180^{\circ}\). The third angle can be found as \(C = 180^{\circ}\) - A - B. This should be computed for both possible triangles (if two solutions exist).
05

Calculate side c for both possible triangles

Finally, compute for the third side \(c\) for both possible triangles (if two solutions exist) by applying the Law of Sines again. This can be done by rearranging the formula to solve for c as follows: \(c = b\cdot \frac{sinC}{sinB}\). Do this for both possible sets of angles if two solutions exist.
06

Round the solutions

Round the resulting answer to two decimal places, to follow the instruction given in the problem. Remember to perform rounding only at the very end to avoid accumulating round-off errors.

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