Solve each triangle. Round to the nearest tenth, if necessary.

1.

SOLUTION:

Because two angles are given, A = 180 (110 + 38 ) or 32 . Use the Law of Sines to find a and b.

Therefore, , a 11.2,andb 19.8.

2.

SOLUTION:

Because two angles are given, H = 180 (53 + 112 ) or 15 . Use the Law of Sines to find f and g.

Therefore, , f 55.5,andg 64.5.

3.

SOLUTION:

Because two angles are given, K = 180 (40 + 58 ) or 82 . Use the Law of Sines to find j and .

Therefore, , j 16.2,and 21.4.

4.

SOLUTION:

Because two angles are given, S = 180 (62 + 56 ) or 62 . Use the Law of Sines to find r and s.

Therefore, , r 6.6,ands 7.

5.

SOLUTION:

Because two angles are given, T = 180 (12 + 148 ) or 20 . Use the Law of Sines to find t and u.

Therefore, t 29.0,andu 17.7.

6.

SOLUTION:

Because two angles are given, D = 180 (25 + 72 ) or 83. Use the Law of Sines to find band c.

Therefore, b 12.8,andc 28.7.

7.GOLF Agolfermissesa12-foot putt by putting 3 off course. The hole now lies at a 129 angle between the ball and its spot before the putt. What distance does the golfer need to putt in order to make the shot?

SOLUTION:Draw a diagram of a triangle with two angle

measures of 3 and 129 and an included side length of 12 feet.

(not drawn to scale) Because two angles are given, the missing angle is

180 (3 +129)or48.UsetheLawofSinestofind x.

Therefore, the distance the golfer needs to putt is about 0.85 ft.

8.ARCHITECTUREAnarchitects client wants to build a home based on the architect Jon Lautners Sheats-Goldstein House. The length of the patio will be 60 feet. The left side of the roof will be at a 49 angle of elevation, and the right side will be at an 18 angle of elevation. Determine the lengths of the left and right sides of the roof and the angle at which they will meet.

SOLUTION:Draw a diagram of a triangle with two angle

measures of 49 and 18 and an included side length of 60 feet.

Because two angles are given, is180 (49 + 18)or113.UsetheLawofSinestofindx and y .

Therefore, the left and right sides of the roof are about 20.1 and 49.2 feet, respectively, and the angle

at which they meet is about 113 .

9.TRAVELFortheinitial90milesofaflight,thepilotheads 8 off course in order to avoid a storm. The pilot then changes direction to head toward the destination for the remainder of the flight, making a

157 angle to the first flight course. a. Determine the total distance of the flight. b. Determine the distance of a direct flight to the destination

SOLUTION:a. Draw a diagram to model the situation.

(Not drawn to scale) Because two angles are given, the missing angle is

180 (157 +8)or15.UsetheLawofSinestofind x.

Therefore, the distance of the flight is 90 + 48.4 or about 138.4 miles. b. Find the length of the side opposite the 157 angle.

Therefore, the distance of a direct flight is about 135.9 miles.

Find all solutions for the given triangle, if possible. If no solution exists, write no solution. Round side lengths to the nearest tenth and angle measures to the nearest degree.

10.a = 9, b = 7, A = 108

SOLUTION:Draw a diagram of a triangle with the given dimensions.

(Not drawn to scale) Notice that A is obtuse and a > b because 9 > 7. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C 180 (108 + 48 ) or about 24 . Apply the Law of Sinesto find c.

Therefore, the remaining measures of are

B 48 , C 24 , and c 3.9.

11.a = 14, b = 15, A = 117

SOLUTION:

A is obtuse and a < b because 14 b because 18 > 12. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C 180 (27 + 17.62) or about 135.38. Apply the Law of Sines to find c.

Therefore, the remaining measures of ABC are

B 18 , C 135 , and c 27.8.

13.a = 35, b = 24, A = 92

SOLUTION:Draw a diagram of a triangle with the given dimensions.

Notice that A is obtuse and a > b because 35 > 24. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C 180 (92 + 43.26 ) or about 44.74 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B 43 , C 45 , and c 24.7.

14.a = 14, b = 6, A = 145

SOLUTION:Draw a diagram of a triangle with the given dimensions.

Notice that A is obtuse and a > b because 14 > 6. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C 180 (145 + 14.23 ) or about 20.77 . Apply the Law of Sines to find c.

Therefore, the remaining measures of are

B 14 , C 21 , and c 8.7.

15.a = 19, b = 38, A = 30

SOLUTION:Draw a diagram of a triangle with the given dimensions.

Notice that A is acute and b > a because 38 > 19. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C =180 (30 + 90 ) or 60 . Apply the Law of Sines to findc.

Therefore, the remaining measures of are

B = 90 , C = 30 , and c 32.9.

16.a = 5, b = 6, A = 63

SOLUTION:Notice that A is acute and a < b because 5 < 6. So, this problem has no solution, one solution, or two

solutions. Find h.

Because a < h, no triangle can be formed with sides

a = 5, b = 6, and A = 63 . Therefore, this problem has no solution.

17.a = 10, b = , A = 45

SOLUTION:Draw a diagram of a triangle with the given dimensions.

Notice that A is acute and b > a because >10. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, C 180 (45 + 90 ) or 45 . Apply the Law of Sines to findc.

Therefore, the remaining measures of are

B =90 , C = 45 , and c =10.

18.SKIINGAskiliftrisesata28 angle during the first 20 feet up a mountain to achieve a height of 25 feet, which is the height maintained during the remainder of the ride up the mountain. Determine the length of cable needed for this initial rise.

SOLUTION:

In this problem, A = 28 , a = 25 ft, and b = 20 ft. So,A is acute and a > b. Therefore, one solution exists. Apply the Law of Sines to find B.

Because two angles are now known, the angle

opposite x is 180 (28 + 22.06 ) or about 129.94 . Apply the Law of Sines to find x.

Therefore, the length of cable needed for the initial rise is about 41 feet.

Find two triangles with the given angle measure and side lengths. Round side lengths to the nearest tenth and angle measures to the nearest degree.

19.A = 39 , a = 12, b = 17

SOLUTION:

A is acute, and h = 17 sin 39 or about 10.7. Notice that a < b because 12 < 17, and a > h because 12 > 10.7. Therefore, two different triangles are possible with the given angle and side measures. Angle B willbe acute, and angle B' will be obtuse, as shown below.

Make a reasonable sketch of each triangle and applythe Law of Sines to find each solution. Start with thecase in which B is acute. Solution 1 B is acute.

Find B.

Find C.

Apply the Law of Sines to find c.

Solution 2 B is obtuse.

Note that . To find B', find an obtuse angle with a sine that is also 0.8915. To do this, subtract the measure given by your calculator tonearest degree, 63, from 180. Therefore, B' is approximately 117. Find C.

Apply the Law of Sines to find c.

Therefore, the missing measures for acute

are

B 63 , C 78 , and c 18.7,whilethemissingmeasures for obtuse are B' 117 , C 24 , and c 7.8.

20.A = 26 , a = 5, b = 9

SOLUTION:

A is acute, and h =9 sin 26 or about 3.9. Notice that a < b because 5 < 9, and a > h because 5 > 3.9.Therefore, two different triangles are possible with the given angle and side measures. Angle B will be acute, and angle B' will be obtuse, as shown below.

Make a reasonable sketch of each triangle and applythe Law of Sines to find each solution. Start with thecase in which B is acute. Solution 1 B is acute.

Find B.

Find C.

Apply the Law of Sines to find c.

Solution 2 B is obtuse.

Note that . To find B', find an obtuse angle with a sine that is also 0.2630. To do this, subtract the measure given by your calculator to

nearest degree, 52 , from 180 . Therefore, B' is

approximately 128 . Find C.

Apply the Law of Sines to find c.

Therefore, the missing measures for acute

are B 52 , C 102 , and c 11.2,while the missing measures for obtuse are B'

128 , C 26 , and c 5.0.

21.A = 61 , a = 14, b = 15

SOLUTION:

A is acute, and h = 14sin 61 or about 12.2. Notice that a < b because 14 < 15, and a > h because 14 > 12.2. Therefore, two different triangles are possible with the given angle and side measures. Angle B willbe acute, and angle B' will be obtuse, as shown below.

Make a reasonable sketch of each triangle and applythe Law of Sines to find each solution. Start with thecase in which B is acute. Solution 1 B is acute.

Find B.

Find C.

Apply the Law of Sines to find c.

Solution 2 B is obtuse.

Note that . To find B', find an obtuse angle with a sine that is also 0.9371. To do this, subtract the measure given by your calculator to

nearest degree, 70 , from 180 . Therefore, B' is

approximately 110 . Find C.

Apply the Law of Sines to find c.