#### Solved Examples and Worksheet for Law of Sines and Cosines to Solve Triangles

Q1In triangle ABC, if a = 4, b = 3, and c = 5, then find the measure of the angle opposite to the longest side to the nearest degree.

A. 0o
B. 45o
C. 90o
D. 60o

Step: 1
The length c is longer side and the angle opposite to it is angle C.
Step: 2
Cos C = 42+32-522(4)(3)
[Use law of cosines:
Cos C = a2+b2-c22ab.]
Step: 3
Cos C = 024 = 0
Step: 4
C = 90o
Q2In triangle DEF, if d = 6, e = 15, and f = 22, then find the measure of angle D to the nearest degree.
A. 70o
B. 60o
C. 80o
D. not possible to find

Step: 1
d2 = e2 + f2 - 2ef cos D.
[Law of Cosines.]
Step: 2
Cos D = e2+f2-d22ef.
Step: 3
Cos D = 152+222-622(15)(22)
Step: 4
Cos D 1.019696
Step: 5
D is not possible. Since the cosine of an angle cannot be greater than one.
Step: 6
Triangle cannot be drawn with the given sides.
Correct Answer is :   not possible to find
Q3In ΔDEF, if d = 19, e = 19 and f = 20, then find F to the nearest degree. A. 26o
B. 66o
C. 68o
D. 64o

Step: 1
Cos F = 192+192 -2022(19)(19)
[Use law of cosines:
Cos F = d2+e2-f22de.]
Step: 2
Cos F 0.446
Step: 3
F 64°
Q4In ΔPQR, if Q = 75o, p = 12, q = 17, then find the measure of the angle R to the nearest degree. A. 75o
B. 137o
C. 43o
D. 62o

Step: 1
Sin 75o17 = Sin P12
[Use law of Sines: Sin Qq = Sin Pp .]
Step: 2
Sin P = 12 Sin 75o17
Step: 3
Sin P = 0.68
Step: 4
P = 43o (or) 137o
Step: 5
The measure of 137o is not possible, since 137o + 75o = 212o and 212o is greater than 180o.
Step: 6
So, P = 43o.
Step: 7
R = 180 - (75o + 43o)
[Sum of the angle measures in a triangle is 180o.]
Step: 8
R = 62o
Q5A point A is c cm from B and b cm from C as shown in the figure. If ABC is xo, then find the distance between B and C.
[b = 14, c = 20, x = 50,y = 30.] A. 21.7 cm
B. 16.1 cm
C. 34.3 cm
D. 15.5 cm

Step: 1
BC2 = AB2 + AC2 - 2(AB)(AC)) cosBAC
[Use law of Cosines.]
Step: 2
BC2 = 142 + 202 - 2(14)(20) cos 50o30'
Step: 3
BC2 239.796320
[Simplify.]
Step: 4
BC = 15.5
[Simplify.]
Step: 5
The distance between B and C is 15.5 cm.
Correct Answer is :   15.5 cm
Q6In ΔABC, if B = 64o, C = 64o, and b = 26 units, then find the length of a.

A. 23 units
B. 26 units
C. 20 units
D. 30 units

Step: 1
A = 180o - (64o + 64o)
[Sum of the measures of angles in a triangle is 180o.] Step: 2
A = 52o
Step: 3
sin 52oa = sin 64o26
[Use law of sines: sin Aa = sin Bb.]
Step: 4
a = 26 sin 52osin 64o = 23 units, to two significant digits.
[Simplify.]
Correct Answer is :   23 units
Q7In triangle DEF, if D = 45°, f = 11, and e = 9, then find the length of d.[Round it to the nearest whole number]. A. 21
B. 10
C. 13
D. 8

Step: 1
d2 = 112 + 92 - 2(11)(9) cos 45°
[Using law of cosines:
d2 = e2 + f2 - 2ef cos D.]
Step: 2
d2 62.014
Step: 3
d = 8.
[Simplify using calculator.]
Q8In ΔDEF, if d = 28, e = 28, and f = 35, then find F to the nearest degree. A. 13°
B. 77°
C. 81°
D. 83°

Step: 1
Cos F = 282+282-3522(28)(28)
[Use law of cosines:
Cos F = d2+e2-f22de.]
Step: 2
Cos F 0.2188
Step: 3
F 77°
Q9In ΔABC, if ∠A = 62°, a = 14 in., b = 8 in., then find the length of c to the nearest two significant digits. A. 14 in.
B. 24 in.
C. 17 in.
D. 16 in.

Step: 1
sin 62°14 = sin B8
[Use law of Sines: sin Aa = sin Bb.]
Step: 2
sin B = 8 sin 62°14
Step: 3
sin B = 0.50
[Simplify.]
Step: 4
∠B = 30° or 150°
Step: 5
The measure of 150° is not possible, since 150° + 62° = 212° and 212° is greater than 180°.
Step: 6
So, ∠B = 30°.
Step: 7
∠C = 180° - (62° + 30°) = 88°
Step: 8
sin 88°c = sin 62°14
[Use law of Sines: sin Cc = sin Aa.]
Step: 9
c = 14 sin 88°sin 62° = 16 in.
Correct Answer is :   16 in.
Q10In ΔPQR, if Q = 68°, p = 17 units, and q = 23 units, then what is the measure of R to the nearest degree? A. 72°
B. 44°
C. 61°
D. 68°

Step: 1
Sin P17 = Sin 68°23
[Use law of Sines : sin Pp = sin Qq .]
Step: 2
Sin P = 17 Sin 68°23
Step: 3
Sin P = 0.7
[Simplify.]
Step: 4
P = 44° (or) P = 136°
Step: 5
The measure of 136° is not possible, since 136° + 68° = 204° and 204° is greater than 180°.
Step: 6
So, P = 44°.
Step: 7
R = 180 - (68o + 44o)
[Sum of the angle measures in a triangle is 180o.]
Step: 8
R = 68o
Q11In ΔABC, if C = 73o, a = 11 units, c = 17 units, then find B to the nearest degree.
A. 38o
B. 59o
C. 142o
D. 69o

Step: 1
sin A11 = sin 73o17
[Use law of Sines: sin Aa = sin Cc .] Step: 2
sin A = 11 sin 73o17
Step: 3
sin A = 0.62
Step: 4
A = 38o (or) 142o
Step: 5
The measure of 142o is not possible, since 142o + 73o = 215o and 215o is greater than 180o.
Step: 6
So, A = 38o.
Step: 7
B = 180 - (38o + 73o)
[Sum of the angle measures in a triangle is 180o.]
Step: 8
B = 69o
Q12In triangle DEF, D = θ = 35°44′, e = 12.5 units, and f = 17.4 units. What is the length d to four significant digits? A. 15.42
B. 10.29
C. 10.2637
D. 20.412

Step: 1
d2 = 12.52 + 17.42 - 2(12.5)(17.4) cos 35°44′
[Use law of cosines:
d2 = e2 + f2 - 2ef cos D.]
Step: 2
d2 ≈ 105.901
Step: 3
d = 10.29, to four significant digits