The law of sines, also known as the sine rule, relates the ratios of sides and opposite angles in any triangle. Given any two elements of a triangle (side or angle), the law of sines can be used to calculate the remaining unknown elements. The formula is a/sinA = b/sinB = c/sinC, where a, b, c are the sides and A, B, C are the opposite angles. The document provides examples of using the law of sines to solve for unknown sides and angles in various triangles. It also includes practice problems for students to work through applying the law of sines.

1. LAW OF SINES

2. 42
In trigonometry, the law of sine's, sine
law,
sine formula, or sine rule is an equation
relating the lengths of the sides of an
arbitrary triangle to the sine of its
angles.
According to the law,

3. where a, b, and c are the lengths of the sides of a
triangle, and A, B, and C are the opposite angles
(see the figure to the right), and D is the diameter
of the triangle's circumcircle. When the last part of
the equation is not used, sometimes the law is
stated using the reciprocal;

4. EXAMPLE 1
In ∆ ABC, side a = 8, m<A = 30˚ and m<C = 55˚. Find side c to the nearest
tenth of an integer.
42
Since this problem refers to two angels and two sides, use
the Law of Sines.
B
a =8 c=?
(sin 30˚)•C = 8(sin 55˚)
55˚ 30˚
C A
This answer makes sense, since the larger side is opposite the larger angle.

5. In the figure on the right, we are given side b and angle B, which EXAMPLE 2
opposite each other, so we can use them to calculate the 'Law of Sine's
ratio (s) for this particular triangle:
We are also given the length of side c. Because we know the Law of Sine's
42
ratio, we can find the opposite angle C. A
12 14
c b
40˚ a
B C
We now know both angles B and C, so using the fact that the interior angles of a
triangle add up to 180°, we can find the third angle A:
A = 180 – (40 + 33.4) = 106.6˚
Using the same principle as above we know that,
so we solve this for a, the last unknown side: a = 21.78 sin (106.6˚) = 20.87

6. EXAMPLE 3
Calculate side “c” :
B
42
35˚
c
105˚
A
Now we use our algebra to rearrange and solve:
C 7
Calculate: c = 11.8 (to 1 decimal place)

7. EXAMPLE 4
Calculate angle B :
42
A
5.5 4.7
63˚
B C
Calculate: sin B = 0.7614…
B = 49.6˚

8. EXAMPLE 5
Calculate angle R :
The first thing to notice Is that this triangle has different labels: PQR 42
instead of ABC. But that’s not a problem. We just use P,Q and R
instead of A, B and C in The Law of Sines.
Calculate : sin R : 0.9215…
R = 67.1˚

9. SEATWOR HOMEWORK
K
42
1 1
2 2
3 3
4 4
5 5

10. Seatwork question #1 :
42
Use the formula for law of sines to determine the measure of angle b below
ANSWER :
11
20
62
29

11. Seatwork question #2 :
Is it possible to use the law of sines to calculate x pictured in the 42
triangle below?
ANSWER :
34
Yes, first you must remember that the
x sum of the interior angles of a triangle is
180 in order to calculate the measure of
the angle opposite of the side of length
116
19.
Now that we have the measure of that
19 angle, use the law of sines to find value
30
of x
Set up the equation:
then cross multiply to calculate x.
x = 30.5

12. Seatwork question #3 :
42
In ∆PQR, sin P - 0.2,
sin R = 0.4 and r = 22.
Find the length of p. ANSWER :
Q
p
22
R
P

13. Seatwork question #4 :
In ∆ABC, 42
m<B = 80º,
m<C = 34º and a = 16.
Find the length of b to the nearest tenth. ANSWER :
A Since this problem uses sides a and b, we need to know
the size of <A.
m<A = 180 - (80 + 34) = 66
b=?
80˚ 34˚
B C
16

14. Seatwork question #5 :
In ∆RST, m<R = 105º, r = 12, and t = 10. Find 42
the m<S, to nearest degree.
ANSWER :
Since this problem uses sides r and t, we
S need to work with <T.
?
r = 12
t = 10
105
T R

15. Homework question #1 :
Points A & B are on one side of a river, 100'
apart, with C on the opposite side. The angles
A and B measure 70º and 60º respectively.
What is the distance from point A to point C, to ANSWER :
nearest foot?
Since this problem uses side c,
we need to find m<C = 50º.
C
x
70 60
A 100 B

16. Homework question #2 :
42
In ∆ABC, m<A = 50º, m<B = 35º, and a = 12. Find the missing sides and
angle. (nearest tenth, nearest degree)
ANSWER :
C m<C = 180 - (50 + 35) = 95º.
Side c: Side b:
b a = 12
50 35
A c B

17. Homework question #3 :
In an isosceles triangle, the vertex angle is 30º 42
and the base measures 12 cm. Find the
perimeter of the triangle to the nearest integer.
ANSWER :
B
30
Find the base angles:
2x + 30 = 180
2x = 150
x = 75.
each base angle = 75
degrees
Perimeter:
A 12 C

18. Homework question #4 :
42
A triangle ABC has angle A = 106 o, angle
B = 31 o and side a = 10 cm. Solve the
ANSWER :
triangle ABC by finding angle C and sides
b and c.
1.Use the fact that the sum of all three angles of a
triangle is equal to 180 o to write an equation in C.
(round answers to 1 decimal place).
A + B + C = 180 o
2. Solve for C. 4. Solve for b.
C = 180 o - (A + B) = 43 o b = a sin (B) / sin(A) = (approximately) 5.4 cm
3. Use sine law to write 5. Use the sine law to write an equation in c.
an equation in b. a / sin(A) = c / sin(C)
a / sin(A) = b / sin(B)
6. Solve for c.
c = a sin (C) / sin(A) = (approximately) 7.1 cm

19. Homework question #5 :
42
A triangle ABC has side a = 12 cm, side b
= 19 cm and angle A = 80 o (angle A is ANSWER :
opposite side a). Find side c and angles B
and C if possible. 1. Use sine law to write an equation in sin(B).
(round answers to 1 decimal place). a / sin(A) = b / sin(B)
2. Solve for sin(B).
sin (B) = (b / a) sin(A) = (19/12) sin(80) = (approximately) 1.6
3. No real angle B satisfies the equation
sin (B) = 1.6
4. The given problem has no solution.

20. Submitted by :
42
John Marion G. De Guzman
2nd year BTTE AUTOMOTIVE
TRIGONOMETRY 121