6.1 Law of Sines

Math 150: 6.1 The Law of Sines and Area
The following formulas are valid for all (right and non-right) triangles.
Area of any Triangle: Draw and label a non-right triangle with sides of a, b and c and
angles of A, B and C. The area of any triangle (and rearrangements) is:
Area =
1
2
bc sin A
or
Area =
1
2
ab sin C
Area =
or
1
2
ac sin B
Exercise 1: Show that the altitude (or height) h of a triangle of base b is c sin A , and
that the formula Area =
1
2
bc sin A reduces to the formula for the area given the
base b and height h of Area =
1
2
bh .
Law of Sines: This formula is used to determine one missing piece of information
from: 2 sides and the 2 angles that are opposite those sides. The formula (and
rearrangements) is:
sin A
a
=
sin B
b
or
sin B
b
=
sin C
c
or
sin A
a
=
sin C
c
.
In the event that two angles and one side are known (AAS or ASA), then the triangle
can be uniquely determined. Refer to Examples 1 and 2 in the textbook (p. 429). But
in the event that two sides and one angle are known (SSA), the information may
represent no solution (it is impossible to make a triangle with the given information),
1 solution (only 1 triangle can satisfy the given information) or 2 solutions (2
different triangles satisfy the given information).
© Raelene Dufresne 2013
1 of 2
Math 150: 6.1 The Law of Sines and Area
Exercise 2: For triangle PQR, the angle Q is 30o , the length q is 25 and angle R is
10o . Solve the triangle.
Exercise 3: For triangle PQR, the acute angle Q is 30o , the length q is 25. For each
different value of the length r, set up an equation to solve for R, interpret
graphically, solve for R (if possible) and sketch the triangle.
a) r = 40
b) r = 50
c) r = 60
Exercise 4: Solve the triangle given the following information, if possible.
a) Show that there is no triangle ABC for which a = 15 , b = 25 and ∠A = 85o .
b) For triangle PQR, ∠Q = 32º , ∠R = 39º and the length r is 40.
c) Solve for two triangles such that a = 12 , b = 31 and ∠A = 20.5o
© Raelene Dufresne 2013
2 of 2
Math 150: 6.2 The Law of Cosines
The following formulas are valid for all (right and non-right) triangles.
Law of Cosines: This formula is used to determine one missing piece of information
from: 3 sides and 1 angle. The formula (and rearrangements) is:
c 2 = a 2 + b 2 − 2ab ⋅ cos C
or a 2 = b 2 + c 2 − 2bc ⋅ cos A
or b 2 = a 2 + c 2 − 2ac ⋅ cos B
Exercise 1: A racing committee wants to lay out a triangular course with a 40o angle
between the two sides of 3.5 miles and 2.5 miles. What will be the perimeter of the
course? Could you use the Sine Law in this case? Explain.
Exercise 2: What famous formula does the Cosine Law simply to if one of the angles
is 90o ? Prove this fact.
© Raelene Dufresne 2011
1 of 2
Math 150: 6.2 The Law of Cosines
Heron’s Area Formula: Another formula for the area A of any triangle, called Heron’s
Formula, is given by
(
)(
)(
)
A = s ⋅ s − a ⋅ s − b ⋅ s − c , where s =
a +b +c
2
is the triangle’s half-perimeter.
Exercise 3: If the units of the side lengths of the triangle are meters, what are the
units of (a) s and (b) Area? Explain.
Exercise 4: Determine the area of a triangle with side lengths of 1,
© Raelene Dufresne 2011
1
2
and
3
4
.
2 of 2