Law of sine and cosines

Right Triangle Problems, Law of Sines, Law of Cosines
& Problem Solving
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Right Triangle Problems
Any situation that includes a
right triangle, becomes solvable
with trigonometry.
Angle of Elevation
Angle of Depression
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This diagram has an inscribed triangle whose
hypotenuse is also the diameter of the circle.A
B C
Given: = 40o, AB = 6 cm
Determine the area of the shaded
region.
Solution: Sin 40o = 6/AC; AC = 6/ sin40o = 9.33 cm
AC/2 = radius of the circle = 4.67 cm; BC = 6/tan 40o =7.15 cm
Ao = r2 = 68.4 cm2,
A = ½bh = 21.5 cm2;
A = 46.9 cm2

This diagram has a right triangle within a cone
that can be used to solve for the surface area
and volume of the cone.h
r
l
Given: = 55o, l = 6 cm
Determine the area and volume of
the cone.
Solution: Sin 55o = h/l; h = 6 x sin55o = 4.91 cm
tan 55o = h/r; r = h/ tan 55o = 4.91/ tan 55o = 3.44 cm
Acone = r(r + l) = (3.44)(3.44 + 6)cm2; Acone = 102 cm2
Vcone =⅓ r2h = ⅓ (3.44)2(4.91)cm3; Vcone = 60.8 cm3

The angle of elevation of a ship
at sea level to a neighboring
lighthouse is 2o . The captain
knows that the top of
lighthouse is 165 ft above sea
level. How far is the boat from
the lighthouse?
h
x
Solution: tan 2o = h/x;
1 mile = 5280 ft, so the ship is .89 mile away from the lighthouse.
x = h/tan 2o
x = 165/ tan 2o
= 4725 ft;

Function Values – Unit Circle
The unit circle is a circle with a radius of 1 unit.
x
y
Sin = y; Cos = x; Tan = y/x
 The coordinate of P will lead to the value of x and y which in
turn leads to the values for sine, cosine, and tangent.
 Use the reference angle in each quadrant and the coordinates
to solve for the function value.

The reference angle is always measured
in its quadrant from the x – axis.
 Quadrant I – Angle =
 Quadrant II – Angle = (180 - )
 Quadrant III – Angle = (180 + )
 Quadrant IV – Angle = (360 - )
 Quadrant I – P (x, y)
 Quadrant II – P (-x, y)
 Quadrant III – P (-x, -y)
 Quadrant IV – P (x, -y)
With the values of changing from (+) to (-) in each quadrant, and
with the functions of sine, cosine, and tangent valued with ratios
of x and y, the functions will also have the sign values of the
variables in the quadrants.

Degrees 0 30 45 60 90 120 135 150
Sine 0 ½ 2/2 3/2 1 3/2 2/2 ½
Degrees 180 210 225 240 270 300 315 330
Sine 0 -½ - 2/2 - 3/2 -1 - 3/2 - 2/2 -½
Cosine 1 3/2 2/2 ½ 0 -½ - 2/2 - 3/2
Tangent 0 3/3 1 3 - - 3 -1 - 3/3
Cosine -1 - 3/2 - 2/2 -½ 0 ½ 2/2 3/2
Tangent 0 3/3 1 3 - - 3 -1 - 3/3

Example: Cos 240 = _______
60
P (-½, - 3/2)
- ½
Example: Sin 240 = _______
Example: Tan 240 = _______ 3
- 3/2
240

Law of Sines
Law of Sines – For a triangle with angles A, B, C and sides of
lengths of a, b, c the ratio of the sine of each angle and its
opposite side will be equal. Sin A = Sin B = Sin C
a b c
A
B
C
c
b
a
h
Proof:
Sin A = h/b; Sin B = h/a
h = b Sin A, h = a Sin B
b Sin A = a Sin B; Sin A = Sin B
a b

Proof, continued:
A
B
C
c
b
a
kk is the height of the same
triangle from vertex A
Sin C = k/b and Sin B = k/c
k = b Sin C and k = c Sin B
b Sin C = c Sin B;
Sin B = Sin C
b b c
Conclusion: Sin A = Sin B = Sin C
a b c

Law of Sines
Let’s take a closer look:
Suppose A < 90o
A
c
a
If a = c Sin A, one
solution exists
a
A
c
c Sin A
If a < c Sin A, no
solution exists
Suppose A < 90o
A
ac
If a > c Sin A and a
c, one solution
exists
c Sin A
a
A
c
c Sin A
If c Sin A < a < c ,
two solution exists
a

Suppose A 90o
A
a
c
If a < c, no solution exists
A
a
If a > c , one
solution exists
c

Law of Cosines
A
B CD
b
a
c
a -x x
h
b2 = h2 + x2; h2 = b2 - x2
Cos C = x/b ; x = b Cos C
c2 = h2 + (a-x)2; c2 = h2 + a2 –2ax + x2
Substitute and we get:
c2 = (b2 - x2)+ a2 –2a(bCos C) + x2
c2 = b2 + a2 – 2abCos C
For any triangle given two sides
and an included angle,
a2 = b2 + c2 – 2bcCos A
b2 = a2 + c2 – 2acCos B
c2 = b2 + a2 – 2abCos C
PROOF:

Law of Cosines - example
Given: A = 50o; AB = 8 cm, AC = 14 cm
x 2 = 8 2+ 14 2 – (8)(14)Cos50
Find x.
A
B
C
x
x 2 = 260 – 72; x 2 = 188
x = 13.7 cm

Law of Cosines – Hero’s Formula
For any triangle, the area of the triangle can be determined with
Hero’s Formula.
s = ½(a + b + c) where s = semi-perimeter of the triangle.
A = s(s – a)(s – b)(s – c)
Given: ABC; a = 7, b = 24, c = 25
A = 28(28 – 7)(28 – 24)(28 – 25) = 84 cm 2
7cm
25 cm
24 cm
S = ½ (7 + 24 + 25) = 28
Check: A = ½(b)(h) = ½(7)(24) = 84 cm 2
Example

The End
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Law of sine and cosines

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