Download Solving an SAA triangle

The Law of Sines
PREREQUISITE
What is a triangle?
 Closed figure consisting of three line segments
linked end-to-end.
 A 3-sided polygon.
There are three types of triangles that are common
Acute
All angles less than 90°
Right Triangle
One angle 90°.
One angle greater than 90°
Obtuse
a. Acute angles
These are angles with degree measured 0o<θ<90o
b. Right angles
These are triangles with one angle degree measured θ=90o
c. Obtuse angles
These are angles with degree measured θ>90o
Pythagorean Theorem
In any right triangle, the area of the square whose side is the hypotenuse (the
side opposite the right angle) is equal to the sum of the areas of the squares
whose sides are the two legs (the two sides that meet at a right angle). Always
understand that the Pythagorean Theorem relates the areas of squares on the
sides of the right triangle.
The Pythagorean Theorem formula can be written as:
a2+b2=c2
where
a and b are the sides that are adjacent, also called the legs to the right angle.
c is the hypotenuse, side that is opposite of the right angle
Using the
Pythagorean Theorem
Let’s try an example.
If ABC is a right triangle, with a being 3 units long,
and b being 4 units long, what is the length of c?
Step 1
When I solve a problem like this, first I like to write down the
formula that I am to use:
a2 +b2 =c2
Step 2
Then I write the information that are given
a=3
b=4
c=?
Step 3
Then I “plug them in” to the formula:
32+42=c2
Step 4
Simplifying the expression, we now have
9 + 16 = c2
25 = c2
Step 5
Remember to take the square root of both sides!
5= c
So the length of the hypotenuse c is 5 units.
Trigonometric Functions
The trigonometric functions are functions of an angle.
They are used to relate the angles of a triangle
to the lengths of the sides of a triangle.
Trigonometric functions are important
in the study of triangles and modeling
periodic phenomena, among many
other applications
Facts you need to remember
The hypotenuse is the side opposite the right angle. The
hypotenuse is always the longest side of a right-angled
triangle. The opposite side is the side opposite to the
angle we are interested in. The adjacent side is the side
that is in contact with both the angle we are interested
in and the right angle.
The most familiar trigonometric functions are the sine, cosine, and tangent.
The sine of an angle is the ratio of the length of the opposite side to the length
of the hypotenuse.
The six trigonometric functions can also be defined in terms of the unit circle,
the circle of radius one centered at the origin. It relies on right triangles for
most angles. From the Pythagorean Theorem the equation for the unit circle is:
x2 +y2 = 1
(cos, sin)
r
Y
The sine function takes an angle and tells the length of the y-component (rise)
of that triangle
Degrees and radians
In everyone's experience it is usual to measure angles in degrees but for
mathematicians it is usual to measure angles in radians.
The size of a radian is determined by the requirement that there are 2∏ radians
in a circle. Thus 2∏ radians = 360o. This means that:
and
The reason for this is that so many formulas become much easier to write and
to understand when radians are used to measure angles.
A radian is a central angle, for which lengths of its arc and radius are equal
AB = r
A
r
B
So, a radian measure of any angle is a ratio of a length of an arc drawn by an
arbitrary radius and concluded between sides of this angle to the arc radius.
Converting radians to degrees:
Converting degrees to radians:
Sign of Trigonometric Function
Figuring out the sine for different quadrants is pretty basic.
That's because sine corresponds to the y-axis. So:
if y is positive, sine is positive
and
if y is negative, sine is negative
Y>0
sin +
Y>0
sin +
Y<0
sin -
Y<0
sin -
the graph of sine function
sin +
sin -
Using the Calculator
Often not knowing how to use your calculator
is the hardest thing about trig problems.
Our example is based on TI- 84.
You can use any update or old version of this calculator. This is a special
calculator that helps you with mathematical problems.
So, how to solve trigonometric function with the TI- 84?
First make sure it is in degree mode, by pressing the mode button, and then
press the arrows and highlight- Degree.
Now you are ready to use your calculator
1. Press the sin button, when it shows on your screen
2. Enter the value that is giving in the problem,
3. Press enter and your answer will show up.
How to remember?
SOH CAH TOA
SOH
This is short for Sine equals Opposite side over Hypotenuse.
CAH
This is short for Cosine equals Adjacent side over Hypotenuse.
TOA
This is short for the Tangent equals the Opposite side over the Adjacent side
ALL STUDENTS TAKE CALCULUS
In Quadrant I
A for All
all trigonometric functions are positive in this quadrant.
In Quadrant II
S for Sine
sine functions are positive in this quadrant.
In Quadrant III
T for Tangent
tangent functions are positive in this quadrant.
In Quadrant IV
C for Cosine
cosine functions are positive in this quadrant
The Law of Sines
Oblique Triangle
A triangle that does not contain a right angle is called an oblique triangle. You can
make an oblique triangle two ways.
By 3 acute angles
Or by 1 obtuse angle and 2 acute angles
To label angles and sides of a triangle we use letters.
We use A, B and C to label angles of the triangle
and a, b and c to label the sides opposite to each angle
The Law of Sines
If A, B and C are the measures of the angles of a triangle, and a, b and c are the lengths
of the sides opposite these angles, then
The ratio of the length of the side of any triangle to the sine of the angle
opposite that side is the same for all three sides of the triangle. The Law of Sines is the
relationship between sides and angles in any triangle.
We can prove the Law of Sines
B
c
A
h
D
a
C
b
First we draw an altitude from one of the vertices of the triangle to the opposite
length to form two similar, right triangles. So draw an altitude BD from the vertex B to
the length AC to form ABD and BCD triangles. Since these two triangles are right
triangles, we can use the definition of the sine of an angle which is
So,
h= c sinA
solve each equation for h.
h = a sinC
Since we have two expressions for h, we can set these two equations equal to each
other.
c sinA = a sinC
Divide both sides by sinA sinC.
Now when we proved one part of the Law of Sines, let’s prove the other part of it.
Consider different altitude from a different vertex.
Draw an altitude h from the vertex A to the length BC to form ABE and ACE right and
similar triangles.
B
D
c
A
h
b
a
C
Draw an altitude h2 from the vertex A to the length BC to form ABE and ACE right and
similar triangles.
Now
h = c sinB
Solve for h
and
h = b sinC
Since we have two expressions for h, we can set these two equations equal to each
other
c sinB = b sinC
Divide both side by sinB sinC
When do I use it?
You can use Law of Sines if you already know
1. One side and it's opposite angle ( SAA and ASA triangles)
2. Two sides and one non-included angle ( SSA triangle)
To use law of Sines, you need to know the measure of at least one angle in the triangle.
When don’t I use it?
You cannot use law of sines when you know only all three side of the triangle
and when you know two sides and one included angle (SAS triangle)
Before you start using the Law of Sines:
Facts you need to remember:
1. In a triangle, the sum of the interior angles is 180º.
2. No triangles can have two obtuse angles.
3. The sine function has a range of
4. If the sin
the
= positive decimal < 1,
can lie in the first quadrant (acute angle)
or in the second quadrant (obtuse angle).
Solving oblique triangles
Finding the length of the sides and the measurements of angles of an oblique triangle
is called solving an oblique triangle. We consider three situations to solve oblique
triangles. In each case there are three known measurements and we solve for other
three unknown measurements.
1. Solve SAA triangle ( side, angle, angle)
2. Solve ASA (Angle ,side, angle)
3. SSA case (Ambiguous case)
Solving an SAA triangle
Let’s Solve the triangle where
a= 100 inches,
A= 48o
B= 37o
C
a= 100
A=48o
B=37
o
We know two angles of the triangle. Since we know the fact that the sum of the
measures of interior angles of a triangle is 180, we can easily find the other angle.
A+B+C = 180o
Substitute the given values A= 48, B= 37
48o+ 37o+ C= 180o
85o+ C = 180o
C = 95o
Use Law of Sines to find missing sides
Use the Law of Sines again to find b.
The solution is C=95, a ≈ 81.03 inches, and c ≈ 134.05 inches.
Solving the ASA triangle
Solve the triangle where A= 85,
B= 35 and
B=35o
c =50
A=85o
c= 50 inches.
First find the unknown third triangle C
A+B+C = 180o
85o+ 35o+ C= 180o
120o+ C = 180o
C = 60o
Use Law of Sines to find a
Use the Law of Sines again to find b.
The solution is C=60, a ≈ 28.88 inches, and b ≈ 70.24 inches.
The ambiguous case SSA
By definition, the word ambiguous means open to two or more interpretations.
In math, this case is called ambiguous case because the given information may
result in one triangle, two triangles, or no triangle at all.
a > h and a > b
one triangle exists.
a=h
one right angle triangle exists.
a > h and a < b
two triangles exists.
a < h and a < b
no such triangle exists.
Solving the SSA triangle when there
is one solution
Let’s solve the triangle where a > h and a > b
In the triangle ABC
A= 45o
Begin with the sketch of the given triangle
a= 20
b= 40
C
b=15
a = 20
A=40o
B
Apply the Law of Sines to find the missing angle B
=
=
There are two angles possible:
B=29o and B1=180o-29o= 151o
but since 40o+151o= 191o, B1 is impossible
Now find the missing angle C using A and B
C=180o-(A+B)
C= 180o-(40o+29o) = 180o- 69o=111o
Use the Law of Sines to find side c
There is one triangle and the solution is B=29o, C=111o, and c=29
Let’s solve the triangle where a = h
In the triangle ABC
A= 62
a= 30
b= 34
Begin with the sketch of the given triangle
B
a=30
A=62o
b=34
C
Apply the Law of Sines to find the missing angle B
There are two angles possible
B=90o and B1=180o-90o=90o since B=B1 we can disregard B1
Now find the missing angle C using A and B
C= 180o-(A+B)
C=180o-(62o+90o) =180o-152o=28o
Use the Law of Sines to find side c
There is one triangle and the solution is B=90o, C=28o, and c=16
Solving the SSA triangle when there is no solution a < h and a < b
In the triangle ABC
A=30o
a=10
b=40
Begin with the sketch of the given triangle
C
b=40
a=10
A=30o
B
Apply the Law of Sines to find the missing angle B
Because the sine can never exceed 1, there is no angle B for which sinB=2.
Solving triangle with given measurements is impossible. There is no such
triangle.
We can guess that the picture of this non existing triangle may look like this
C
b=40
A= 30o
a=10
B
Solving the SSA triangle when there are two solutions
The so-called ambiguous case arises from the fact that an acute angle and an obtuse
angle have the same sine.
For example, when sinx = , we would have x = 45° or x = 135°.
In the following example, we will see how this ambiguity could arise
In the triangle ABC
A= 37o
a=12
b=16.1
Begin with the sketch of the given triangle
C
b=16.1
a=12
A=37o
B
Apply the Law of Sines to find the missing angle B
There are two angles possible:
B1= 54o and B2= 180o-54o= 126o
Finding C will help you make sure that indeed there are two triangles
Now find the missing angle C using A and B
C= 180o-(A+B1)
C1=180o-(37o+54o) =180o-91o=89o
C= 180o-(A+B2)
C2=180o-(37o+126o) =180o-163o=17o
Use the Law of Sines to find side c1 and c2
The solution is
The solution is
B1=54o, C1=89o and c1=19.9
B2=126o, C2=17o and c2=5.8
You can draw a picture of the solutions
C
C2=17o
128o
A
C2=5.8
C1=89o
54o
B2
B1
C1=19