Download Angles in Standard Position

Pre-Calculus 11
2.1 – Angles in Standard Position
Let’s begin with an x-y coordinate plane:
Draw the following angles, starting from the positive x-axis:
a)
  45
b)
  180
c)
  90
d)
  210
Angles in Standard Position
I
II
P(x, y)
r
y
 is the angle that
starts at the x – axis
and ends at the line
OP(terminal arm)

O
III
STANDARD POSITION :
Circle centred at (0, 0)
x
IV
+  counter clockwise
-  clockwise
Reference Angle: Angle made between the terminal arm and the nearest x-axis.
Ex.
a) Show   150 in standard position.
b) Show the reference angle (  R ).
Ex. a) Show   60 in standard position.
b) Show the reference angle (  R ).
Ex. List all the angles in standard position that have  R  20 , 0    360 .
Triangles in Standard Position
First, let’s review our TRIG BASICS:
SOH
Now, let’s use  R  60 in each quadrant to set up triangles:
CAH
TOA
Ex. Determine the angle in standard position when an angle of 40 is reflected:
a) in the x-axis
b) in the x-axis and then in the y-axis
Pre-Calculus 11
2.2 – Trigonometric Ratios and Special Right Triangles
Given any  in standard position:
P(x, y)
r
y

O
We know:
x2  y 2  r 2
We know:
sin  
Now:
sin  

opposite
hypotenuse
Remember, by “ASTC”:
x
Therefore:
cos  
r
adjacent
hypotenuse
tan  
cos 
tan  
sin   0 (Quadrants
&
)
cos  0 (Quadrants
&
)
tan   0 (Quadrants
&
)
opposite
adjacent
Ex.
Ex.
3
Suppose  is an angle in standard position with terminal arm in quadrant III, and cos    .
4
What are the exact values of sin  and tan  ?
Determine the values of sin  , cos , and tan  when the terminal arm of quadrantal angle 
coincides with the positive y-axis,   90 .
SPECIAL RIGHT TRIANGLES
There are two right triangles where we know the ratios. We can use these ratios whenever we come
across a reference angle of 30 , 45 , or 60 .
45-45-90 Triangle
30-60-90 Triangle
Ex. Determine the exact value of cos135 .
A quick trig review (with your calculator): Round to 4 decimal places.
Ex. sin 40 
Ex. sin   0.75
Ex. Solve for  , where sin   0.5 , 0    360 .
Ex. Given cos  0.6753 , where 0    360 , determine the measure of  , to the nearest tenth of a
degree.
Pre-Calculus 11
2.2 (Con’d) – More on Trigonometric Ratios
Ex.
Determine the exact value of sin 240 .
Ex.
The point (6, k) is 10 units from the origin. If P lies on the terminal arm of an angle,  , in standard
position, 0    360 , determine (a) the measure(s) of  and (b) the ratios of sin  , cos , and
tan  .
Ex. If sin   
3
and tan   3 , determine the exact value of cos .
10
Ex. Does cos135  sin 225 ? Verify without using a calculator.
Pre-Calculus 11
2.3 - The Sine Law (Part I)
C
64
57
59
B
A
Which of the following sides is the longest side? Explain why.
The Sine Law relates the sides to the opposite angles:
C
A
B
sin A sin B sin C


a
b
c
Solve the following triangle – solve for all unknown sides and angles.
Ex. 1
C
85
26 cm
37
58
A
B
Ex. 2
C
7.25 cm
46
A
7.83 cm
B
Ex. 3
12
31
129
20
Ex 4. Calculate the height of the tree (to 1 decimal place) if the distance from A to B is 10 metres.
A
33
10 m
50
B
Pre-Calculus 11
2.3 - The Sine Law (Part II): “The Ambiguous Case”
Sometimes, we are given a description of a triangle, rather than a diagram.
Ex. In a triangle, a side of length 1.8 cm is contained by angles of 61 and 88 . Solve the triangle.
The above example was straightforward. We were given the measures of two angles and one side
(ASA).
THE AMBIGUOUS CASE: Given one angle, a side, and the opposite side.
Ex. Draw a triangle given  A & sides a and b, where b > a.
(Hint: Sketch the opposite side last!)
- There are 2 possible
triangles
- If a = h, then we have
only a right .
- If a < h, we have no ’s
Definition: Oblique Triangle: A triangle that does not contain a right angle.
Ex. Given a triangle in which A  30 , a  24 cm , and b  42 cm , how many possible triangles are
there? Solve for all unknown sides and angles. (p. 107 Example 3)
Number of Possible Triangles (Angle-Side-Side)
Depending on the information given, we can limit the number of
possible triangles.
I) In this case the length of the opposite side of 70 is too short and
will not make a triangle
a=4
b= 10
A=70
- When a < h  no possible triangle
II) In this case length of the opposite side of 40 can occupy 2
positions (ambiguous case)
a’ = 8
b= 10
A = 40
- When h < a < b  two possible triangles
- When a > b  only one possible triangle
III) For obtuse A: If a  b, then there is no triangle
a = 14
b= 10
A = 120
- We must have: a > b  one possible triangle
a=8
Pre-Calculus 11
2.4 – The Cosine Law
Ex. In a triangle, an angle of 17 is contained by sides of 9 cm and 14 cm.
Can we solve the triangle using the Sine Law in this case?
Explain.
The above triangle is a SAS case. It cannot be solved using the Sine Law. In which other case are we not
able to use the Sine Law?
The Cosine Law
Given SAS or SSS, we use the Cosine Law to solve the triangle:
C
a 2  b 2  c 2  2bc cos A
b 2  a 2  c 2  2ac cos B
A
B
c 2  a 2  b 2  2ab cos C
Ex.
Ex.
Let’s find the missing side length from our first example. Since it is SAS, we must use the Cosine
Law.
Solve  ABC. Round angles to the nearest degree.