DATE: ______________________________
MHF 4U1 – Trigonometric Ratios for Obtuse Angles
Previously, we defined trigonometric ratios of angles in acute right angle triangles (SOH CAH TOA). This
definition can be extended to obtuse angles.
The trigonometric ratios of an acute angle,  , can be defined in terms of the coordinates x, y  of a
point P on a grid, where  is the measure of the angle between the positive x  axis and the line
segment OP . The length of OP is r .
Therefore,
sin   Y
csc  My
cos  
sec 
r
I
X
f
r
Y
X
70
X
tan 
cot  
I
Since Px, y  is in the first quadrant for 0   

2
, where x and y are positive, then all trigonometric
ratios of acute angles are positive.
The above definitions can be extended to obtuse angles. If

2
    , then Px, y  is in the second
quadrant, where x is negative. Hence, any trigonometric ratio of an obtuse angle whose ratio is
described using x will be negative.
Therefore,
sin   Y
csc 
cos  
sec 
j
-
X
m
r
Y
m
Y
-
X
~
tan  Y
-
X
cot  
-
X
xy
CAST RULE:
Trigonometric functions take on a positive or negative sign depending on which quadrant angle 
terminates. Positive sign for trigonometric functions are summarized in the CAST Rule:
Quadrant I
y
r
x
cos  
r
y
tan 
x
sin  
Quadrant II
Quadrant III
+
I
C:
cos  is positive
A:
all ratios are positive
S:
sin  is positive
T:
tan is positive
-
=
2
3
+
S
T
A
C
b) cos
Q II
5
4
c) tan
QIII
5
PIV
3
negative
negative
positive
F
-
ex: 1. Use the CAST Rule to state the sign of each of the following.
a) sin
Quadrant IV
meow
abdefghijkimnop
ex: 2. Determine the primary and secondary ratios for angle  in standard position determined by the
point P6,5 .
sing
g
X
-
S
Y
a
·
(b -5)
,
Sc
=
cost
=
seef-
tand
=
cot o
=
< 90
·
(E)
180 %)
>
ex: 3. If  is an obtuse angle with cot   
3
, find sin  and cos  .
4
sing
=
cost
=
-
&
4
20
-
3
ex: 4. Determine the value(s) of  for y  sin  when y  0.3 given 0    2 . Round your answers
to 2 decimal place.
-B
J
+
B
A
Sing
-0 3
=
sinB
T
51
- =
+
25
25
-
be
B
Let
.
0 305
.
the
R A A.
.
.
0 3
=
.
0 305
B =
.
0 305
=
.
B
=
-
5 98
3 45
.
.
,
C
ex: 5. Determine the value(s) of  for y  tan when y  1.45 given 0    2 . Round your
answers to 2 decimal place.
tant =
-1 45
B be
Let
.
tanB
~
S
~
St
E
T
=
5
25
C
0 =
-
0 97
.
-
2 1
.
B
=
=
the
RA A
.
1 45
.
0 97
.
0 97
.
,
5 32
.
ex: 6. Determine the value(s) of  for y  sec when y  1.456 given 0    2 . Round your answers
to 2 decimal place.
sect
=
1 450
.
ex: 7. Determine the value(s) of  for y  cos  when y  0.75 given 0    4 . Round your answers
to 2 decimal place.
ex: 8. Determine the value(s) of  for y  csc when y  3.125 given  3    3 . Round your
answers to 2 decimal place.