Trigonometry Reviewer 2nd Term – PART 1
I. Solving for Right Triangles
 All triangles add up to 180°
 Not all triangles are found on the Cartesian plane
 SOHCAHTOA can be used to find the missing sides/angles
SOHCAHTOA and
Pythagorean theorem can only
be used for right triangles.
 Complementary angles: angles that add up to 90°
 Cofunctions: sin & cos, tan & cot and csc & sec
For any acute angle A,
sin A = cos(90° - A)  cos A = sin(90° - A)
tan A = cot(90° - A)  cot A = tan(90° - A)
csc A = sec(90° - A)  sec A = csc(90° - A)
*cofunction values of complementary angles are equal
* if cot A = tan B, then A + B = 90
Examples:
Write each function in terms of its cofunction (assuming that all angles in which an unknown appears as acute angles)
2. sin 25.4°
1. cot73°
sin 25.4° = cos(90° - 25.4°)
cot 73° = tan(90° - 73°)
= cos 64.6°
= tan 17°
A. Given One Side and One Angle
Ex. Given ABC is a right triangle with side lengths of a, b & c and right angle at <C. If <A = 36° and c = 10cm, find
the unknown sides and angle.
1. Use the formula 180 – (90 + A) to find the third angle.
180 – (90 + 36°) = 54 <B = 54°
2. Use SOHCAHTOA to find the missing sides.
a
opp
 sin 36 
 (sin 36°)(10) = 5.88 a = 5.88cm
sin A 
hyp
10
cos A 
therefore:
<A = 36°
<B = 54°
<C = 90°
total: 180°

a = 5.88cm
b = 8.09cm

c = 10cm
b
adj
 cos36 
 (cos36°)(10) = 8.09 b = 8.09cm
hyp
10

Remember that angles and sides are proportional.
Longest side = largest angle
If side a is the longest side, <A should be the largest angle.

B. Given 2 sides
Ex. Solve the right triangle ABC given that side c = 25 cm and side b = 24 cm.
1. Use the Pythagorean theorem to find side a.
Pythagorean theorem: a 2  b 2  c 2
Transpose so that,

a2  c 2  b2
to find <A, cos A  24
25
24  25  0.96
press shift/2nd function
a 2  25 2  24 2
a 2  625  576

a  49
2
To find <B,
180 – (90 + 16.26) = 73.74°
<B = 73.74°
a7


A = 16.26΅
& cos
II. Special Angles
A. Equilateral triangles: all sides & angles are equal
 There is no right angle.
So if there’s no right angle, how do you find the missing parts?
Split the triangle in the middle, so that you end up with 2 triangles and now there are right angles! 
Now that you have a right triangle, you can use SOHCAHTOA and the Pythagorean theorem.
to find the missing side (x),
1  x 2
2
2
use pythagorean theorem
SOHCAHTOA
2
sin 30 
x 2  2 2 12
x 2  4 1
3
cos 30 
2
1
3
3
tan 30 


3
3
3
x2  3
x 3

csc30  2
1
2
12 12 

60°


2
3
2
2
1
2

1
2
2


2
2
2
csc45 
cos 45 
1
2
2


2
2
2
sec45΅ = 2
cot
sec
3
3
3
2 3
3
2
1
1
3 
3

3
2
tan45΅ = 1
tan


3
sin 45 


cot 30 
2
tan 45  1

2
3 2 3


3
3
3

B. Isosceles Right Triangle: 2 sides (legs) are equal

table of values for special angles 30°, 45° and 60°
sin
cos

1
30°
3 
2
2
45°
2
2
sec30 
csc
2
2
2




2 3
3
III. Reference Angles
 a positive acute angle associated with every nonquadrantal angle in standard position.
 Reference angle for an angle  (written as  ’) is the positive acute angle made by the terminal side of an angle 
and the x-axis
Q1


Q2
*find the coterminal angle for angles larger than 360°

to find  ’ :
QI:  ’ = 
QII:  ’ = 180° - 
QIII:  ’ =  - 180
 QIV:  ’ = 360° - 
 


Examples:
 1. 120°, Q2  180° - 120° = 60°

 120° is the same as 60° (except for the signs)
Therefore,
Recall: QII  sin & csc are positive
Q3
3
2
3
sin120 
2
sin 60 
Q4
cos60 
1
2
1
cos120  
More Examples:
2
1. 45°, QI   ’ = 45
2. 210°, QIII   ’ = 210 – 180 = 30°
3. 315°, QIV   ’ = 360 – 315 = 45°
4. 480°, QII  480 – 360 = 120° (coterminal)
 180 – 120 = 60°

5. -510°,
QIII  -510 – 720 = 210° (coterminal)  210 – 180 = 30°

tan 60  3
tan120   3

Elevation and Depression
V. Angles of
 always acute (less than 90°)
 formed by a horizontal line & the line of sight
 tangent is usually used
Example:
An airplane took-off and flew at an angle of 18.26°
with the horizontal runway and gained a distance of
598.53 feet above the ground. What is the horizontal
distance that the airplane has traveled?

 x
 =18.26°


  598.53
tan18.26
x
x = 1,814.03 ft.
598.53 ft.