This document discusses trigonometric ratios and solving trigonometric equations. It provides the definitions of sine, cosine, and tangent ratios relating the opposite, adjacent, and hypotenuse sides of a right triangle. An example problem is worked through to solve for the unknown side of a triangle using the sine ratio, substituting known values and rearranging the equation by multiplying both sides by the variable. A second example is presented using the tangent ratio to solve another trigonometric equation for the unknown side.

1. i
Th
s
o
pp
so
i
e
th
ite
s
.
le
ng
a
This is opposite the right-angle
θ
There are three ratios that you need to
learn:
opp
adj
sin θ =
cos θ =
hyp
hyp
opp
tan θ =
adj
This is next to the angle
Where are the hypotenuse, adjacent and opposite lengths.

2. You need to label all the lengths of the
triangle.
The adjacent is not needed in this
question; the question is only using
opposite and hypotenuse.
x
10
cm
30o

3. You need to label all the lengths of the
triangle.
The adjacent is not needed in this
question; the question is only using
opposite and hypotenuse.
x
10
cm
30o

4. sin θ =
opp
adj
cos θ =
hyp
hyp
tan θ =
We need to use the sine ratio and substitute the
values/variables that we know.
sin θ =
opp
hyp
x
sin 30 =
10
opp
adj
x
10
cm
30o

5. We need to solve the equation (*).
opp
sin θ =
hyp
x
( *) sin 30 =
10
We can think of this as:
OR
10
cm
x
30o
x → ÷10 → sin 30
5 ¬ ×10 ¬ sin 30
rearrange the equation (*) by
multiplying both sides by x.
x
10 × sin 30 = × 10
10
⇒
10 × sin 30 = x
5= x

6. These two triangles are very different.
⇒
The variable x will be on the denominator
This is an example of where we have to solve
an equation with the variable on the
denominator, so multiply both sides by x.
10
x × sin 30 = × x
x



opp
sin θ =
hyp
10
sin 30 =
x
still need to get x on its own, so
x × sin 30 = 10
divide both sides by sin30
10
10
x=
=
= 20
sin 30 0.5

7. 15cm
opp
tan θ =
adj
x
30o
15
x × tan 30 = × x
x
x × tan 30 = 15
tan 30 tan 30

8. 15cm
opp
tan θ =
adj
x
30o
15
tan 30 =
x
15
⇒ x × tan 30 = × x
x
⇒ x × tan 30 = 15
15
x=
⇒
tan 30