Name: _______________________
Date: __________________
The Tangent Ratio
What kind of triangle is ABC? _________________________
In ABC, what is the name of the side AB? _____________________
B
A

C
Let BAC =  (the lowercase, Greek letter theta)
BC is referred to as the opposite side to .
A
AC is referred to as the adjacent side to .
i)
Label the sides of the triangle using the first three letters of the names of
the sides.
ii)
Each of the triangles on the next page has  = 40. Explain why the triangles are
similar.
iii)
Complete the table below using the triangles on the next page.
Side Opposite 
Name
Measure
ABC
DEF
GHJ
KLM
PON
Side Adjacent to 
Name
Measure
Ratio
opp/adj
Ratio
3 decimals
A
D
F

B
C

E
G
L

H
J
P
K
N


O
M
iv)
How are the ratios in the last column related?
v)
Draw 4 similar right triangles with one angle of 30. Create and complete a table like
the one on the first page. How are the ratios in the last column related?
vi)
If the measure of one angle of a right-angled triangle is kept constant, then what can
you conclude about the following ratio:
length of side opposite angle
length of side adjacent angle
Tangent  = length of side opposite angle
length of side adjacent angle
or more commonly, tan  = opp
adj
To find the ratio of opp:adj, given the angle, use your calculator; i.e., tan 40 = 0.8391.
Important Note: Ensure your calculator is in “degree” or “deg” mode.
Determine the ratios for the following angles:
a)
tan 30 =
b)
tan 130 =
c)
d)
tan 289 =
tan 195 =
To find the angle, given the ratio, use your calculator by pressing 2nd function key before the
tan key; i.e., if tan  = 0.8391, then to determine  press 2nd tan 0.8391 = 40.
Determine the angle for the following ratios:
a)
tan  = 1.4826
b)
tan A = -0.4663
c)
d)
tan C = -1.732
tan B = 0.1763
Solving Problems Using the Tangent Ratio
Remember: tan  
opp
adj
To solve problems using the tan ratio, you must know:
i)  an angle and either the opposite or adjacent side length to find the adjacent or
opposite side, or
ii)
the opposite and adjacent side lengths to find the angle.
Example:
Maria places a ladder on level ground 3 m from a vertical wall so that the ladder makes a 70
angle with the ground. How far up the wall does the ladder reach?
Solution:
Let x represent the distance up the wall in metres.
tan 70 =
x
3
x = 3 tan 70
x
 3(2.7475)
x
x  8.24
 The ladder reaches 8.24 m up the wall.
70
3m
Problems
1.
a)
Determine the unknown quantity for each of the following right triangles (round to one
decimal place).
b)
x
x
60
12
40
17 cm
c)
4 cm

15 cm
c
m
2.
Determine the height of a tree casting a 20 m shadow at the same time of day as the
sun’s rays make an angle of 35 with the ground. Include a labeled diagram with your
solution.
3.
How tall is a flagpole if it casts a shadow 15.5 m long when the sun’s rays make a 25
angle with the ground? Include a labeled diagram with your solution.
4.
Commercial airplanes fly at about 10 km above the ground. If the landing approach is to
make a 5 angle with the ground how far from the airport must the pilot begin the
descent? Include a labeled diagram with your solution.
5.
From the top of a cliff, the angle of depression of a hut is 46. If the cliff is 500 m
high, how far is the hut from the base of the cliff?
horizontal
Angle of depression
6.
From the top of a cliff 120 m above the water, the angle of depression of a boat on the
water is 18. How far is the boat from the base of the cliff? Include a labeled diagram
with your solution.
7.
A tower 115 m high casts a shadow 24 m long. Find the angle of elevation of the sun.
Angle of elevation
horizontal
8.
A 10 story building (each story is 3 m high) casts a shadow of 55 m. What is the angle
of elevation of the sun?