1. Right Triangle Trigonometry
Trigonometry is based upon ratios of the sides of
right triangles.
The six trigonometric functions of a right triangle,
with an acute angle , are defined by ratios of two sides
of the triangle.
θ
opphyp
adj
The sides of the right triangle are:
the side opposite the acute angle ,
the side adjacent to the acute angle ,
and the hypotenuse of the right triangle.
2. A
A
The hypotenuse is the longest side and is always
opposite the right angle.
The opposite and adjacent sides refer to another angle,
other than the 90o
.
Right Triangle Trigonometry
3. S O H C A H T O A
The trigonometric functions are:
sine, cosine, tangent, cotangent, secant, and cosecant.
opp
adj
hyp
θ
sin = cos = tan =
csc = sec = cot =
opp
hyp
adj
hyp
hyp
adj
adj
opp
opp
adj
hyp
opp
Trigonometric Ratios
4. Finding the ratios
The simplest form of question is finding the decimal
value of the ratio of a given angle.
Find using calculator:
1) sin 30 =
sin 30 =
2) cos 23 =
3) tan 78 =
4) tan 27 =
5) sin 68 =
5. Using ratios to find angles
It can also be used in reverse, finding an angle from a ratio.
To do this we use the sin-1
, cos-1
and tan-1
function keys.
Example:
1. sin x = 0.1115 find angle x.
x = sin-1
(0.1115)
x = 6.4o
2. cos x = 0.8988 find angle x
x = cos-1
(0.8988)
x = 26o
sin-1
0.1115 =
2nd sin
( )
cos-1
0.8988 =
2nd cos
( )
6. Calculate the trigonometric functions for ∠θ .
The six trig ratios are 4
3
5
θsin =
5
4
tan =
3
4
sec =
3
5
cos =
5
3
cot =
4
3
csc =
4
5
cos α =
5
4
sin α =
5
3
cot α =
3
4
tan α =
4
3
csc α =
3
5
sec α =
4
5
What is the
relationship of
α and θ?
They are
complementary
(α = 90 – θ)
Calculate the trigonometric functions for ∠α.
α
7. Finding an angle from a triangle
To find a missing angle from a right-angled triangle we
need to know two of the sides of the triangle.
We can then choose the appropriate ratio, sin, cos or tan
and use the calculator to identify the angle from the
decimal value of the ratio.
Find angle C
a) Identify/label the names of
the sides.
b) Choose the ratio that
contains BOTH of the
letters.
14 cm
6 cm
C
1.
8. C = cos-1
(0.4286)
C = 64.6o
14 cm
6 cm
C
1.
h
a
We have been given the
adjacent and hypotenuse so
we use COSINE:
Cos A = hypotenuse
adjacent
Cos A =
h
a
Cos C =
14
6
Cos C = 0.4286
9. Find angle x2.
8 cm
3 cm
x
a
o
Given adj and opp
need to use tan:
Tan A = adjacent
opposite
x = tan-1
(2.6667)
x = 69.4o
Tan A =
a
o
Tan x =
3
8
Tan x = 2.6667
10. 3.
12 cm
10 cm
y
Given opp and hyp
need to use sin:
Sin A = hypotenuse
opposite
x = sin-1
(0.8333)
x = 56.4o
sin A =
h
o
sin x =
12
10
sin x = 0.8333
11. Cos 30 x 7 = k
6.1 cm = k
7 cm
k
30o
4. We have been given
the adj and hyp so we
use COSINE:
Cos A =
hypotenuse
adjacent
Cos A =
h
a
Cos 30 =
7
k
Finding a side from a triangle
12. Tan 50 x 4 = r
4.8 cm = r
4 cm
r
50o
5.
Tan A =
a
o
Tan 50 =
4
r
We have been given the opp
and adj so we use TAN:
Tan A =
13. Sin 25 x 12 = k
5.1 cm = k
12 cmk
25o
6.
sin A =
h
o
sin 25 =
12
k
We have been given the opp
and hyp so we use SINE:
Sin A =
14. x =
x
5 cm
30o
1. Cos A =
h
a
Cos 30 =
x
5
30cos
5
x = 5.8 cm
4 cm
r
50o
2.
Tan 50 x 4 = r
4.8 cm = r
Tan A =
a
o
Tan 50 =
4
r
3.
12 cm
10 cm
y
y = sin-1
(0.8333)
y = 56.4o
sin A =
h
o
sin y =
12
10
sin y = 0.8333
15. Example: Given sec θ = 4, find the values of the
other five trigonometric functions of θ .
Solution:
Use the Pythagorean Theorem to solve
for the third side of the triangle.
tan θ = = cot θ =
1
15
15
1
15
sin θ = csc θ = =
4
15
15
4
θsin
1
cos θ = sec θ = = 4
4
1
θcos
1
15
θ
4
1
Draw a right triangle with an angle θ such
that 4 = sec θ = = .
adj
hyp
1
4
16. A surveyor is standing 115 feet from the base of the
Washington Monument. The surveyor measures the angle
of elevation to the top of the monument as 78.3°.
How tall is the Washington Monument?
Applications Involving Right Triangles
Solution:
where x = 115 and y is the height of the monument. So, the height of the
Washington Monument is
y = x tan 78.3°
≈ 115(4.82882) ≈ 555 feet.
17. Fundamental Trigonometric Identities
Co function Identities
sin θ = cos(90− θ ) cos θ = sin(90− θ )
sin θ = cos (π/2− θ ) cos θ = sin (π/2− θ )
tan θ = cot(90− θ ) cot θ = tan(90− θ )
tan θ = cot (π/2− θ ) cot θ = tan (π/2− θ )
sec θ = csc(90− θ ) csc θ = sec(90− θ )
sec θ = csc (π/2− θ ) csc θ = sec (π/2− θ )
Reciprocal Identities
sin θ = 1/csc θ cos θ = 1/sec θ tan θ = 1
cot θ = 1/tan θ sec θ = 1/cos θ csc θ = 1/sin θ
Quotient Identities
tan θ = sin θ /cos θ cot θ = cos θ /sin θ
Pythagorean Identities
sin2
θ + cos2
θ = 1 tan2
θ + 1 = sec2
θ cot2
θ + 1 = csc2
θ