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Trigonometry of Right Triangles
Right Triangles
Tamara Kucherenko
Tamara Kucherenko
Right Triangles
Trigonometry of Right Triangles
Trigonometric Ratios
We define sine, cosine, tangent, cosecant, secant, cotangent as
hypotenuse
opposite
θ
adjacent
The Trigonometric Ratios
sin θ =
opposite
hypotenuse
cos θ =
adjacent
hypotenuse
tan θ =
opposite
adjacent
csc θ =
hypotenuse
opposite
sec θ =
hypotenuse
adjacent
cot θ =
adjacent
opposite
The trigonometric ratios depend only on the angle θ and not on the size of the
triangle. The reason is that any two right triangles with angle θ are similar and the
ratios of their corresponding sides are the same.
Tamara Kucherenko
Right Triangles
Trigonometry of Right Triangles
Trigonometric Ratios
Example 1. Find the six trigonometric ratios of the angle θ.
√
13
2
θ
3
Tamara Kucherenko
Right Triangles
Trigonometry of Right Triangles
Trigonometric Ratios
Example 1. Find the six trigonometric ratios of the angle θ.
2
3
2
Solution. sin θ = √
cos θ = √
tan θ =
3
13
13
√
√
13
13
3
csc θ =
sec θ =
cot θ =
2
3
2
Tamara Kucherenko
Right Triangles
√
13
2
θ
3
Trigonometry of Right Triangles
Trigonometric Ratios
Example 1. Find the six trigonometric ratios of the angle θ.
2
3
2
Solution. sin θ = √
cos θ = √
tan θ =
3
13
13
√
√
13
13
3
csc θ =
sec θ =
cot θ =
2
3
2
Example 2. If sin θ = 47 , sketch a right triangle with acute
angle θ.
Tamara Kucherenko
Right Triangles
√
13
2
θ
3
Trigonometry of Right Triangles
Trigonometric Ratios
Example 1. Find the six trigonometric ratios of the angle θ.
2
3
2
Solution. sin θ = √
cos θ = √
tan θ =
3
13
13
√
√
13
13
3
csc θ =
sec θ =
cot θ =
2
3
2
√
2
θ
3
Example 2. If sin θ = 47 , sketch a right triangle with acute
angle θ.
Solution. Since sin θ is dened as the ratio of the opposite side
to the hypotenuse, we sketch a triangle with hypotenuse of
length 7 and a side of length 4 opposite to θ.
Tamara Kucherenko
Right Triangles
13
7
4
θ
Trigonometry of Right Triangles
Trigonometric Ratios
Example 1. Find the six trigonometric ratios of the angle θ.
2
3
2
Solution. sin θ = √
cos θ = √
tan θ =
3
13
13
√
√
13
13
3
csc θ =
sec θ =
cot θ =
2
3
2
√
2
θ
3
Example 2. If sin θ = 47 , sketch a right triangle with acute
angle θ.
Solution. Since sin θ is dened as the ratio of the opposite side
to the hypotenuse, we sketch a triangle with hypotenuse of
length 7 and a side of length 4 opposite to θ. To find the
adjacent side we use the Pythagorean Theorem:
√
(adjacent)2 = 72 − 42 = 33, so adjacent = 33.
Tamara Kucherenko
Right Triangles
13
7
4
θ
√
33
Trigonometry of Right Triangles
Special Triangles
We can use the special triangles below to calculate the trigonometric ratios for angles
with measures 30◦ , 45◦ , and 60◦ .
√
2
1
2
30◦
√
45◦
1
Tamara Kucherenko
60◦
1
Right Triangles
3
1
Trigonometry of Right Triangles
Special Triangles
We can use the special triangles below to calculate the trigonometric ratios for angles
with measures 30◦ , 45◦ , and 60◦ .
√
2
30◦
2
1
√
45◦
60◦
1
1
3
1
θ in degrees
θ in radians
sin θ
cos θ
tan θ
csc θ
sec θ
30◦
π
6
π
4
π
3
1
√2
2
√2
3
2
3
√2
3
2
1
2
3
3
2
√
2
√
45◦
60◦
√
√
1
√
3
√
2 3
3
√
2 3
3
2
2
cot θ
√
3
1
√
3
3
You do not need to, and should not, memorize this table. Rather, memorize the two
red triangles and be able to quickly calculate all trig functions of 30, 45, or 60 degrees.
Tamara Kucherenko
Right Triangles
Trigonometry of Right Triangles
Applications of Trigonometry of Right Triangles
To solve a triangle means to determine all of its parts from the information known
about the triangle, that is, to determine the lengths of the three sides and the
measures of the three angles.
Example. Solve the right triangle
B
5
A
π
3
C
Tamara Kucherenko
Right Triangles
Trigonometry of Right Triangles
Applications of Trigonometry of Right Triangles
To solve a triangle means to determine all of its parts from the information known
about the triangle, that is, to determine the lengths of the three sides and the
measures of the three angles.
Example. Solve the right triangle
B
5
A
π
3
π π
π
− = .
2
3
6
To find AC, we look for an equation that relates
AC to the lengths and angles we already know. In
AC
this case, we have sin π3 = AB
= AC
5 . Thus,
Solution. First, ∠A =
√
π
5 3
AC = 5 sin =
.
3
2
C
Similarly, BC = 5 cos
Tamara Kucherenko
Right Triangles
π
5
= .
3
2