This document contains a lesson on right triangle trigonometry from Lehman College's Department of Mathematics. It includes definitions of the tangent, sine and cosine ratios, examples of solving right triangles using trigonometric ratios, and a derivation of the Pythagorean identity that relates sine and cosine functions. Key concepts covered are the definitions of trigonometric ratios in right triangles, using trigonometric functions to solve for unknown side lengths, and deriving trigonometric identities from the Pythagorean theorem.

1. 𝐏𝐓𝐒 𝟑
Bridge to Calculus Workshop
Summer 2020
Lesson 12
Right Triangle
Trigonometry
“A person who never made a
mistake never tried anything new."
– Albert Einstein -

2. Lehman College, Department of Mathematics
Right Triangle Definitions (1 of 2)
Consider the right triangle given in the figure below,
where the lengths of the sides are denoted by the
positive numbers 𝑎, 𝑏 and 𝑐:
How many ratios of different side lengths can we get?

3. Lehman College, Department of Mathematics
Right Triangle Definitions (2 of 2)
The possible ratios are:
And their reciprocals:
A trigonometric ratio is a ratio of the lengths of two
different sides of a right triangle.
The tangent of an acute angle of a right triangle is the
ratio of the length of the side opposite the angle to the
length of the side adjacent to the angle
𝒂
𝒃
𝒃
𝒄
𝒂
𝒄
𝒃
𝒂
𝒄
𝒃
𝒄
𝒂
tan 𝜽 =
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐨𝐩𝐩𝐨𝐬𝐢𝐭𝐞 𝐚𝐧𝐠𝐥𝐞 𝜽
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐚𝐝𝐣𝐚𝐜𝐞𝐧𝐭 𝐭𝐨 𝐚𝐧𝐠𝐥𝐞 𝜽
=
𝒂
𝒃
(i) (ii) (iii)
(iv) (v) (vi)

4. Lehman College, Department of Mathematics
The Tangent Ratio (1 of 2)
Example 1. Answer the following questions based on
right triangle 𝐴𝐵𝐶 below:
a) Identify the leg opposite ∠ 𝐴.
b) Identify the leg adjacent to ∠ 𝐴.
c) Which angle of ⊿𝐴𝐵𝐶 has a tangent of
Solution.
12
5
a) 𝐵𝐶 b) 𝐴𝐶 c) ∠ 𝐴

5. Lehman College, Department of Mathematics
The Tangent Ratio (2 of 2)
Example 2. Find tan 𝐴 for each right triangle 𝐴𝐵𝐶:
Solution. a) tan 𝐴 =
7
24
b) c)
a) b)
c)
tan 𝐴 =
15
8
tan 𝐴 =
4
3

6. Lehman College, Department of Mathematics
Angle of Elevation (1 of 3)
DEFINITION. An angle formed by a horizontal line and
the line of sight to an object that is above the horizontal
line is called the angle of elevation. The angle formed
by the horizontal line and the line of sight to an object
that is below the horizontal line is called the angle of
depression.

7. Lehman College, Department of Mathematics
Angle of Elevation (2 of 3)
Example 3. A man stands 50 meters from the trunk of a
tree. The angle of elevation from eye level to the top of
the tree is 52°. The distance from the man’s eye level to
the ground is 1.5 meters. Given that tan 52° = 1.28.
a) Let 𝑑 be the distance from the
man’s eye level to the top of the
tree. Write an expression for the
height of the tree in terms of 𝑑.
b) Find 𝑑 to the nearest tenth of
a meter
c) Find the height of the tree to
the nearest tenth of a meter

8. Lehman College, Department of Mathematics
Angle of Elevation (3 of 3)
Solution.
b) Express tan 52° through 𝑑:
tan 52° =
It follows that: 𝑑 = 50 tan 52°
𝑑 = 50 1.28 = 64 m
a) Height of tree = 𝑑 + 1.5
c) Height of tree = 𝑑 + 1.5
= 64 + 1.5
= 65.5 m
𝑑
50

9. Lehman College, Department of Mathematics
Tapeworm Head under Microscope

10. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (1 of 6)
The sine of an acute angle of a right triangle is the ratio
of the length of the side opposite the angle to the length
of the hypotenuse:
sin 𝜽 =
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐨𝐩𝐩𝐨𝐬𝐢𝐭𝐞 𝐚𝐧𝐠𝐥𝐞 𝜽
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐡𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞
=
𝒂
𝒄

11. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (2 of 6)
The cosine of an acute angle of a right triangle is the
ratio of the length of the side adjacent to the angle to
the length of the hypotenuse:
cos 𝜽 =
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐬𝐢𝐝𝐞 𝐚𝐝𝐣𝐚𝐜𝐞𝐧𝐭 𝐭𝐨 𝐚𝐧𝐠𝐥𝐞 𝜽
𝐥𝐞𝐧𝐠𝐭𝐡 𝐨𝐟 𝐡𝐲𝐩𝐨𝐭𝐞𝐧𝐮𝐬𝐞
=
𝒃
𝒄

12. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (3 of 6)
Example 4. For triangle 𝐽𝐾𝐿 below, find the sine, cosine
and tangent of ∠ 𝐿 :
Solution.
8
17
sin 𝐿 =
cos 𝐿 =
15
17
8
15
tan 𝐿 =

13. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (4 of 6)
Example 5. For triangle 𝐷𝐸𝐹 below, find the value of 𝑥
to the nearest tenth, given that sin 41° = 0.656 and
cos 41° = 0.755:
Solution.
cos 41° =
𝑥
16
Step 1. Express 𝑥 and the known
length of 16 through a trigonometric
ratio of 41°. Which one?
Step 2. Solve for 𝑥:
𝑥 = 16 cos 41° = 16 0.755 = 12.1

14. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (5 of 6)
Example 6. You set a 24-foot ladder against a building.
You want the angle between the ladder and the ground
to be 75°. To the nearest foot, how far from the building
should you place the bottom of the ladder, given that
sin 75° = 0.966, cos 75° = 0.259, and tan 75° = 3.732?
Solution.
Let 𝑑 be the distance from
the base of the building to
the bottom of the ladder.
Write an expression for the
length of the ladder in terms
of 𝑑 and the angle 75°.

15. Lehman College, Department of Mathematics
The Sine and Cosine Ratios (6 of 6)
Solution (cont’d).
Write an expression for the
length of the ladder in terms
of 𝑑 and the angle 75°.
cos 75° =
𝑑
24
𝑑 = 24 cos 75° = 24 0.259 = 6.2 ft
It follows that:

16. Lehman College, Department of Mathematics
The Pythagorean Theorem (1 of 3)
Example 7. For triangle 𝐷𝐸𝐹 below, find the value sin 𝐷:
Solution.
Triangle 𝐷𝐸𝐹 is called a 3-4-5 triangle.
Step 1. To determine sin 𝐷, we
need to know the length of side 𝐷𝐹.
Step 2. Evaluate sin 𝐷 :
𝑐2
=
Let the length of side 𝐷𝐹 be 𝑐.
Then, by the Pythagorean theorem:
32
+ 42
= 9 + 16 = 25
It follows that: 𝑐 = 5
sin 𝐷 =
3
5

17. Lehman College, Department of Mathematics
The Pythagorean Theorem (2 of 3)
Example 8. For triangle 𝐴𝐵𝐶 below, find the value tan 𝐴:
Solution.
Triangle 𝐴𝐵𝐶 is called a 5-12-13 triangle.
Step 1. To determine tan 𝐴, we
need to know the length of side 𝐴𝐶.
Step 2. Evaluate tan 𝐴 :
132
=
Let the length of side 𝐴𝐶 be 𝑏.
Then, by the Pythagorean theorem:
52
+ 𝑏2
25 + 𝑏2
It follows that: 𝑏2 = 144
tan 𝐴 =
5
12
169 =
and 𝑏 = 12

18. Lehman College, Department of Mathematics
The Pythagorean Theorem (3 of 3)
Example 8. For the right triangle below, find the
unknown length 𝑐 in simplest form:
Solution.
𝑐2
= Pythagoras’ theorem82
+ 142
= 64 + 196
260 Add𝑐2
=
Take positive square roots of either side𝑐 = 260
Simplify
= 4 ⋅ 65
= 4 ⋅ 65
= 2 65
Find highest perfect square factor of 260
Split square roots
Evaluate

19. Lehman College, Department of Mathematics
Trigonometric Identities (1 of 2)
Consider the right triangle 𝐴𝐵𝐶 below:
We know that:
Let us look at the ratio:
sin 𝜃 =
𝑎
𝑐
÷
and that: cos 𝜃 =
𝑏
𝑐
sin 𝜃
cos 𝜃
=
𝑎
𝑐
𝑐
𝑏
=
𝑎
𝑏
=
𝑏
𝑐
=
𝑎
𝑐
⋅ tan 𝜃

20. Lehman College, Department of Mathematics
Trigonometric Identities (1 of 2)
Write the Pythagorean Theorem for triangle 𝐴𝐵𝐶 below:
𝑐2
𝑎2
+ 𝑏2
= Pythagorean theorem
Divide either side by 𝑐2
𝑎2
+ 𝑏2
𝑐2
=
𝑐2
𝑐2
𝑎2
𝑐2
+
𝑏2
𝑐2
= 1 Split fraction on left-hand side
𝑎
𝑐
2
+
𝑏
𝑐
2
= 1 Property of exponents

21. Lehman College, Department of Mathematics
Trigonometric Identities (2 of 2)
From the previous slide, we had:
𝑎
𝑐
2
+
𝑏
𝑐
2
= 1
What trigonometric
functions of 𝜃 are the ratios:
𝑎
𝑐
= and
𝑏
𝑐
=sin 𝜃 cos 𝜃
It follows that: sin 𝜃 2
+ cos 𝜃 2 = 1
We write that as: sin2
𝜃 + cos2 𝜃 = 1
The above is called a Pythagorean trigonometric identity.