TRIGONOMETRY
Lesson 2: Solving Right Triangles
Todays Objectives
• Students will be able to develop and apply the
primary trigonometric ratios (sine, cosine,
tangent) to solve problems that involve right
triangles, including:
• Solve right triangles, with or without technology
Using Trigonometry to solve for a side
• In order to find an unknown side measure in a right triangle using trig
ratios, the length of one other side and the measure of one of the
acute angles is required.
• Example: Solve for the side length x to the nearest tenth of a
centimeter
62°
10 cm
hypotenuse
adjacent
x
opposite
• Solution: First, identify the positions of the side lengths relative to the
acute angle whose measure is known
• Since the length of the hypotenuse and opposite side is required (side
x), the opposite-hypotenuse ratio is used. This is the sine ratio
• 𝑠𝑖𝑛62 =
𝑜𝑝𝑝
ℎ𝑦𝑝
𝑥
= 10 , 10𝑠𝑖𝑛62 = 𝑥,
𝑥 ≅ 8.8 𝑐𝑚
Example
• Solve for side length x to the nearest tenth of a meter
x
37°
46 m
• Solution: 𝑐𝑜𝑠37 =
•𝑥=
46
𝑐𝑜𝑠37
• 𝑥 ≅ 57.6 𝑚
46
𝑥
Example
• Triangle ABC has ‫ﮮ‬C = 90º, ‫ﮮ‬A = 55º, and AC = 50 in. Solve
for side length AB to the nearest inch.
• Solution: First, draw and label a sketch of a representative
B
triangle.
x
A
55° 50 in.
C
• Since the length of the adjacent side relative to ‫ﮮ‬A is known
(50 in) and the length of the hypotenuse (AB) is required, the
adjacent-hypotenuse ratio is used. This is the cosine ratio.
• 𝑐𝑜𝑠55 =
𝑎𝑑𝑗
ℎ𝑦𝑝
=
50
,
𝐴𝐵
𝐴𝐵 =
50
,
𝑐𝑜𝑠55
𝐴𝐵 = 87 𝑖𝑛.
Using Trigonometry to Solve for an Angle
• In order to find the measure of one of the acute angles in
a right triangle when the measure of each acute angle is
unknown, the lengths of two of the three sides must be
known
15 cm
6 cm
𝜃
• Solve for 𝜃 to the nearest tenth of a degree
Example
• Solution: Label the given sides relative to the angle whose
measure you are trying to determine.
hypotenuse
opposite
adjacent
𝜃
• Since the lengths of the opposite side and the hypotenuse are
known, the opposite-hypotenuse ratio is used. This is the sine
ratio.
• 𝑠𝑖𝑛𝜃 =
6
15
• Use the inverse since function, sin-1, to solve for 𝜃
• 𝜃 = sin−1
6
15
, 𝜃 = 23.6°
Example (You do)
• Solve for 𝜃 to the nearest tenth of a degree.
𝜃
30 m
40 m
• Solution: Since the lengths of the adjacent side and
opposite side are known, apply the tangent ratio.
• 𝑡𝑎𝑛𝜃 =
𝑜𝑝𝑝
𝑎𝑑𝑗
=
40
30
• Use the inverse tan function, tan-1, to solve for 𝜃
•𝜃=
40
−1
tan
30
= 53.1°
Solving Triangles
• Often, you will need to determine all the unknown
measures of the sides and angles of a triangle.
This is referred to as solving the triangle.
• In order to solve a triangle, it is common to use
one or more of the following:
• Pythagorean theorem
• Sum of angles in a triangle
• Trigonometric ratios
Example
• Solve the following triangle. Give side lengths to the
nearest tenth of a centimeter.
𝜃
12 cm
y
x
35°
• Solution: If it has not already been done for you, label the
unknowns that you are to find
• The measure of angle 𝜃 can be found first
• 90° + 33° + 𝜃 = 180°, 𝜃 = 57°
Example
• Next, the length of the side labeled x can be determined using
a trig ratio.
• tan 33 º =
• tan 33 º =
𝑜𝑝𝑝
𝑎𝑑𝑗
12
𝑥
• 𝑥𝑡𝑎𝑛 33º = 12, 𝑥 =
12
tan33°
= 18.5 𝑐𝑚
• The final side, y, can be determined either using the
Pythagorean theorem or a trig ratio. Let’s use the trig ratio,
sine.
• sin 33 º =
𝑜𝑝𝑝
ℎ𝑦𝑝
=
12
𝑦
• 𝑦𝑠𝑖𝑛 33º = 12
• 𝑦 =
12
sin 33°
= 22.0 𝑐𝑚
Example (You do)
• Solve the following triangle. Give the unknown side length to
the nearest tenth of a centimeter, and give unknown angles to
the nearest tenth of a degree.
C
B
8 cm
10 cm
A
• Solution: Side AC = 6 cm, Angle A = 53.1°, Angle B = 36.9°
Homework
• Exercises #3-16, pg. 111-112
• Begin to add the chapter 2 vocabulary words to your
vocabulary books. These are the words in BOLD in your
textbook (the first one is angle of inclination….the
second one is tangent ratio, etc.)