MATHepedia

(Grade 9_Q4_W5-B)

Competency: The learner uses trigonometric ratios to solve real-life

problems involving right triangles (M9GE-IVe-1)

Concept:

In order to create an accurate illustration presented by the information in the

Direction:
given word problemRead the given
read and problem
understand theand complete
problem and the
findtable below.
the key pieces of
information needed. The following are the trigonometric ratios we can use as the
A building is 31 m high. At a certain distance away from the building, an
formula in solving word problems
observer involving
determines right
that the triangles.
angle of elevation to the top of it is 39°. How far is
the observer from the base of the building?
opposite hypotenuse
1. sin θ = 4. csc θ =
Solution:
hypotenuse opposite
adjacent hypotenuse
2. cos θ = 5. sec θ =
hypotenuse adjacent
opposite adjacent
3. tan θ = 6. cot θ =
adjacent opposite
Example A

Make an illustration. 31 m

390
x

31 m high building
What is/are given?
390 angle of elevation to the top of the building
Exercise
The distance of the observer from the base of the
A
What is to be determined building. It can be represented by any letter of the
Direction: Read the given problem andEnglish
completealphabet such
the table as x.
below.
Tangent ratio involves the two legs of the triangle
A coconut tree is 19 m high. Away from that tree, a boy determines that the
as part of the problem, while sine or cosine ratio
angle of elevation to the top of it is 41°. How far is the boy from the base trunk of
involves the hypotenuse as part of the problem. In
the coconut
Formulatree?
to be used this problem, the height of the building and the
distance of the observer from the base of the
building represent the legs of the triangle, so it is
a tangent ratio. Hence, the formula to be used is
opposite
tan θ = .
adjacent
Author: Gerwayne M. Palomar 1
 Make an illustration.

What is/are given?

What is to be determined

Formula to be used

Example B
Direction: Read the given problem and complete the table below.
Joshua placed a ladder 11 meters long leaning against the wall of a
building. The foot of the ladder makes an angle of 610 with the ground, how far is
the base of the ladder from the wall?

Solution:

11 m
Make an illustration.
610
x

11 m long ladder
What is/are given?
610 angle of elevation

The distance of the base of the ladder from

the wall of the building. It can be represented
What is to be determined
by any letter of the English alphabet such as
x.
Tangent ratio involves the two legs of the
triangle as part of the problem, while sine or
cosine ratio involves the hypotenuse as part
of the problem. In this problem, the length of
the ladder is the hypotenuse of the right
Formula to be used triangle while the distance of the base of the
ladder from the wall of the building is the
adjacent side, so it is a cosine ratio. Hence,
the formula to be used is
adjacent
cos θ = .
hypotenuse
Author: Gerwayne M. Palomar 2
Exercise B
Direction: Read the given problem and complete the table below.
Joshua placed a ladder 11 meters long leaning against the wall of a
building. The foot of the ladder makes an angle of 610 with the ground, how far
is the base of the ladder from the wall?

Make an illustration.

What is/are given?

What is to be determined

Formula to be used

Example C
4 cm 3 cm
Direction: Given cos β = and tan β = , draw and label the right triangle, then
5 cm 4 cm
find
Solution: the other trigonometric ratios of the angle β.
The given trigonometric ratios are cos β and tan β,
D
adjacent 5 cm
we know that cos β = from the given cos β =
hypotenuse
4 cm opposite 3 cm
, while tan β = and from the given tan β = β
5 cm adjacent
E F
3 cm 4 cm
. The sides of the right triangle involved based from
4 cm
the given are 3 cm, 4 cm, and 5 cm. 5 cm is the longest
opposite the right angle of the triangle, 3 cm will be the shortest side, and 4 cm will be
the other side of the triangle. Hence, the right triangle can be illustrated as the figure
at the right. We can label the angles of the right triangle with any letters of the English
3 cm opposite
alphabet such as D, E, and F. From the given, tan β = and tan β =
4 cm adjacent
means that angle β can be found opposite to the side 3 cm and adjacent to the side
4cm. Hence, the angle β can be found at angle F.
The other trigonometric ratios can now be found using the right triangle DEF at
the right.
opposite side of θ 3 cm
1. sin β = =
hypotenuse 5 cm
hypotenuse 5 cm
2. csc β = =
opposite side of θ 3 cm
hypotenuse 5 cm
3. sec β = =
adjacent side of θ 4 cm
Author: Gerwayne M. Palomar
adjacent side of θ 4 cm 3
4. cot β = =
opposite side of θ 3 cm
Exercise C

15 cm 8 cm
Direction: Given cos θ = and tan θ = , draw and label the right triangle,
17 cm 15 cm
the
find the other trigonometric ratios of the angle θ.

Author: Gerwayne M. Palomar 4