W4-5
Learning Area
Quarter
Mathematics
4th Quarter
Grade Level
Date
9
I. LESSON TITLE
WORD PROBLEMS INVOLVING RIGHT TRIANGLES
II. MOST ESSENTIAL LEARNING
Uses trigonometric ratios to solve real-life problems involving right triangles.
COMPETENCIES (MELCs)
M9GE-IVe-1
III. CONTENT/CORE CONTENT
IV. LEARNING PHASES AND LEARNING ACTIVITIES
I. Introduction (Time Frame: ___5 minutes_____)
This lesson shall focus on solving real-life problems about angles of elevation and depression using the six trigonometric ratios
you have learned from the previous weeks. We will learn to see the world mathematically by reimagining objects into their
triangular counterparts and then solving that triangle by finding the lengths of the sides or the measure of the angles of th e
triangle.
Steps in problem solving involving trigonometric ratios:
1. Draw a diagram. Transform it into a geometric triangle and label each part correctly.
2. Determine the given and the question that needs to be answered.
3. Identify the corresponding trigonometric formula to be used.
4. Show your solution.
5. Present the conclusion.
D. Development (Time Frame: ___60 minutes_____)
Example#1
Problem
A tower is 15.24 m high. At a certain distance away from the tower, an observer determines that
the angle of elevation to the top of it is 41°. How far is the observer from the base of the tower?
Pictures below from MATHEMATICS GRADE 9 Learner’s Material, DepEd-BLR, First Editon, 2014, page
460
Draw the diagram
H
B
What
are
the
given?
What is to be
determined?
Formula used
O
height of tower 15.24 m and the angle of elevation 410.
Distance from the observer to the base of the tower
tan 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
IV. LEARNING PHASES AND LEARNING ACTIVITIES
Solution
Statement
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
15.24𝑚
tan 410 =
𝑥
15.24𝑚
0
𝑥 tan 41 = (
)𝑥
𝑥
𝑥 tan 410 15.24𝑚
=
tan 410
tan 410
15.24𝑚
𝑥=
tan 410
15.24𝑚
𝑥=
0.8693
𝑥 = 17.53𝑚
Conclusion
Example#2
Problem
Reason
Given
Substitution
Multiplication Property of Equality (MPE)
Multiplication Property of Equality (MPE)
Division Property
Computing for the value of tan 410
Division Property
Therefore, the distance between the observer to the base of the tower is 17.53 meters.
An airplane is flying at a height of 4 kilometers above the ground. The distance along the ground
from the airplane to the airport is 6 kilometers. What is the angle of depression from the airplane to
the airport?
Pictures below are from MATHEMATICS GRADE 9 Learner’s Material, DepEd-BLR, First Editon, 2014,
page 461
Draw the diagram
O
6km
B
α
4km
4km
6km
What
are
the
given?
What is to be
determined?
Formula used
A
height of airplane 4 km and the distance of the airplane along the ground to the airport is 6 km.
Angle of Depression
tan 𝜃 =
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Solution
Statement
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
4𝑘𝑚
tan 𝜃 =
6𝑘𝑚
2
tan 𝜃 = ( )
3
Reason
Given
Substitution
Simplification of fraction
𝜃 = 𝑡𝑎𝑛−1 ( )
Arctangent of an angle
33.69𝑂
Solving for arctan of 2/3
2
3
𝜃=
Conclusion
Example#3
Problem
Therefore, the airport has an angle of depression of 33.69𝑂 from the airplane
An observer in a lighthouse 48.8 m above sea level saw two vessels moving directly towards the
lighthouse on the same path. He observed that the angles of depression are 42° and 35°. Find the
distance between the two vessels, assuming that they are coming from the same side of the
tower.
Draw the diagram
O
350
D
Let
𝛼 = 42°
𝛽 = 35°
E
420
48.8 m
48.8 m
48.8 m
B
What are the
given?
What is to be
determined?
Formula used
Height of lighthouse 48.8 m and the angle of depression is 420 and 350 respectively.
Distance between two boats
tan 𝜃 =
Solution
Statement
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
tan 𝜃 =
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
̅̅̅̅
𝐸𝐶
𝑂𝐸
tan 𝛽 = ̅̅̅̅
̅̅̅̅ =
𝑂𝐸
Given
̅̅̅̅
𝐷𝐵
𝑂𝐷
̅̅̅̅
𝐷𝐵
𝑂𝐷
tan 𝛼 = ̅̅̅̅ and tan 𝛼 = ̅̅̅̅
̅̅̅̅
𝐸𝐶
tan 𝛽
̅̅̅̅ =
and 𝑂𝐷
𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒
𝑎𝑑𝑗𝑎𝑐𝑒𝑛𝑡
Reason
̅̅̅̅
𝐸𝐶
tan 𝛼
48.8𝑚
̅̅̅̅ =
𝑂𝐸
tan 350
48.8𝑚
̅̅̅̅
𝑂𝐸 =
0.7002
̅̅̅̅
𝑂𝐸 = 69.69𝑚
48.8𝑚
̅̅̅̅ =
𝑂𝐷
tan 420
48.8𝑚
̅̅̅̅ =
𝑂𝐷
0.9004
̅̅̅̅ = 54.20𝑚
𝑂𝐷
̅̅̅̅
𝐷𝐸 = ̅̅̅̅
𝑂𝐸 − ̅̅̅̅
𝑂𝐷
̅̅̅̅
𝐷𝐸 = 69.69 − 54.20𝑚
̅̅̅̅ = 15.49𝑚
𝐷𝐸
Conclusion
C
Substitution
Substitution
Substitution
Computing for the value of tan 350
Division property
Substitution
Substitution
Division property
Segment Subtraction Postulate
Substitution
Subtraction Property
Therefore, the distance between two boats is 15.49 meters.
IV. LEARNING PHASES AND LEARNING ACTIVITIES
E. Engagement (Time Frame: ___20 minutes_____)
Solve for the missing value in each problem.
1. Find the measure of ∠𝐴.
2. Find the height of the flagpole. .
A. Assimilation (Time Frame: __30 minutes_______)
Solve the following word problems using the method above.
1. A ladder 8 meters long leans against the wall of a building. If the foot of the ladder makes an angle of 68° with the
ground, how far is the base of the ladder from the wall?
2. A man, 1.5 m tall, is on top of a building. He observes a car on the road at an angle of 75°. If the building is 30 m high,
how far is the car from the building?
3. An airplane took off from an airport and travelled at a constant rate and angle of elevation. When the airplane
reached an altitude of 500 m, its horizontal distance from the airport was found to be 235 m. What was the angle
when the airplane rose from the ground?
V. ASSESSMENT (Time Frame: _30 minutes_____)
(Learning Activity Sheets for Enrichment, Remediation, or Assessment to be given on Weeks 3 and 6)
Solve the following word problems using the method above.
1. Obiwan is standing at a distance of 15 m from the base of a tree. From where he is standing, he can see the top of
the tree. If the tree is 15 m high and Obiwan is 1 m tall, what is the angle of elevation of the top of the tree?
2. A bamboo pole is leaning against a tree. If the height of the tree is 12.2 meters and the angle made by the pole and
the ground is 40°, what is the length of the pole?
3. 3. According to a lighting specialist for an art gallery, for best illumination of a piece of art, it is recommended that a
ceiling-mounted light be 1.8 m from the piece of art and that the angle of depression of the art piece be 38°. How
far from the wall should the light be placed so that the recommendations of the specialist are met? Give your answer
to the nearest tenth of a meter.
VI. REFLECTION (Time Frame: ____10 minutes_____)

Communicate your personal assessment as indicated in the Learner’s Assessment Card.

The learner, in their notebook, will write their personal insights about the lesson using the prompts below.
I understand that ___________________.
I realize that ________________________.
I need to learn more about __________.
Personal Assessment on Learner’s Level of Performance
Using the symbols below, choose one which best describes your experience in working on each given task. Draw it
in the column for Level of Performance (LP). Be guided by the descriptions below:
 - I was able to do/perform the task without any difficulty. The task helped me in understanding the target content/ lesson.
 - I was able to do/perform the task. It was quite challenging, but it still helped me in understanding the target
content/lesson.
? – I was not able to do/perform the task. It was extremely difficult. I need additional enrichment activities to be able to
do/perform this task.
Learning Task
LP
Learning Task
LP
Learning Task
LP
Learning Task
LP
Number 1
Number 3
Number 5
Number 7
Number 2
Number 4
Number 6
Number 8
VII. REFERENCES
Prepared by:
MATHEMATICS GRADE 9 Learner’s Material, DepEd-BLR, First Editon, 2014
Wilson Ray G. Anzures
Checked by:
MA. FILIPINA M. DRIO
LAILA R. MALOLES
HENRY P. CONTEMPLACION
GINALYN D. BELTRAN