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Republic of the Philippines Department of Education Region IV-A (CALABARZON) Division of Cavite ALFONSO NATIONAL HIGH SCHOOL Telefax:(094)8630796; Tel. No.: (094)DAILY LESSON PLAN IN MATHEMATICS 9CONTENT STANDARD : Demonstrate understanding of the basic concept of trigonometry. PERFORMANCE STANDARD ...

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Republic of the Philippines Department of Education Region IV-A (CALABARZON) Division of Cavite ALFONSO NATIONAL HIGH SCHOOL Telefax:(094)8630796; Tel. No.: (094)8630942 DAILY LESSON PLAN IN MATHEMATICS 9 CONTENT STANDARD: Demonstrate understanding of the basic concept of trigonometry. PERFORMANCE STANDARD: Is able to apply the concepts of trigonometric ratios to formulate and solve real problem with precision and accuracy. LEARNING COMPETENCY: illustrate the six trigonometric Ratios; sine, cosine, tangent, secant, cosecant, and cotangent M9GE-IVd-1 I. Objective 1. Determine the trigonometric ratios involving special angles; 2. Compute the numerical values of trigonometric expressions involving special angles. 3. Appreciate trigonometric ratios of special angles in solving real life problems. II. LEARNING CONTENT Topic: Trigonometric Ratios of Special Angles Sub Topic:Special Triangle and Exact Values Material: activity sheet, chalk and blackboard, laptop and projector Reference: grade 9 mathematics learner’s material pg. III. LEARNING STRATEGIES Teacher Activity A. Daily Routine 1. Prayer Good day everyone. Before we start we all stand up and pray. Anyone who wants to volunteer? Yes Elaine please led the prayer.

Student’s activity

Good morning ma'am/teacher.

Elaine will lead the prayer. 2. Classroom Management Ok class before you sit down, kindly pick up the pieces of paper and trash 3. Greetings Once again, good morning class. How are you today?

The student will pick up the pieces of paper and trash, then they will sit down.

Good morning ma’am/ teacher Different response

Okay good. So now, are they any absent today? The student will tell who the absences are. B. PRIMING

1. Recall Before we begin to our lesson, let’s have a recap on our past lesson. Anyone who can still remember our past lesson yesterday. What are the trigonometric Ratio?

What is the opposite of SOHCAHTOA?

Thank you, student.

Our past lesson is about Trigonometric Ratio The six Trigonometric Ratios are sine, cosine, tangent, cosecant, secant and cotangent. The opposite of SOHCAHTOA is the CHOSHACAO where hypotenuse cosecant = opposite side hypotenuse secant = and adjacent side adjacent side cotangent = opposite side

2. Motivation Class before we proceed to our lesson, I have here a video. Let watch the video.

C. ACTIVITY What is the video all about? What have you noticed about the lengths of the sides of each triangle?

What have you observed about the measures of the angles of each triangle? Very good class. Now triangle having those angles will be called a Special one and that is our topic for today. How can we find that the triangle is a Special one?

The video is about the Special Triangle. The length of the side of each triangle have a radicals. In a triangle having 30 ° , 60 ° and 90 ° , the hypotenuse is two times the shorter leg and the longer leg (opposite the 60 ° angle) have the measures √ 3 times the shorter leg while in 45 ° , 45 and 90 ° The triangles is in 30°,60°,90° and 45°, 45°, 90° in angle.

D. ANALYSIS Class our lesson for today is about special right triangle. Can someone please draw a right triangle in front The right triangle is

Very good class. And that is an example of a right triangle, now we said before that a right triangle is said to be a right triangle it must have one angle measure 90°. This right triangle can be special if it has a 30°,60°,90° and 45°, 45°, 90° in angle. Class I have here two right triangles.

What have you observe?

Very good class. This right triangles are special because of the measure of an angle they have, why do you think so that the angle is the reason why the triangle is called special one? So class let proceed first to 45° 45° 90° special right triangle.

We noticed that One triangle have 30°,60°,90° in angle and the other triangle have 45°, 45°, 90° in angle

The right triangle is special since their angle have 30°,60°,90° and 45°, 45°, 90°.

Class read the definition. So class the 45° 45° 90°right triangle is also called isosceles triangle, again class. What is isosceles triangle?

Special 45° 45° 90° right triangle is a triangle whose angles are 45°, 45° and 90°. It is also called isosceles triangle. Isosceles triangle is a triangle with two equal sides.

Very good class. Class isosceles triangle also have two equal angle. Now in 45° 45° 90° special Right Triangle, the two equal angle obviously are the 45° angled. In 45° 45° 90° Special Right Triangle Theorem, The legs are congruent and the length of the hypotenuse is equal to shorter leg times √ 2 . I have here a triangle

What have you noticed? What kind of triangle is this? What will be the value of the other leg? How did you find that the value of the two legs is congruent? How about the hypotenuse? What will be the value of the hypotenuse? How did you find the value of the hypotenuse?

Very good class. Now let’s proceed to the other kind of the Special Right Triangle which is called the In 30° 60° 90° Special Right Triangle.

The triangle is in 45° -45°- 90° angled. The triangle is 45° -45°- 90° Right triangle The value of the shorter leg will be 8 too. Since in 45°-45°-90° Right Triangle theorem the two legs are always congruent. The hypotenuse is equal to 8 ∙ √2 =8 √2 Since in 45° -45° -90° Right Triangle theorem the value of the hypotenuse is equal to the shorter legs times √ 2 .

In 30° 60° 90° Special Right Triangle Theorem, The length of the longer leg is equal to shorter leg times √ 3 and the hypotenuse is twice the shorter leg. I have here a triangle

What have you noticed? What kind of triangle is this? What will be the value of the shorter leg? How did you find that the value of the shorter leg? What will be the value of the longer leg? How did you find that the value of the longer leg?

Very good class. F. ABSTRACTION What have you learned? What are the two kinds of special right triangle?

What is the difference of 45°-45°-90° Right

The triangle is in 30° -60°- 90° angled. The triangle is 30° -60°- 90° Right triangle The value of the shorter leg will be 20. Since in 30°-60°-90° Right Triangle 1 theorem the shorter legs are always 2 of hypotenuse. The longer leg is equal to 20 ∙ √3 =20 √3. Since in 30° -60° -90° Right Triangle theorem the value of the longer leg is equal to the shorter legs times √ 3 .

Triangle Theorem and 30° 60° 90° Special Right Triangle Theorem.

Very good class. E. APPLICATION.

We learn the two kinds of Special Right Triangle. The two kinds of Special Right Triangle are 45°-45°-90° Right Triangle and 30° 60° 90° Special Right Triangle. In 45° -45°-90° Special Right Triangle Theorem, The legs are congruent and the length of the hypotenuse is equal to shorter leg times √ 2 while in 30° 60° 90° Special Right Triangle Theorem, The length of the longer leg is equal to shorter leg times √ 3 and the hypotenuse is twice the shorter leg.

José has locked himself out his house. Fortunately, he did leave an upstairs window open and does have access to an extension ladder. The ladder, when fully extended safely, will extend to 24 feet. For optimal safety reasons, he wants to maintain a 60-degree angle with the ground. If he extends the ladder to 18 feet, a. how far will the base have to be away from the wall and b. How high up on the wall will the ladder reach? G. EVALUATION Activity: What Makes You Special? Given the angles of the triangles below, find the values of the six trigonometric ratios then answer the questions that follow.

a. Let a be the leg of a 45 °

– 45 °

– 90

Answer: a. If Jose extends the ladder 18 ft. the base of the ladder away from the wall will be 6 ft. b. The ladder will be high up the wall by 6 √ 3 ft. high.

° Triangle

sin 45 ° =_______ cos 45 ° = _______ _______ tan 45o = _______

sec 45 ° = _______ csc 45 ° = cot 45 = _______

b. Let a be the shorter leg of a 30 ° – 90 ° Triangle

sin 30 ° = _______ _______ cos30 ° = _______ tan 30 ° = _______ _______

– 60 °

sec 30 °

=

csc 30 ° = _______ cot 30 ° =

H. ASSIGNMENT Find the values of each variable used in the figures.

Republic of the Philippines Department of Education Region IV-A (CALABARZON) Division of Cavite ALFONSO NATIONAL HIGH SCHOOL Telefax:(094)8630796; Tel. No.: (094)8630942 DAILY LESSON PLAN IN MATHEMATICS 9 CONTENT STANDARD: Demonstrate understanding of the basic concept of trigonometry. PERFORMANCE STANDARD: Is able to apply the concepts of trigonometric ratios to formulate and solve real problem with precision and accuracy. LEARNING COMPETENCY: Illustrate angles of elevation and angle of depression. I. Objective At the end of the lesson the student should be able to: 1. Identify the concept of angle of elevation and depression. 2. To solve word problem dealing with trigonometric ratio and elevation. 3. To apply trigonometric ratio to solve problems in real life. II. LEARNING CONTENT Topic: angle of elevation and depression Material: activity sheet, chalk and blackboard, laptop and projector Reference: grade 9 mathematics learner’s material pg. 457-460 III. LEARNING STRATEGIES Teacher Activity Student’s activity A. Daily Routine 1. Prayer Good day everyone. Before we start we all Good morning ma'am/teacher. stand up and pray. Anyone who wants to volunteer? Yes student A please led the prayer. Student A will led the prayer. 2. Classroom Management Ok class before you sit down, kindly pick up the pieces of paper and trash The student will pick up the pieces of paper and trash, then they will sit down. 3. Greetings Once again, good morning class. How are you today?

Good morning ma’am/ teacher Different response

The student will tell who the absences are. Okay good. So now, are they any absent today? President tell me who are the absentees. B. PRIMING Our past lesson is about the Angle of 1. Recall Before we begin to our lesson, let’s have a Elevation and Depression recap on our previous lesson. Anyone who can An angle is formed. still remember our past lesson yesterday. Angle of elevation is above the horizontal What is formed when you look up or down to line of sight of the observer while the Angle of Depression is below the Horizontal line. an object above or below your eyes level. What is the difference between the two? Thank you student. 2. Motivation Okay class do you want to hear a music? Okay class our game called “Pass me and Answer me” Are you ready? Now I will give this bag to the first student in the right side. He/she will pass this bag besides him/her. While passing the bag, I will play a music and as the music stop, the student having the bag, he or she will get a colored paper (red violet, or green) inside the bag. If he/ she will get a red colored paper, he/she will answer my provided question. If he/she will get a violet colored paper, the man in his/ her left side will answer and if he/ she will get a green colored paper, the man in his/ her right side will answer my question. Understood?

C. ACTIVITY From our game, what can you noticed? Class what is your idea about the trigonometric Ratio. So class how can you connect the words trigonometric Ratio and Angle of Elevation to the word Right triangle?

Different response.

Yes ma'am.

The game is about trigonometric ratio, right triangle and angle of elevation. It is the SOHCAHTOA and CHOSHACAO. I think we can use the trigonometric Ratio to solve a right triangle involving the Angle of elevation and Depression.

Very good class and that is our topic for today. Yes ma’am D. ANALYSIS So let's start our discussion. Our lesson for today is word problems involving right triangle. Now, I have a picture here.

What have you noticed? Now if I connect the line of sight, the distance of the observer away from the waterfalls and the height of the tower to each other, what figure formed?

Very good class. Now the question is, how far is the observer from the base of the waterfall? If the height of the waterfall is 15.24m high and at the certain distance away from the waterfall, an observer determines that the angle of elevation to the top of it is 41°. Now how can we solve for the distance of the observer given that we have 15.24m (the height of the tower) as the side opposite to the angle of elevation which is 41°. Now we Let b =distance of the observer away from the waterfall. Then what will we do? Then what is the trigonometric ratio of

We notice a man is looking at the waterfall. If we connect the line of sight, the distance of the observer away from the waterfalls and the height of the tower to each other, we can formed a Right Triangle.

We can solve for the distance of the observer by the use of trigonometric ratio of tangent.

Find the b using trigonometric ratio of tangent. opposite side Tangent θ= adjacent side

tangent? Very good class. Now we substitute the value we will have? We have a fraction, what will we do? Then we will have, b tan 41 °=¿ 15.25 we solve for b. But what will we do now?

Multiply both side by tan 41 ° to left the only b on the left side. 15.25 The answer will be b ¿ or 18.17 tan 41 °

What will be our answer?

Very good class. Now I have here another example. Now, I have a picture

If we substitute the value we will have 15.25 tan 41 °= b Multiply both side by b to get 15.25 on its own.

here.

What have you noticed? Now I will draw horizontal line, the distance of the cat from the tip of tree and line of sight here. What figure is formed again?

We noticed that the man is looking up at the cat in the tree.

When we connect the horizontal line, the distance of the cat from the tip of tree and line of sight of the observer we can form a Right Triangle.

Very good class. Yes ma’am. We have another right triangle formed, then we can use trigonometric ratio again, right? Then given the distance of a 6 ft. man away from the base of tree is 12 ft. and the angle of elevation which is 76 ° . About how tall is

the tree? So how can we solve for the height of the tree? First we let a as the distance from the horizontal line of observer to the top of the tree. Then what will we do?

Then what is the trigonometric ratio of tangent, again? opposite side Then tan76 °= adjacent side We have a fraction, what will we do?

Since we have the distance of a 6 ft. man away from the base of tree is 12 ft. and the angle of elevation which is 76 ° we can use the trigonometric ratio of tangent. opposite side Tangent θ= adjacent side

Find a using trigonometric ratio of tangent. a tan 76 °= 12 Multiply both side by 12 to get a on its own

Then. 12 tan 76 °=¿ a 12(4.0108) = a 48.1294 = a or 48.13 Are we already get the height of the tree? Then what will we do?

No. Add the answer to the height of the man which is 6ft then 48.13+6 = 54.13

Very good. ABSTRACTION What have you learned? APPLICATION. Anthony is on the ground looking at the top of Answer: the new constructed building in their barangay at an angle of elevation of 300. After moving 50 feet closer, the angle of elevation is now 400. How far is Anthony from the building on his new location?

oppositeside adjacent side a Tan 46 ° = 400 400 ∙ tan 46 ° =a a= 414.2 m.

Tan 46 ° =

EVALUATION Solve for the following problem.

Assignment Solve the following problem 1. Find the angle of elevation of the sun when a 7.6 m flag pole casts an 18.2 m shadow. 2. . Olivia is in a lighthouse on a cliff. The top of the cliff is 110 feet above the water and Olivia is in a place in the lighthouse 85 feet above the top of the cliff. She observes two sailboats due east of the lighthouse. The angle of depression to the two boats is 33 degrees and 57 degrees. Find the distance between the two boats.

Republic of the Philippines Department of Education Region IV-A (CALABARZON) Division of Cavite ALFONSO NATIONAL HIGH SCHOOL Telefax:(094)8630796; Tel. No.: (094)8630942 DAILY LESSON PLAN IN MATHEMATICS 9 CONTENT STANDARD: Demonstrate understanding of the basic concept of trigonometry. PERFORMANCE STANDARD: Is able to apply the concepts of trigonometric ratios to formulate and solve real problem with precision and accuracy. LEARNING COMPETENCY: I. OBJECTIVE At the end of the lesson the student should be able to: 1. 2. 3. II. LEARNING CONTENT Topic: Word Problems Involving Oblique Triangle Subtopic: Word Problem Involving Law of Sine in SAA Case Material: activity sheet, chalk and blackboard, laptop and projector Reference: grade 9 mathematics learner’s material pg. 470-474 III. LEARNING STRATEGIES Teacher Activity A. Daily Routine 1. Prayer Good day everyone. Before we start we all stand up and pray. Anyone who wants to volunteer? Yes student A please led the prayer.

Student’s activity

Good morning ma'am/teacher. Student A will led the prayer.

2. Classroom Management Ok class before you sit down, kindly pick up the pieces of paper and trash.

The student will pick up the pieces of paper and trash, and then they will sit down.

3. Greetings Once again, good morning class. How are you today?

Good morning ma’am/ teacher Different response

Okay good. So now, are they any absent today? President tells me who the absentees are.

The student will tell who the absences are.

B. PRIMING 1. Recall Before we begin to our lesson, let’s have a recap on our previous lesson. Anyone who can still remember our past lesson yesterday.

Word problem involving Right Triangle

Thank you student.

2. MOTIVATION Class before we proceed to our lesson, let’s have first the game. I will group you into four. I have here 10 different sine values. I will give this each group. Each group need to look for the pair of each sine values. Example: sin 45 ° = sin 135 ° The first group to pair all the sine values will be the winner. C. ACTIVITY From our game, what can you noticed? Now if they have a pair then can we pair them to the other sine values? And based to our game, who could guess our topic for today? Very good. ANALYSIS So let's start our discussion. Our lesson for today is about law of sine. Now class Exactly class, all triangles in our game, they don’t have any right angle, and they are called oblique triangle. Class I have here a picture of the different kind of triangle. Now who can define what oblique triangle is?

All the sine values have their pair. Yes, ma’am

Yes, ma’am.

A triangle which no angle is measured exactly 90 ° .

Very good student. A Right triangle is formed.

Now let just first to know what is Law of sine This is the formula use in law of sine SAA Case or the side-angle-angle. sin A sin B sin C = = a b c Now I have here an example.

If ∠A=70 ° a=25 b= 14.2 ∠B=32.2 ° c= 26 ∠C=77.8 ° What is the value of sin 70 °÷ 25 What is the value of sin 32.2 °÷ 14.2 What is the value of sin 77.8 °÷ 26 Then we have, the value of sin 32.2 °÷ 14.2 Which is sin70 ° sin 32.2° sin 77.8° = = 14.2 25 26 0.04 = 0.04 = 0.04 Then our law of sine formula is true. Now class, try to remember that we can use the law of sine in a triangle that two angles and any side are given (AAS or the Angle-Angle-Side and ASA or the Angle-Side-Side) and the triangle that the two sides and an angle are given (SSA or Side-Side-Angle). Now I have here triangle with missing parts.

The value of sin 70 °÷ 25 is 0.9397 ÷ 25 =0.04 The value of sin 32.2 °÷ 14.2 is 0.5329 ÷ 14.25 =0.04 The value of sin...