OVERVIEW Use Trigonometry & Pythagorean Theorem to Solve G.SRT.8

G.SRT.8
Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. / No surprises here…. Use trigonometry and the Pythagorean Theorem to solve problems. /
Apply these concepts to solve problems.
(1) The student will be able interpret the word descriptions into lengths and angles of a right triangle so that they can diagram the relationship.
(2) The student will be able to solve trigonometry and Pythagorean Theorem problems based on written descriptions. / The emphasis here is on the use of these two very powerful relationships. Students need to see a variety of different situations and be able to simplify them into a trigonometry or Pythagorean problem. / 1 – Vocabulary…. Lots of the issues with these problems are the use of ‘new’ vocabulary - the foot of the ladder, the angle of elevation, the altitude of the plane, and so on. Help students gain a good understanding of the ‘typical’ questions.

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NOTES Use Trigonometry & Pythagorean Theorem to Solve G.SRT.8

CONCEPT 1 – Use trigonometric ratios and the Pythagorean Theorem to solve right triangles in applied problems. *(Modeling Standard)

In this objective the principles that have been learned concerning similarity, trigonometry and the Pythagorean Theorem are applied to solve ‘real’ world problems. While they are called real world problems many of them have been greatly simplified so that the mathematics at this level can be applied. Real world problems have many variables and are often too complex to approach with this simple level of mathematics but as students’ progress through mathematics more detail is added to the problems making them ‘more real’. At this stage students’ must gain proficiency in the skill within a context to see where these mathematical tools apply.

These real world problems are often called word problems by students. Yes, that is often true… Real world problems do come with context and so, yes, they look and feel different than the standard symbol based problems that are often done. It also shows the need to gain proficiency with language as well as mathematics to succeed with these types of problems. Most of these notes will be discussing the context of these problems and the translation of word to mathematics and not the solving of them. That skill was discussed in an earlier objective. Start with how angles are referred to.

THE ANGLE OF ELEVATION

To elevate is the move in an upward direction. Thus the angle of elevation is an upward angle from the horizon. The angle of elevation is the reference angle from the position in an upward direction from the horizon.
The stranded person on the island looks up
to the plane at an ANGLE OF ELEVATION. /

Very few students struggle with this concept – The name and the placement seem to come easily. This is not true concerning the next angle, many students struggle with its placement and understanding.

THE ANGLE OF DEPRESSION

To be depressed is to be down, thus the angle of depression is the reference angle from the position, in a downward direction, from the horizon. Sounds simple, like the angle of elevation but it isn’t quite as easy.

Example #1 / Example #2 / Example #3

In example #1, the angle of depression (1) has been placed correctly, as the angle from the airplanes’ horizon looking down. In example #2, the most common error has taken place; the angle of depression (1) was PLACED IN THE WRONG LOCATION. Student’s make this error because no horizon has been drawn and so they see this as the only angle from that reference point. Finally, in example #3, the angle of depression (1) has been correctly drawn and located - placing it from the horizon down.

Level 2 Difficulty -- THE USE OF TWO TRIANGLES

The use of two triangles just simply means that two different values must be found to gain the final answer. Really this problem isn’t an increase of difficulty it is simply an increase of work. It often gains difficulty because it does require more contexts to explain what is going on.

Example Problem

A flagpole is at the top of a building. 300 ft from the base of the building, the angle of elevation of the top of the pole is 32° and the angle of elevation of the bottom of the pole is 30°. Determine the length of the flagpole (to the nearest foot).
This problem looks quite complex because of the overlapping angles of elevation but really if we think through the problem we can simplify it by solving two different triangles. /
The way to get the height of just the flagpole would be to calculate the total height from ground to top of flag (x) and then calculate the distance from ground to the bottom of the flagpole (y) and then subtract the two values. / /
( x – y) will gives the flagpole height, 187.46 – 173.21 = 14.25 ft.

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NOTES Use Trigonometry & Pythagorean Theorem to Solve G.SRT.8

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ASSESSMENT Use Trigonometry & Pythagorean Theorem to Solve G.SRT.8

1. The angle of elevation from the car to the girl is:
A) Ð1 B) Ð2 C) Ð3 D) Ð4 /
2. The angle of depression from the helicopter to the girl is:
A) Ð1 B) Ð2 C) Ð3 D) Ð4 /
3. Tommy has caught his kite at the top of a 16 ft tree. From where Tommy is standing the elevation to the top of the tree is 29°, what is the length of string (round to the nearest foot)?
A) 33 ft B) 29 ft C) 18 ft D) 8 ft /
4. A plane at an altitude of 7000 ft is flying in the direction of an island. If angle of depression is 21° from the plane to the island, what is the horizontal distance until the plane flies over the island? (round to the nearest foot)
A) 2,687 ft B) 7,498 ft C) 18,236 ft D) 19,533 ft /
5. A 18 ft tree casts a 15 ft long shadow. What is the angle formed by the sun’s rays and the ground? (round to the nearest degree)
A) 60° B) 50° C) 40° D) 30° /
6. The angle of elevation from the car to the girl at the top of the parking garage is Ð11 & Ð12.
7. The angle of depression from the helicopter to the boy on the third floor of the parking garage is Ð2.
8. The angle of depression from the boy on the third floor of the parking garage to the car is Ð10. / T or F
T or F
T or F
/
9. After reading that “the angle of depression from the girl to the car is 55°”, Tim labeled the diagram as shown. Has he labeled the diagram correctly? Explain. /
10. Jazmin is flying a kite but it got caught at the top of a 24 ft tree. If she let out 100 ft of string, what is the angle of elevation from Jazmin to the top of the tree (round to the nearest degree)? /

Answers:

1)  C

2)  C

3)  A

4)  C

5)  B

6)  T

7)  F

8)  F

9)  No, he hasn’t… the angle of depression is from the horizon down. He has placed the 55° between the building and the line of sight. Instead he should have put it between the girl’s horizon and the line of sight to the car.

10) 

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