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MODULE 4 - APPLICATIONS OF TRIGONOMETRY
OVERVIEW
In this module, we are going to discuss the application of trigonometry, the right and oblique triangles. There are many real-world applications involving these triangles. In some problems, only the sides are given. Sometimes, two sides and an angle are given. It is necessary to identify which of these laws are to be applied in these problems to solve it easily and correctly.
In this module, we are going to discuss the application of trigonometry, the right and oblique triangles. There are many real-world applications involving these triangles. In some problems, only the sides are given. Sometimes, two sides and an angle are given. It is necessary to identify which of these laws are to be applied in these problems to solve it easily and correctly.
LEARNING OBJECTIVES:
At the end of this module, you should be able to:1. Solve right triangles, sides, angles and areas.2. Solve real-life right triangle problems.3. Solve oblique triangles using Law of Sines and Cosines.
At the end of this module, you should be able to:1. Solve right triangles, sides, angles and areas.2. Solve real-life right triangle problems.3. Solve oblique triangles using Law of Sines and Cosines.
SOLUTIONS OF RIGHT TRIANGLES
A triangle is composed of six parts, the three sides and three angles. To solve a triangle is to find the unknown parts from the parts that are given. In the case of a right triangle, this can always be done if we have given (besides the right angle) two parts, at least one of which is a side (Rider).
In solving a right triangle, we can use the following:
A triangle is composed of six parts, the three sides and three angles. To solve a triangle is to find the unknown parts from the parts that are given. In the case of a right triangle, this can always be done if we have given (besides the right angle) two parts, at least one of which is a side (Rider).
In solving a right triangle, we can use the following:
- Six Trigonometric Functions (sin, cos, tan, csc, sec, and cot)
- Pythagorean Theorem (c2 = a2 + b2, where a and b are the legs and c is the hypotenuse)
- Triangle Sum Theorem or the Relation A + B + C = 180°
- Since it is a right triangle and C = 90°, we can say A + B = 90°.
Example from problem set #4: To measure the height of a building, a surveyor sets up his transit at a distance of 112.2 feet from the building. He finds the angle of elevation of the top of the building to be 48°17′. If the telescope of the transit is 5 feet above the base of the building, how high is the building? To solve the problem, I first illustrated the problem description to better understand the situation. Then, I got three quantities to get the height of the building. I converted the angle of elevation into full degrees first before inserting it to the equation. Obviously, I used tangent because we are dealing with opposite and adjacent relative to the angle of elevation. Then, I derived the value of x on which the 5 feet should be added to get the height of the building. Hence, the building is approximately 131 feet high.
SOLUTIONS OF OBLIQUE TRIANGLES
An oblique triangle is a triangle which does not contain a right angle. Such a triangle contains either three acute angles or two acute angles and one obtuse angle (Moyer).
An oblique triangle is a triangle which does not contain a right angle. Such a triangle contains either three acute angles or two acute angles and one obtuse angle (Moyer).
LAW OF SINES
The Law of Sines is the relationship between the sides and angles of non-right (oblique) triangles. Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle. In any triangle ABC, the ratio of a side and the sine of the opposite angle is constant; i.e.:
The Law of Sines is the relationship between the sides and angles of non-right (oblique) triangles. Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle. In any triangle ABC, the ratio of a side and the sine of the opposite angle is constant; i.e.:
Example from quiz 4: Solve for the sides and missing angle of the triangle where A = 40°, C = 65°, and b = 7.Since the given are two angles and one included side, we must use law of sines. But before that, we can already apply the triangle sum theorem which states that the sum of all the angles of a triangle is equal to 180°. Thus, we can get the third angle through the other two angles which is B = 75°. After that, we can now substitute the given values to the formula to get the other two sides. I simply get one equation in which there is only one missing value. I apply cross multiplication and simplify. Henceforth, a = 4.66 and c = 6.57.
LAW OF COSINES
The Law of Cosines is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known. In either of these cases, it is impossible to use the Law of Sines because we cannot set up a solvable proportion. In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of the included angle (Moyer) ; i.e.:
The Law of Cosines is used to find the remaining parts of an oblique (non-right) triangle when either the lengths of two sides and the measure of the included angle is known (SAS) or the lengths of the three sides (SSS) are known. In either of these cases, it is impossible to use the Law of Sines because we cannot set up a solvable proportion. In any triangle ABC, the square of any side is equal to the sum of the squares of the other two sides diminished by twice the product of these sides and the cosine of the included angle (Moyer) ; i.e.:
Example from quiz 4: Solve the triangle: a = 7, b = 6, and c = 8. Since the given are three sides, then we will use the law of cosines. As you can see in my solution, I just substituted the given sides to get the angles. For instance, if you want to find the angle A, then use the formula of a2. If you want to find the angle B, then use the formula of b2. In the same manner, if you want to find angle C, then use the formula of c2. Just be careful in simplifying and make sure that the calculator is in degree mode. After getting the two angles, I used the triangle sum theorem to easily get the third angle. To sum up, A = 58°, B = 47°, and C = 75°.
SUMMARY
Use law of sines when the given are:
Use law of cosines when the given are:
Use law of sines when the given are:
- two angles and the side opposite one of them (AAS)
- two angles and the included side (ASA)
- two sides and the angle opposite one of them (SSA)
Use law of cosines when the given are:
- two sides and the included angle (SAS)
- three sides (SSS)
Salazar - PS#4 - BSCE1B.pdfPROBLEM SET #4 (45/50)
Salazar - Activity 4 - BSCE1B.pdfACTIVITY #4 (100/100)
Salazar-Quiz4-BSCE1B.pdfQUIZ #4 SOLUTIONS (15/15)
Salazar -Assessment- BSCE1B.pdfMIDTERM ASSESSMENT (65/65)
REFLECTION
In this module, I have learned how to solve for the missing angles and missing sides of a triangle whether a right triangle or an oblique one. In a right triangle, one must know the Pythagorean theorem, triangle sum theorem, the six trigonometric functions, and that one angle is a right angle. In oblique triangles, one must know the law of sines as well as the law of cosines. I learned how to use these two laws depending on the given quantities. If the given are AAS, ASA, and SSA, use the law of sines. If the given are SAS and SSS, use the law of cosines. I was able to apply this knowledge in activity 4 in which we were asked to make triangles out of three balls thrown to the walls. The same as the previous modules, the instructional material for this module is comprehensive and understandable. Thanks to the accurate discussions of our teacher.
In this module, I have learned how to solve for the missing angles and missing sides of a triangle whether a right triangle or an oblique one. In a right triangle, one must know the Pythagorean theorem, triangle sum theorem, the six trigonometric functions, and that one angle is a right angle. In oblique triangles, one must know the law of sines as well as the law of cosines. I learned how to use these two laws depending on the given quantities. If the given are AAS, ASA, and SSA, use the law of sines. If the given are SAS and SSS, use the law of cosines. I was able to apply this knowledge in activity 4 in which we were asked to make triangles out of three balls thrown to the walls. The same as the previous modules, the instructional material for this module is comprehensive and understandable. Thanks to the accurate discussions of our teacher.
REFERENCES:
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