LESSON 10 : RIGHT TRIANGLE TRIGONOMETRY Instructor: Mattheus Marcus Contreras LEARNING OBJECTIVES At the end of the lesson, the learner is able to : • Define the types of triangles • Apply Pythagorean Theorem • Determine Trigonometric Ratios TYPES OF TRIANGLES LESSON 9 : RIGHT AND OBLIQUE TRIANGLE TRIGONOMETRY According to the Length of Sides I. Equilateral Triangle – equal length of sides II. Isosceles Triangle – two sides are equal III. Scalene – no sides are equal According to Internal Angles I. Right Triangle - there exist a 90-degree angle. II. Oblique Triangle - no 90-degree angle. A. Acute Triangle - all angles are less than 90 degrees B. Obtuse Triangle – there exist an angle greater than 90 degrees. One right angle PROPERTIES OF TRIANGLE LESSON 9 : RIGHT AND OBLIQUE TRIANGLE TRIGONOMETRY DETERMINING THE EXISTENCE OF TRIANGLE TRIANGLE SUM THEOREM : The sum of the interior angles of a Euclidean triangle is always 180 degrees. METHODS IN SOLVING TRIANGLES LESSON 9 : RIGHT AND OBLIQUE TRIANGLE TRIGONOMETRY METHODS IN SOLVING TRIANGLES MAP Right Triangle Oblique Triangle Given and finding sides only Given at least one angle and side Given a pair of an angle and side No pair but given 3 elements(either sides/angles) Pythagorean Theorem Trigonometric Ratios Sine Law Cosine Law Note: Sine and Cosine Laws can also be used in Right Triangles PYTHAGOREAN THEOREM LESSON 9 : RIGHT AND OBLIQUE TRIANGLE TRIGONOMETRY PYTHAGOREAN THEOREM THEOREM 8.3 : The sum of the squares of the lengths of the legs of a right triangle (denoted by a and b) is equal to the square of the length of the hypotenuse (c). Formula : π2 + π 2 = π 2 ππ ππ ππ PYTHAGOREAN THEOREM EXAMPLE : Find the missing side. πΉπππ π₯ = βπ¦πππ‘πππ’π π π2 + π2 = π₯ 2 (6)2 + 8 2 = π₯2 36 + 64 = π₯ 2 π₯ 2 = 100 π₯ = 10 PYTHAGOREAN THEOREM EXAMPLE : Find the missing side. πΉπππ π₯ = π πππ π₯ 2 + π2 = π 2 π₯ 2 + (6)2 = (10)2 π₯ 2 + 36 = 100 π₯ 2 = 100 − 36 π₯ 2 = 64 π₯=8 TRIGONOMETRIC RATIOS LESSON 8 : RIGHT AND OBLIQUE TRIANGLE TRIGONOMETRY TRIGONOMETRIC RATIOS TRIGONOMETRIC RATIOS EXAMPLE : Determine all trigonometric ratios of the given triangle. πΉπππ π‘βπ π ππ₯ πππ ππ π‘πππππππππ‘πππ πππ‘πππ . π 8 β 17 sin πΌ = = ≈ 0.47 csc πΌ = = ≈ 2.13 β 17 π 8 π 15 cos πΌ = = ≈ 0.88 β 17 π 8 tan πΌ = = ≈ 0.53 π 15 β 17 sec πΌ = = ≈ 1.13 π 15 π sot πΌ = = 15 ≈ 1.88 π 8 α TRIGONOMETRIC RATIOS EXAMPLE : Find the missing element. πΉπππ π‘βπ πΊπ½. πππ πΉππππ’ππ βΆ sin π = βπ¦π πππ‘ π = 68° π₯ πππ = π₯ = πΊπ½ sin(68°) = 11.4 βπ¦π = 11.4 11.4 sin(68°) = π₯ π₯ ≈ 10.57 TRIGONOMETRIC RATIOS EXAMPLE : Find the missing element. πΉπππ π‘βπ π΄π΅. πππ‘ π = 57° πππ πΉππππ’ππ βΆ tan π = πππ πππ = π₯ = π΄π΅ πππ = 9.8 9.8 tan(57°) = π₯ 9.8 π₯= tan(57°) x tan(57°) = 9.8 π₯ ≈ 6.36 TRIGONOMETRIC RATIOS EXAMPLE : Find the missing element. πΉπππ π‘βπ π. πππ‘ πππ = 34.2 βπ¦π = 38.7 πππ πΉππππ’ππ βΆ sin π = βπ¦π 34.2 sin π = 38.7 −1 π = sin π = 62.09° 34.2 38.7 TRIGONOMETRIC RATIOS EXAMPLE : Find the missing element. πΉπππ π‘βπ πΌ. πππ‘ πππ = 17.1 πππ = 15.6 πππ πΉππππ’ππ βΆ tan πΌ = πππ 17.1 tan πΌ = 15.6 −1 πΌ = tan πΌ ≈ 47.63° 17.1 15.6 END OF LESSON 10 Thank you!
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