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Right Triangle Trigonometry Trigonometry is based upon ratios of the sides of right triangles. The six trigonometric functions of a right triangle, with.

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Presentation on theme: "Right Triangle Trigonometry Trigonometry is based upon ratios of the sides of right triangles. The six trigonometric functions of a right triangle, with."— Presentation transcript:

1 Right Triangle Trigonometry Trigonometry is based upon ratios of the sides of right triangles. The six trigonometric functions of a right triangle, with an acute angle , are defined by ratios of two sides of the triangle. θ opp hyp adj The sides of the right triangle are:  the side opposite the acute angle ,  the side adjacent to the acute angle ,  and the hypotenuse of the right triangle.

2 A A The hypotenuse is the longest side and is always opposite the right angle. The opposite and adjacent sides refer to another angle, other than the 90 o. Right Triangle Trigonometry

3 S O H C A H T O A The trigonometric functions are: sine, cosine, tangent, cotangent, secant, and cosecant. opp adj hyp θ sin  = cos  = tan  = csc  = sec  = cot  = opp hyp adj hyp adj opp adj Trigonometric Ratios

4 Reciprocal Functions sin  = 1/csc  csc  = 1/sin  cos  = 1/sec  sec  = 1/cos  tan  = 1/cot  cot  = 1/tan 

5 Finding the ratios The simplest form of question is finding the decimal value of the ratio of a given angle. Find using calculator: 1)sin 30= sin 30 = 2)cos 23 = 3)tan 78= 4)tan 27= 5)sin 68=

6 Using ratios to find angles It can also be used in reverse, finding an angle from a ratio. To do this we use the sin -1, cos -1 and tan -1 function keys. Example: 1.sin x = 0.1115 find angle x. x = sin -1 (0.1115) x = 6.4 o 2. cos x = 0.8988 find angle x x = cos -1 (0.8988) x = 26 o sin -1 0.1115 = shiftsin () cos -1 0.8988 = shiftcos ()

7 Calculate the trigonometric functions for . The six trig ratios are 4 3 5  sin  = tan  = sec  = cos  = cot  = csc  = cos α = sin α = cot α =tan α = csc α = sec α = What is the relationship of α and θ? They are complementary (α = 90 – θ) Calculate the trigonometric functions for .

8 Cofunctions sin  = cos (90    ) cos  = sin (90    ) sin  = cos (π/2   ) cos  = sin (π/2   ) tan  = cot (90    ) cot  = tan (90    ) tan  = cot (π/2   ) cot  = tan (π/2   ) sec  = csc (90    ) csc  = sec (90    ) sec  = csc (π/2   ) csc  = sec (π/2   )

9 Finding an angle from a triangle To find a missing angle from a right-angled triangle we need to know two of the sides of the triangle. We can then choose the appropriate ratio, sin, cos or tan and use the calculator to identify the angle from the decimal value of the ratio. Find angle C a)Identify/label the names of the sides. b) Choose the ratio that contains BOTH of the letters. 14 cm 6 cm C 1.

10 C = cos -1 (0.4286) C = 64.6 o 14 cm 6 cm C 1. h a We have been given the adjacent and hypotenuse so we use COSINE: Cos A = Cos C = Cos C = 0.4286

11 Find angle x2. 8 cm 3 cm x a o Given adj and opp need to use tan: Tan A = x = tan -1 (2.6667) x = 69.4 o Tan A = Tan x = Tan x = 2.6667

12 3. 12 cm 10 cm y Given opp and hyp need to use sin: Sin A = x = sin -1 (0.8333) x = 56.4 o sin A = sin x = sin x = 0.8333

13 Cos 30 x 7 = k 6.1 cm = k 7 cm k 30 o 4. We have been given the adj and hyp so we use COSINE: Cos A = Cos 30 = Finding a side from a triangle

14 Tan 50 x 4 = r 4.8 cm = r 4 cm r 50 o 5. Tan A = Tan 50 = We have been given the opp and adj so we use TAN: Tan A =

15 Sin 25 x 12 = k 5.1 cm = k 12 cm k 25 o 6. sin A = sin 25 = We have been given the opp and hyp so we use SINE: Sin A =

16 x = x 5 cm 30 o 1. Cos A = Cos 30 = x = 5.8 cm 4 cm r 50 o 2. Tan 50 x 4 = r 4.8 cm = r Tan A = Tan 50 = 3. 12 cm 10 cm y y = sin -1 (0.8333) y = 56.4 o sin A = sin y = sin y = 0.8333

17 Example: Given sec  = 4, find the values of the other five trigonometric functions of . Solution: Use the Pythagorean Theorem to solve for the third side of the triangle. tan  = = cot  =sin  = csc  = = cos  = sec  = = 4 θ 4 1 Draw a right triangle with an angle  such that 4 = sec  = =. adj hyp

18 A surveyor is standing 115 feet from the base of the Washington Monument. The surveyor measures the angle of elevation to the top of the monument as 78.3 . How tall is the Washington Monument? Applications Involving Right Triangles Solution: where x = 115 and y is the height of the monument. So, the height of the Washington Monument is y = x tan 78.3   115(4.82882)  555 feet.

19 Fundamental Trigonometric Identities Co function Identities sin  = cos(90    ) cos  = sin(90    ) sin  = cos (π/2   ) cos  = sin (π/2   ) tan  = cot(90    ) cot  = tan(90    ) tan  = cot (π/2   ) cot  = tan (π/2   ) sec  = csc(90    ) csc  = sec(90    ) sec  = csc (π/2   ) csc  = sec (π/2   ) Reciprocal Identities sin  = 1/csc  cos  = 1/sec  tan  = 1/cot  cot  = 1/tan  sec  = 1/cos  csc  = 1/sin  Quotient Identities tan  = sin  /cos  cot  = cos  /sin  Pythagorean Identities sin 2  + cos 2  = 1 tan 2  + 1 = sec 2  cot 2  + 1 = csc 2 


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