Section 10.6 Right Triangle Trigonometry

Transcription

1 153 Section 10.6 Right Triangle Trigonometry Objective #1: Understanding djacent, Hypotenuse, and Opposite sides of an acute angle in a right triangle. In a right triangle, the otenuse is always the longest side; it is the side opposite of (or "across from") the right angle. The otenuse of the following triangles is highlighted in gray. The acent side of an acute angle is the side of the triangle that forms the acute angle with the otenuse. The acent side is the side of the angle that is "acent" to or next to the otenuse. The acent side depends on which of the acute angles one is using. The acent side of the following angles (labeled ) is highlighted in gray. The opposite side of an acute angle is the side of the triangle that is not a side of the acute angle. It is "across from" or opposite of the acute angle. The opposite side depends on which of the acute angles one is using. The opposite side of the following angles (labeled ) is highlighted in gray.

2 154 For each acute angle in the following right triangles, identify the acent, otenuse, and opposite side: R Ex. 1 B Ex. 2 Q C For, the acent side is C, the opposite side is BC, and the otenuse is B. For B, the acent side is BC, the opposite side is C, and the otenuse is B. X For R, the acent side is QR, the opposite side is QX, and the otenuse is RX. For X the acent side is QX, the opposite side is QR, and the otenuse is RX. Objective #2: Understanding Trigonometric Ratios or Functions. Recall that two triangles are similar if they have the same shape, but are not necessarily the same size. In similar triangles, the corresponding angles are equal. Thus, if BC HET, then m = m H, m B = m E, and m C = m T. The corresponding sides are not equal. However, the ratios of corresponding sides are equal since one triangle is in proportion to the other triangle. Thus, if BC HET, then B E B BC C = = HE ET HT Now, let's consider the proportion B HE = BC ET. If we cross multiply, C we get (B) (ET) = (HE) (BC). Now, H T divide both sides by (BC) (ET) and simplifying, we get: (B) (ET) (HE) (BC) B HE = = (BC) (ET) (BC) (ET) BC ET

3 155 Now consider angles C and T. B E If and H are right angles, then the proportion B BC = HE ET implies that the ratio of the opposite side to the otenuse is the same for any right triangle if the corresponding acute C angles have the same measure. H T We will define this ratio as the Sine Ratio or Sine Function. The abbreviation for the sine function is "sin", but it is pronounced as "sign". Sine Ratio or Function Given an acute angle in a right triangle, then the sine ratio of, is the ratio of the opposite side of to the otenuse. sin = oppositeside otenuse = opp For each acute angle in the following right triangles, find the sine of the angle: 16.1 in Ex. 3 Ex cm 42.0 cm 19.2 in 25.0 in cm sin 35 = opp = 24.1 sin 40 = opp 42.0 = = = sin 55 = opp = 34.4 sin 50 = opp 42.0 = = = Since the opposite is never greater than the otenuse, then the value of the sine function cannot exceed 1. If the opposite side is equal to the otenuse, then the measure of the angle is 90 and the triangle degenerates into a vertical line. This means that sin 90 = 1. Similarly, if the opposite side has zero length meaning the sine ratio is 0, then the measure of the angle is 0 and the triangle degenerates into a horizontal line. Hence, sin 0 = 0. opp

4 We have seen that if the corresponding acute angles of rights triangles have the same measure, the ratio of the opposite side to the otenuse is fixed. We can make the same argument that the ratio of the acent side to the otenuse is also fixed. We will define this ratio as the Cosine Ratio or Cosine Function. The abbreviation for the cosine function is "cos", but it is pronounced as "co-sign". 156 Cosine Ratio or Function Given an acute angle in a right triangle, then the cosine ratio of, is the ratio of the acent side of to the otenuse. cos = acentside otenuse = opp For each acute angle in the following right triangles, find the cosine of the angle: 16.1 in Ex. 5 Ex cm 42.0 cm 19.2 in 25.0 in cm cos 35 = = 34.4 cos 40 = 42.0 = = = cos 55 = = 24.1 cos 50 = 42.0 = = = Since the acent is never greater than the otenuse, then the value of the cosine function cannot exceed 1. If the acent side is equal to the otenuse, then the measure of the angle is 0 and the triangle degenerates into a horizontal line. This means that cos 0 = 1. Similarly, if the acent side has zero length meaning the cosine ratio is 0, then the measure of the angle is 90 and the triangle degenerates into a vertical line. Hence, cos 90 = 0. We have seen that if the corresponding acute angles of rights triangles have the same measure, the ratio of the opposite side to the otenuse is

5 157 fixed and the ratio of the acent side to the otenuse is fixed. We can make the same argument that the ratio of the opposite side to the acent side is also fixed. We will define this ratio as the Tangent Ratio or Tangent Function. The abbreviation for the tangent function is "tan", but it is pronounced as "tangent". Tangent Ratio or Function Given an acute angle in a right triangle, then the tangent ratio of, is the ratio of the opposite side of to the acent side of. tan = oppositeside acentside = opp opp For each acute angle in the following right triangles, find the tangent of the angle: 16.1 in Ex. 7 Ex cm 42.0 cm 19.2 in 25.0 in cm tan 35 = opp = 24.1 tan 40 = opp 34.4 = = tan 55 = opp tan 50 = opp = = = = = When the opposite side has zero length, then the measure of the angle is zero. The tangent ratio is the ratio of 0 to the acent side, which is equal to zero. Hence, tan 0 = 0. Similarly, if the acent side has zero length, then the measure of the angle is 90. The tangent ratio is the ratio of the opposite side to 0, which is undefined. Thus, tan 90 is undefined. Unlike the sine and cosine function, the values of the tangent function can exceed one. There are some memories aids that you can use to remember the trigonometric ratios. If you remember the order sine, cosine, and tangent, then you might find one of these phrases helpful:

6 158 Trigonometric Ratio sin = cos = tan = Ratio of sides opp opp Rated G Phrase oscar had a heap of apples Rated PG-13 Phrase oh h ll another hour of algebra If you want to include the trigonometric functions themselves, you can use the following phrases: s-o-h-c-a-h-t-o-a s o h c a h t o a i p y o d y a p d n p p s j p n p j e o o i a o g o a s t n c t e s c i e e e e n i e t n n n t t n e u t u e t s s e e sinful oscar had consumed a heaping taste of apples sinful oscar had consumed a heaping taste of apples. i p y o d y a p d n p p s j p n p j e o o i a o g o a s t n c t e s c i e e e e n i e t n n n t t n e u t u e t s s e e

7 Objective #3: Evaluating trigonometric ratios on a scientific calculator. On a scientific calculator, SIN key is for the sine ratio, COS key is for the cosine ratio, and TN key is for the tangent ratio. To evaluate a trigonometric ratio for a specific angle, you first want to be in the correct mode. If the angle is in degree, then the calculator needs to be in degree mode. Hit the appropriate trigonometric ratio, type the angle and hit enter. Evaluate the following: 159 Ex. 9a cos 26 Ex. 9b tan 78.3 Ex. 9c sin a) We put our calculator in degree mode, hit the COS key, type 26 and hit enter: cos 26 = b) We put our calculator in degree mode, hit the TN key, type 78.3 and hit enter: tan 78.3 = c) We put our calculator in degree mode, hit the SIN key, type and hit enter: sin 16 2 = Objective #4: Finding ngles using Inverse Trigonometric Ratio. If we know the ratio of two sides of a right triangle, we can use inverse trigonometric ratios or functions to find the angle. The inverse trigonometric function produces an angle for the answer. Inverse Trigonometric Functions: Given a right triangle, we can find the acute opp angle using any of the following: 1) Inverse sine (arcsine) = sin 1 ( opp ) 2) Inverse cosine (arccosine) = cos 1 ( ) 3) Inverse tangent (arctangent) = tan 1 ( opp ) On a scientific calculator, these functions are labeled SIN 1, COS 1 and TN 1. They are usually directly above the SIN, COS, and TN keys. To

8 find the angle on the calculator, first set your calculator in degree mode. Second, hit the 2nd (or INV) and then the appropriate trigonometric ratio key followed by the ratio of the two sides you are given and hit enter. Solve the following. Round to the nearest tenth: Ex. 10a sin x = Ex. 10b cos x = Ex. 10c tan x = 1.82 a) Since sin x = 0.833, then x = sin 1 (0.833). Hit 2nd SIN, type and hit enter: sin 1 (0.833) 56.4 b) Since cos x = 0.528, then x = cos 1 (0.528). Hit 2nd COS, type and hit enter: cos 1 (0.528) 58.1 c) Since tan x = 1.82, then x = tan 1 (1.82). Hit 2nd TN, type 1.82 and hit enter: tan 1 (1.82) 61.2 Objective #5: Finding the missing sides and angles of a right triangle. 160 Given either two sides or a side and one of the acute angles of a right triangle, we can use our trigonometric ratios and the Pythagorean Theorem to solve for the missing sides and angles of a right triangle. Finding the missing sides and angles of the following. Round all sides and angles to the nearest tenth: E Ex. 11 B Ex ft C F 13 in G Since and B are Since E and G are complimentary, then complimentary, then m B = = 67. m E = = 18.5 = 18. We have the otenuse We have the acent side and and the acute angle. We and the acute angle G. We can can use the sine of 23 to use the tangent of 71.5 to find find BC and the cosine of the EF and the cosine of to find C. to find EG.

9 sin 23 = opp = BC 18 tan 71.5 = opp = EF 13 To solve, multiply by 18: To solve, multiply by 13: BC = 18 sin 23 EF = 13 tan 71.5 = = 7.03 = = ft 38.9 in cos 23 = = C cos 71.5 = 18 = 13 EG To solve for BC, multiply To solve for EG, multiply by 161 by 18: EG and then divide by cos 71.5 : C = 18 cos 23 EG cos 71.5 = 13 = = EG = 13/cos 71.5 =13/ ft = in Thus, BC 7.0 ft. C 16.6 ft Hence, EF 38.9 in, EG 41.0 in, and m B = 67. and m E = Ex. 13 C 7 cm Ex. 14 B 5 cm We have the two legs of a We have the leg and the right triangle. We can use otenuse of a right triangle. the inverse tangent function we can use the inverse cosine to to find B:. to find G. tan B = opp = 5 cos G = 7 = Thus, B = tan 1 ( 5 7 ) 35.5 Hence, G = cos 1 ( ) 47.7 Since and B are Since E and G are complimentary, then complimentary, then m = = 54.5 m E = = 42.3 To find the length B, we can To find the length EF, we can use the Pythagorean Theorem: use the Pythagorean Theorem: (B) 2 = (7) 2 + (5) 2 (10.1) 2 + (EF) 2 = (15) 2 (B) 2 = (EF) 2 = 225 (B) 2 = subtract from both sides E F 15 m 10.1 m G

10 162 (B) 2 = 74 (EF) 2 = B = 74 = cm EF = = m Thus, m B 35.5, Hence, m G 47.7, m 54.5, and B 8.6 cm. m E = 42.3, and EF 11.1 m. Objective #6: Solving applications involving right triangles. The key to solving applications involving right triangles is drawing a reasonable diagram to represent the situation. Next, we need determine what information we are given and what information we need to find. Finally, we then use the appropriate trigonometric function to solve the problem. Solve the following: Ex. 15 surveyor measures the angle between her instruments and the top of a tree and finds the angle of elevation to be If her instruments are 4 ft high off of the ground and are 25 feet away from the center of the base of the tree and the ground is leveled, how high is the tree? We begin by drawing the surveyor's instruments and the tree. Since her instruments are 4 ft above the ground, we can draw a horizontal line from her her instruments to four feet above the center of the base of the tree. Drawing a line from the top of the tree to her instruments and a vertical line down from the top of the tree four feet above the center of the base of the tree, we get a right triangle. The acent side is 25 feet and we are looking for the opposite of the triangle. Once we get the opposite side, we will need to add 4 ft to the answer to find the height of the tree. The trigonometric ratio that uses the opposite and acent sides is the tangent function. Hence, tan 65.7 = opp = x feet x = 25 tan 65.7 = = ft Now, add 4 ft to find the height of the tree: = ft So, the tree is approximately 59.4 ft tall. (multiply by 25 to solve for x) x

11 Ex. 16 helicopter flying directly above an ocean liner at a height of 1400 ft, locates another boat at a 23 angle of depression. How far apart are the two ships? Round to the nearest hundred feet. The angle of depression is the angle between a horizontal line and the line of observation. In other words, it is the angle one has to look down to see the observed object, which is a ship in this case. The helicopter is flying directly above the ocean liner. We can draw a vertical line between the helicopter and the ocean liner. Next we can draw a line from the helicopter to the other ship. The angle between this two lines and the angle of depression are complimentary angles. Thus, the angle between the two lines is = 67. Now, drawing a line between the two ships creates a right triangle. The vertical height is the acent side of 67 and we are looking for the opposite side of 67. So, we will need to use the tangent function. tan 67 = opp = x ft (multiply by 1400 to solve for x) x = 1400 tan 67 = = ft 3300 ft So, the ships are approximately 3300 ft apart. x 163 Ex. 17 For a wheel chair accessible ramp to be D compliant, the ratio of the rise to the run of the ramp must be 1 to (12 or greater). 29 ft 9 in long ramp leading into a building climbs 2 ft 5 in. a) Is this ramp D compliant? b) If a new city ordinance states that the incline of a handicapped access can be no more than 4.5, does this ramp comply with the new ordinance? If not, how long does the ramp need to be? Round to the nearest inch. a) We will begin by drawing a picture: 23 Length ngle of Elevation Run Rise

12 164 The Length of the ramp corresponds to the otenuse, and the Rise and Run correspond to the two legs. Thus, we will need convert the measurements into inches and use the Pythagorean theorem to calculate the run of the ramp: Rise = 2 ft + 5 in = 2 ft 1 ( 12in 1ft ( 1 Length = 29 ft + 9 in = 29ft ) + 5 in = 24 in + 5 in = 29 in ) + 9 in = 348 in + 9 in = 357 in 12in 1ft (Run) 2 = (Length) 2 (Rise) 2 = (357) 2 (29) 2 = = Run = = in The ratio of the Rise to the Run is 29 to Dividing by 29, we get: 1 to Since , the ramp is D compliant. b) Since the ramp climbs 29 in, then 357 in the height of the ramp is 29 in. This x 29 in will correspond to the opposite side of the incline. The length of the ramp is 357 in. This will correspond to the otenuse. This means we will need to use the inverse sine function to find the angle. sin x = opp = (use the inverse sine) x = sin 1 ( 29 ) = which is greater than Thus, the current ramp does not comply with the ordinance. Since the ramp climbs 29 in, then y the height of the ramp is 29 in. This in will correspond to the opposite side of the incline. The length of the ramp is length of the ramp which is the otenuse. This means we will need to use the sine function. sin 4.5 = opp = 29 (multiply by y) y y sin 4.5 = 29 y = 29 = 29 sin (divide by sin 4.5 ) = in But, = 30 with a remainder 10, so 370 in = 30 ft 10 in The ramp should be 30 ft 10 in to comply with the ordinance. Note: nother requirement for a ramp to D compliant is that is one 5 ft by 5 ft rest platform is required after a run of 30 feet in a ramp.