2. Right Triangle Trigonometry 2.1 Definition II: Right Triangle Trigonometry 2.2 Calculators and Trigonometric Functions of an Acute Angle 2.3 Solving Right Triangles 2.4 Applications 2.5 Vectors: A Geometric Approaches 1
2.1 Definition II: Right Triangle Trigonometry B If triangle ABC is a right triangle with C = 90, then the six trigonometric functions for A are defined as follows: A hypotenuse c b Side adjacent A a C Side opposite A sina = cosa = tana = cota = seca = csca = side opposite hypotenuse side adjacent hypotenuse side opposite side adjacent side adjacent side opposite hypotenuse side adjacent hypotenuse side opposite A a = c A b = c A a = A b A b = A a c = A b c = A a 2
2.1 Definition II: Right Triangle Trigonometry 1) Three pairs of cofunctions Sine and Cosine are cofunctions, so are tangent and cotangent, and secant and cosecant. 2) Cofunction Theorem A trigonometric function of an angle is equal to the cofunction of the complement of the angle. i.e. cosθ = sin(90 θ ) cotθ = tan(90 θ ) cscθ = sec(90 θ ) 3
2.1 Definition II: Right Triangle Trigonometry θ sinθ cosθ tanθ 30 45 1 3 2 2 1 2 = 2 2 2 22 1 = 3 1 = 1 3 1 60 3 2 TABLE 1. EXACT VALUES (MEMORIZE) 2 3 3 4
2.1 Definition II: Right Triangle Trigonometry (1) ΔABC is a right triangle with C = 90. Find the six trigonometric functions of A if a = 3 and b = 7 [6] (2) Find the following [12] sin A cos A tan A sin B cos B tan B A 5 C 1 B 5
2.1 Definition II: Right Triangle Trigonometry (3) Use the Cofunction theorem to fill in the blanks so that each expression becomes a true statement sin y = cos [26] cot(12 ) = tan [24] (4) Simplify. (sin30 + cos30 ) 2 [35 modified] sin 2 30 + cos 2 30 [36 modified] (5) Find exact value: csc30 [50] cot30 [54] 6
2.2 Calculators and Trig Functions of an Acute Angle 1) Degree, minute and second (DMS) and decimal degree 2) Convert between DMS and decimal degree 3) Add or subtract angles 4) Using calculator 5) Using inverse trig function 7
2.2 Calculators and Trig Functions of an Acute Angle 1) Degree, minute and second (DMS) and decimal degree Expression 52 10 13 24 15 Read as 52 degrees, 10 minutes 13 degrees, 24 minutes, 15 seconds DMS Expression Read as 27.36 27 point 36 degrees Decimal Degree 8
2.2 Calculators and Trig Functions of an Acute Angle 2) Convert between DMS and decimal degree 1 = 60 or 1 = 1 = 60 or 1 = (1) Change to degree and minutes: 18.75 [18] ans. 18 45 (2) Change to decimal points: 21 48 [26] ans. 21.8 1 60 1 60 o 9
2.2 Calculators and Trig Functions of an Acute Angle (3) Add or subtract 63 38 + 24 52 [4] 90 (62 25 ) [8] (4) Use calculate to find. Round the answer to four decimal places. [32, 36] cos 82.9 = tan 81.43 = (5) Use calculate to find (convert to degree decimal). Round the answer to four decimal places. [44, 50] sin 35 10 Sec 84 48 ans. 88 30 ans. 27 35 ans. 0.1236 ans. 0.6357 ans. 0.4760 ans. 11.0336 10
2.2 Calculators and Trig Functions of an Acute Angle (6) Inverse trig function (more later, similar for sine, cosine) tanθ = 3.152. Find θ. [e.g.10] ans. 72.4 11
2.3 Solving Right Triangles 1) Number of significant digits 2) Accuracy of side and accuracy of angle. 3) Solving right triangles 12
2.3 Solving Right Triangles 1) Number of significant digits Definition. The number of significant digits in a number is found by counting all the digits from left to right with the first non-zero digit on the left. When no decimal point is present, trailing zeros are not considered significant. 0.042 has two significant digits 0.005 has one significant digit 20.5 has three significant digits 6.000 has four significant digits 9,200. has four significant digits 700 has one significant digit 13
2.3 Solving Right Triangles 2) Accuracy of side and accuracy of angle. The relationship between the accuracy of the sides of a triangle and the accuracy of the angles in the same triangle. Accuracy of Sides Two significant digits Three significant digits Four significant digits Accuracy of angles Nearest degree Nearest tenth of a degree Nearest hundredth of a degree 14
2.3 Solving Right Triangles (1) ΔABC is a right triangle with C = 90 If B = 16.9 and c = 7.55 cm, find b [6] ans. 2.19 cm If a = 42.3 in and b = 32.4 in, find B [10] ans. 37.5 (2) If C = 26 and r = 19, find x [36] ans. 2.1 C r 26 r B x D A 15
2.3 Solving Right Triangles (3) ΔABC is a right triangle with C = 90 [46] If A = 32, BDC = 48 and AB = 56. Find h, then x. B ans. h = 56sin(32 ) = 30 h ans. x = h/tan(48 ) = 27 A y D x C 16
2.4 Applications 1) Geometry isosceles triangle 2) Angle of elevation and angle of depression 3) Distance and bearing 17
2.4 Applications 1) Geometry isosceles triangle e.g.1 Two equal sides of an isosceles triangle are each 12 centimeters. If each of the two equal angles measures 52, find the length of the base and the altitude. C A 24 52 b h 24 52 B ans. h = 24sin(52 ) = 19 cm ans. b = 2 24cos(52 ) = 30 cm 18
2.4 Applications 2) Angle of elevation and angle of depression An angle measured from the horizontal up is called angle of elevation; an angle measured from the horizontal down is called angle of depression. Angle of elevation horizontal horizontal Angle of elevation 19
2.4 Applications 2) Angle of elevation and angle of depression Angle of elevation. If the angle of elevation of the Sun is 63.4 when a building casts a shadow of 37.5 ft, what is the height of the building? [10] ans. h = 37.5tan(63.4 ) = 74.9 ft h 63.4 37.5 ft 20
3) Distance and bearing 2.4 Applications Definition. The bearing of a line l is the acute angle formed by the north-south line and the line l. The notation used to designate the bearing of a line begins with N or S (North or South), followed by number of degrees in the angle, and ends with E or W (East or West) 21
2.4 Applications 3) Distance and bearing four scenarios N N B W The bearing Of B from A Is N40 E A 40 E W B 65 A The bearing Of B from A Is N65 W E S S N N W The bearing A Of B from A Is S70 E 70 B E W A E The bearing 20 Of B from A Is S20 W B S S 22
3) Distance and bearing 2.4 Applications A man wandering in the desert walks 2.3 miles in the direction of S31 W. He then turns 90 and walks 3.5 mile in the direction of N59 W. At the same time, how far is he from his starting point, and what is his bearing from his starting point? [18] x θ 31 ans. x = 4.2 mi 59 ans. θ = 88 23
2.5 Vectors: A Geometric Approach 1) Vector 2) Notation of vector 3) Equality for vectors 4) Addition and subtraction of vectors 5) Horizontal and vertical components of a vector 6) applications 24
1) Vector 2.5 Vectors: A Geometric Approach Magnitude direction 2) A zero vector is a vector with its magnitude equals to zero 3) Two vectors are equal if they have the same magnitude and the same direction. 25
2.5 Vectors: A Geometric Approach Notation The quantity is v a vector (boldface) v a vector (arrow above the variable) AB a vector x a scalar v the magnitude of vector v, a scalar 26
2.5 Vectors: A Geometric Approach 4) Addition and Subtraction of Vectors: given u and v find u + v and u v (Parallelogram Rule) v v u u + v v v v u u v 27
2.5 Vectors: A Geometric Approach 5) Horizontal and vertical components of a vector Vector v is in standard position if the tail of the vector is at the origin. y v y is the vertical component of v v y Vector v in standard position v x v x v x is the horizontal component of v 28
2.5 Vectors: A Geometric Approach 6) Application Draw a vector representing a velocity of 50 cm/sec at N30 E [6] N v 30 W A E S 29
2.5 Vectors: A Geometric Approach 6) Application A person is riding in a hot air balloon. For the first hour, the wind current is a constant 9.50 mph at N37.5 E. Then the wind changes to 8.00 mph and heads the balloon in the direction S52.5 E. This continues for 1.5 hours. How far is the balloon from its starting position? [9] 37.5 θ 52.5 d ans. d = 15.3 mi ans. θ = arctan(12/9.5) = 51.6 ans. bearing = 51.6 + 37.5 = 89.1 30
2.5 Vectors: A Geometric Approach 6) Application A bullet is fired into the air with an initial velocity of 1,800 ft/sec at 60 from the horizontal. Find the magnitudes of the horizontal and vertical components of the velocity. [24] v y N v ans. v x = 1800cos(60 ) = 900 ft/sec W A 60 v x E ans. v y = 1800sin(60 ) = 1600 ft/sec S 31
2.5 Vectors: A Geometric Approach 6) application A ship travels in direction S12 E, for 68 miles and then changes its course to S60 E and travels another 110 miles. Find the total distance south and the total distance east that the ship traveled. [32] distance south: 12 60 68cos(12 ) + 110cos(60 ) = 120 mi distance east: 68sin(12 ) + 110sin(60 ) = 110 mi 32