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APPS 4 US MATHS Laia Erra, Irene Lama, Aina Gràcia 1BAT1 Theory: TRIGONOMETRY -HOW TO MEASURE THE ANGLES Degree (angle) 1º= 60’ 1’=60’’ Gradiant 1º= 100’ 1‘=100’’ Radians (PI)radians= 180 degrees What is a radian? A radian is the angle wich its single arc is equal to its radiant. 1 radian= 57’296º Arc Radian Radius Equivalences 0º = 0 30º = π/6 45º = π/4 60º = π/3 90º = π/2 120 = 2π/3 180 = π 270 = 3π/2 360º = 2π Internal angles Degrees of a triangle: 180º Degrees: 360º 5 sides 360º/5 = 72º 180º - 72º= 108º 108/2 = 54º 54º x 10 external angles = 540º Trigonometric reasons Trigonometric reason sen a Definition opposite ray to a/hypotenuse cos a contiguous ray to a/hypotenuse tang a opposite ray to a/contiguous ray to a ctg a 1/tang a sec a cosc a 1/cos a 1/sen a Sen and Cos theorem Signs of the trigonometric reasons Trigonometric reason Sign Complementary angles Supplementary angles Addition and subtraction of angles Double angle Half angle REMEMBER Pythagoras sin2 a + cos2 a = 1 1 + tag2 a = sec2 a 1 + ctg2 a = cosc2 a Sines theorem A/sin a = B/sin b = C/sin c Cosines theorem A2 = B2 + C2 – 2BC x cos a B2 = A2 + C2 – 2AC x cos b C2 = A2 + B2 – 2AB x cos c Activities cos(t)+sin(t) cos(2t) ------------- = ------------cos(t)-sin(t) 1- sin(2t) sin(a)+sin(b)+sin(c) --------------------- = tan(b) cos(a)+cos(b)+cos(c) In the figure we see a triangle and a hexagon inscribed in the same circle. Calculate the ratio of the areas of the two figures. Solution cos(2t) ---------- = 1- sin(2t) sin(b-v)+sin(b)+sin(b+v) ------------------------- = tan(b) <=> cos(b-v)+cos(b)+cos(b+v) cos2(t)-sin2(t) --------------- = 1-2sin(t)cos(t) 2sin(b)cos(v)+sin(b) --------------------- = tan(b) <=> 2cos(b)cos(v)+cos(b) (cos(t)-sin(t))(cos(t)+sin(t)) ------------------------------------------ = cos(t)cos(t)-2sin(t)cos(t)+sin(t)sin(t) (cos(t)-sin(t))(cos(t)+sin(t)) ------------------------------- = (cos(t)-sin(t))(cos(t)-sin(t)) cos(t)+sin(t) -----------cos(t)-sin(t) sin(b)(2cos(v)+1) ----------------- = tan(b) <=> cos(b)(2cos(v)+1) sin(b) ------- = tan(b) cos(b) 1. 2. First Method: The ratio of the areas of the two figures is independent of the radius of the circle. We can therefore assume that the radius is 1. 3. Area of the hexagon: 4. Connect the center of the circle with the vertices. There arise 6 equilateral triangles with side 1. The area of the hexagon is 6.(1/2).1.1.sin(60°) = 3.sin(60°) 5. Area of the triangle: 6. Connect the center of the circle with the vertices. There arise 3 isosceles triangles. The area of the triangle is 3.(1/2).1.1.sin(120°) = (3/2) sin(60°) 7. We see that the ratio of the areas of the two figures is 2. 8. Second method : 9. The ratio of the areas of the two figures is independent of the mutual position of the polygons. We rely on this property to put the triangle in a favorable position. We connect the center with the vertices of the triangle. 10. There are 6 congruent triangles. Then it is immediately clear that the ratio of the areas of the two figures is 2. Activities A person is 100 meters from the base of a tree, he observes that the angle between the ground and the top of the tree is 18 degrees. Estimate the height (h) of the tree. ● ● ● ● . SOLUTION: Use the tangent tan(18o) = h / 100 Solve for h to obtain h = 100 tan(18o) = 32.5 meters. Activities The angle of elevation of a hot air balloon, climbing vertically, changes from 25 degrees at 10:00 am to 60 degrees at 10:02 am. The point of observation of the angle of elevation is situated 300 meters away from the take off point. What is the upward speed, assumed constant, of the balloon? Give the answer in meters per second and round to two decimal places. ● ● ● ● ● ● ● ● ● ● ● ● ● Use the tangent to write tan(25o) = h1 / 300 and tan(60o) = (h1 + h2) / 300 Solve for h1 and h2 h1 = 300 tan(tan(25o)) and h1 + h2 = 300 tan(60o) Use the last two equations to find h2 h2 = 300 [ tan(60o) - tan(25o) ] If it takes the balloon 2 minutes (10:00 to 10:02) to climb h2, the the upward speed S is given by S = h2 / 2 minutes = 300 [ tan(60o) - tan(25o) ] / (2 * 60) = 3.16 m/sec Activities BH is perpendicular to AC. Find x the length of BC. ABC is a right triangle with a right angle at A. Find x the length of DC. Solution 1. BH perpendicular to AC means that triangles ABH and HBC are right triangles. Hence 2. tan(39o) = 11 / AH or AH = 11 / tan(39o) 3. 4. 1. Since angle A is right, both triangles ABC and ABD are right and therefore we can apply Pythagora's theorem. 2. 142 = 102 + AD2 , 162 = 102 + AC2 3. Also x = AC - AD 4. = sqrt( 162 - 102 ) - sqrt( 142 - 102 ) = 2.69 (rounded to 3 significant digits) HC = 19 - AH = 19 - 11 / tan(39o) Pythagora's theorem applied to right triangle HBC: 112 + HC2 = x2 5. solve for x and substitute HC: x = sqrt [ 112 + (19 - 11 / tan(39o) )2 ] 6. = 12.3 (rounded to 3 significant digits) Activities From the top of a 200 meters high building, the angle of depression to the bottom of a second building is 20 degrees. From the same point, the angle of elevation to the top of the second building is 10 degrees. Calculate the height of the second building. Solution 1. 2. 3. 4. 5. 6. 7. The diagram below show the two buildings and the angles of depression and elevation. tan(20o) = 200 / L L = 200 / tan(20o) tan(10o) = H2 / L H2 = L * tan(10o) = 200 * tan(10o) / tan(20o) Height of second building = 200 + 200 * tan(10o) / tan(20o) Activities In a right triangle ABC with angle A equal to 90o, find angle B and C so that sin(B) = cos(B). SOLUTION: 1.Let b be the length of the side opposite angle B and c the length of the side opposite angle C and h the length of the hypotenuse. 2.sin(B) = b/h and cos(B) = c/h 3.sin(B) = cos(B) means b/h = c/h which gives c = b 4.The two sides are equal in length means that the triangle is isosceles and angles B and C are equal in size of 45 o. Activities Calculate the length of the side x, given that tan θ = 0.4 Calculate the length of the side x, given that sin θ = 0.6 Solution: Solution: Using Pythagoras’ theorem: Examples The angle of elevation is always measured from the ground up. Think of it like an elevator that only goes up. It is always INSIDE the triangle. In the diagram at the left, x marks the angle of elevation of the top of the tree as seen from a point on the ground. You can think of the angle of elevation in relation to the movement of your eyes. You are looking straight ahead and you must raise (elevate) your eyes to see the top of the tree. Examples The angle of depression is always OUTSIDE the triangle. It is never inside the triangle. In the diagram at the left, x marks the angle of depression of a boat at sea from the top of a lighthouse. You can think of the angle of depression in relation to the movement of your eyes. You are standing at the top of the lighthouse and you are looking straight ahead. You must lower (depress) your eyes to see the boat in the water. Activities You are walking up a 500. meter high hill. The trail has an incline of 12 degrees. How far will you walk to get to the top? Activities Calculating radius of the outer core The S wave shadow zone is caused by the outer core not transmitting S waves. It crosses an arc of 105 degrees on the Earth (see the diagram on the left). Estimate the radius of the outer core. The radius of the entire Earth is 6370 km. A= cos(52.5)*6370km so A= 3877 km Activities The angle of repose for sand is typically about 35°. What is the sine of this angle? SOLUTION: 1.Type 35 into your calculator 2.press the sin button. 3.Your calculator should read 0.574. When driving, a steep hill is typically only 12°. What is the cosine of this angle? SOLUTION: 1.Type 12 into your calculator