Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Trigonometry Pythagoras Theorem & Trigo Ratios of Acute Angles Pythagoras Theorem a 2 + b2 = c 2 where c is the hypotenuse while a and b are the lengths of the other two sides. c a b Trigo Ratios of Acute angles P hypotenuse opposite O adjacent Q Hypotenuse = side opposite right angle/longest side Adjacent = side touching theta Opposite= side opposite theta Trigo Ratios of Acute angles Hypotenuse = AB A Adjacent = AC Opposite= BC B C X Hypotenuse = XZ Y Adjacent = XY Opposite= YZ Z Trigo Ratios of Acute angles P hypotenuse opposite O adjacent Tangent ratio tan Q Cosine ratio Sine ratio cos sin Trigo Ratios of Acute angles P hypotenuse opposite O adjacent Q opposite PQ tan adjacent OQ adjacent OQ cos hypotenuse OP opposite PQ sin hypotenuse OP Trigo Ratios of Acute angles P hypotenuse opposite O adjacent TOA CAH SOH Q opposite PQ tan adjacent OQ adjacent OQ cos hypotenuse OP opposite PQ sin hypotenuse OP Exercise 1 13 5 sin opposite 5 hypotenuse 13 cos adjacent 12 hypotenuse 13 tan 12 opposite 5 adjacent 12 opposite 3 sin hypotenuse 5 5 3 4 adjacent 4 cos hypotenuse 5 opposite 3 tan adjacent 4 Exercise 1 sin opposite 12 4 hypotenuse 15 5 cos adjacent 9 3 hypotenuse 15 5 12 9 15 tan opposite 45 15 sin hypotenuse 51 17 24 51 opposite 12 4 adjacent 9 3 45 adjacent 24 8 cos hypotenuse 51 17 opposite 45 15 tan adjacent 24 8 Exercise 2 3 sin A 5 5 A cos A 4 5 tan A 3 4 3 4 8 cos B 17 17 B sin B 15 17 tan B 15 8 sin C 21 29 cos C 20 29 15 8 21 tan C 20 29 C 20 21 Exercise 3 sin 60 0.866 cos 60 0.5 tan 45 1 sin 0.5 0.00873 cos 30 0.866 tan 23 0.424 Exercise 4 sin (30° 2) = 0.2588… sin ( 2) = sin 2 cos 2 = 2 cos sin 30° 2 = 0.25 cos (2× 30°) = 0.5 2× cos 30° = 1.732… tan (A + B) = tan A + tan B tan (10° + 30°) = 0.839… tan 10° + tan 30° = 0.75 Exercise 5 sin = 0.4537 = sin-1 0.4537 = 26.981≈27.0° cos = 0.3625 = cos-1 0.3625 = 68.746≈68.7° tan = 4.393 = tan-1 4.393 = 77.176≈77.2° Exercise 5 sin = 0.8888 = sin-1 0.8888 = 62.722≈62.7° cos = 0.9999 = cos-1 0.9999 = 0.8102≈0.8° tan = 0.5177 = tan-1 0.5177 = 27.370≈27.4° Further Examples 1 In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate B 7 cm A C 8 cm angle CED, angle DCB, angle BAC, the length of ED, the length of AE, 54.8° D 8 cm E Further Examples 1 In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate B 7 cm A C 8 cm angle CED? 54.8° D 8 cm angle CED = 180° − 90° − 54.8° = 35.2° E Further Examples 1 In the diagram, BCE is a straight line, angle ECD = 54.8° and angle CDE = angle ACB = 90°. BC = 7 cm and AC = CE = 8 cm. Calculate B 7 cm A C 8 cm angle DCB? 54.8° D 8 cm angle DCB = 180° − 54.8° = 125.2° E Further Examples 1 angle BAC? B Let angle BAC be . opposite 7 sin hypotenuse ? 7 cm A C 8 cm adjacent 8 cos hypotenuse ? 54.8° D 8 cm opposite 7 tan adjacent 8 E 7 tan 41.185... 41.2 8 1 Further Examples 1 the length of ED? B opposite ED sin 54.8 hypotenuse 8 adjacent CD cos 54.8 hypotenuse 8 7 cm A C 8 cm 54.8° D 8 cm opposite ED tan 54.8 adjacent CD ED sin 54.8 8 ED 8 sin 54.8 6.537... 6.54cm E Further Examples 1 the length of AE? B opposite sin? hypotenuse adjacent cos? hypotenuse 7 cm A C 8 cm 54.8° D 8 cm opposite tan? adjacent AE 8 8 2 2 2 AE 128 11.3137... 11.3cm E Further Examples 2 A 16 m ladder is leaning against a house. It touches the bottom of a window that is 12 m above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be . sin 12 16 cos ? 16 12 tan ? 16 m 12 m Further Examples 3 A 16 m ladder is leaning against a house. It touches the bottom of a window that is 12 m above the ground. What is the measure of the angle that the ladder forms with the ground? Let the angle be . sin 12 16 12 sin 48.590... 48.6o 16 1 16 m 12 m Exercise 6 In the diagram, angle ADC = 30°, angle ACB = 50°, angle ABD = 90° and BC = 4 cm. Calculate (a) angle DAC A D 50° 30° C 4 cm B Applications – Angle of elevation and Angle of depression Applications – Angle of elevation and Angle of depression Example 1 opposite PQ adjacent OQ adjacent OQ cos hypotenuse OP opposite PQ sin hypotenuse OP tan c tan 35 200 c 200 tan 35 140 .04... 140 m Example 2 A surveyor is 100 meters from the base of a dam. The angle of elevation to the top of the dam measures . The surveyor's eye-level is 1.73 meters above the ground. Find the height of the dam. h 100 tan 26 1.73 50.5m Trigonometric Ratios of Special Angles: 30°, 45° and 60°. 2 2 1 60 45 o 1 cos 45 2 1 o sin 45 2 o tan 45 1 o 3 1 1 o 30 o 1 cos 60 2 o 3 sin 60 2 tan 60o 3 o 1 3 cos 30 2 1 o sin 30 2 1 o tan 30 3 o Trigonometric Ratios of Complementary Angles. Q c b P R a b cos(90 ) c a o sin(90 ) c a o tan(90 ) b adjacent a cos hypotenuse c opposite b sin hypotenuse c opposite b tan adjacent a o 1 cos(90 ) sin , sin(90 ) cos , tan(90 ) tan o o o At the point P, a boat observes that the angle of elevation of the cliff at point T is 32o, and the distance PT is 150m. It sails for a certain distance to reach point Q, and observes that the angle of elevation of the point T becomes 48o. T 150m 48o 32o P Q R (i) Calculate the height of the cliff. (ii) Calculate the distance the boat is from the cliff at point Q. (iii) Calculate the distance travelled by the boat from point P to point Q. T 150m 48o 32o P Q Let the height of the cliff = TR TR sin 32 150 TR 150 sin 32 79.5m R T 150m 48o 32o P Q R Let the distance the boat is from the cliff at point Q = QR TR tan 48 QR TR QR tan 48 TR QR 71.6m tan 48 T 150m 48o 32o P Q R Let the distance travelled by the boat from point P to point Q = PQ TR tan 32 PR TR PR tan 32 TR PR tan 32 PQ PR QR 127.207 71.571 55.6m Q2 3m 1.3 m Let the angle be . sin 1.3 3 1.3 sin 3 25.679... 1 25.7o Q3 In 15 Secs, distance travelled = 140 x 15 = 2100 m 2100 m 10 ° Plane Let the altitude be a. a sin10 2100 a 364.6611.. 365m o altitude Q4 37° 65 m Let the height of cliff be h. h tan 37 65 h 48.9810... 49.0m o Q5 tower Let the height of cliff be h. h tan 53 60 o h 60 tan 53 o cliff 65° 53° 60 m Let the height of cliff and tower be x. x tan 65 60 o x 60 tan 65 o Let the height of tower be t. t 60 tan 65 60 tan 53 49.047725... o 49.0m o Q6 kite 30 m 67° Let the height of kite be h. h sin 67 30 h 27.615... 27.6m o h Q7 balloon d 75° Danny Let the distance be d. 30 sin 75 d d 31.0582... 31.1m o 30 m Q8 23° d 23° Buoy Let the distance be d. 25 sin 23 d d 63.9826... 64.0m o 25 m Q9 h 50o 1000m Let the height be h. h o tan 60 x h tan 50 1000 x o 60o x o x tan 60 tan 50o 1000 x 1000 tan 50o x tan 50o x tan 60o 1000 tan 50o x(tan 60o tan 50o ) 1000 tan 50o x tan 60o tan 50o 1000 tan 50o o h tan 60 tan 60o tan 50o 3820.448... 3820m Q10 h 18 m 46° x 58° h x sin 58 18 o x sin 46 18 o Let the height be h. o h 18sin 46 sin 58o 18 18sin 58o h 18sin 46o h 18sin 58 18sin 46 2.31674... 2.32m o o Q11 h 22o 20 m Let the height be h. h o tan 33 x h tan 22 20 x o 33o x o x tan 33 tan 22o 20 x 20 tan 22o x tan 22o x tan 33o 20 tan 22o x(tan 33o tan 22o ) 1000 tan 22o x tan 33o tan 22o 20 tan 22o o h tan 33 tan 33o tan 22o 21.385... 21.4m Q12 h 39o 35 m A Let the height be h. h o tan 58 x h tan 39 35 x o 58o B x o x tan 58 tan 39o 35 x 35 tan 39o x tan 39o x tan 58o 35 tan 39o x(tan 58o tan 39o ) 35 tan 39o x tan 58o tan 39o 35 tan 39o o h tan 58 tan 58o tan 39o 57.3744... 57.4m B Q13(a) xo 40 xo 60 E Let the angle of depression be x. 40 tan x 60 x 33.690.. o 33.7o Q13(b) A B D FC cos 70 60 o FC 60 cos 70 100 F E o DF 100 60 cos 70o EF sin 70 60 o EF 60sin 70 o 40 C 70o 60 DE 2 DF 2 EF 2 100 60cos 70 60sin 70 o 2 o 2 97.44618... AD tan 30 DE AD (97.44618) tan 30o 56.260... o 56.3m