Module 18 Oblique Triangles (Applications) Florben G. Mendoza

CASE 1: One side and two angles are known (SAA or ASA). Law of Sines FOUR CASES CASE 1: One side and two angles are known (SAA or ASA). Law of Sines CASE 2: Two sides and the angle opposite one of them are known (SSA). Law of Sines CASE 3: Two sides and the included angle are known (SAS). Law of Cosines CASE 4: Three sides are known (SSS). Law of Cosines Florben G. Mendoza

Law of Sines Law of Cosines a2 = b2 + c2 – 2bc cos A b2 = a2 + c2 – 2ac cos B a sin A = b sin B c c2 = a2 + b2 – 2ab cos C cos A = b2 + c2 - a2 2bc cos B = a2 + c2 - b2 2ac cos C = a2 + b2 - c2 2ab Florben G. Mendoza

Practical applications of trigonometry often involve determining distances that cannot be measured directly. In many applications of trigonometry the essential problem is the solution of triangles. If enough sides and angles are known, the remaining sides and angles as well as the area can be calculated, and the triangle is then said to be solved. Problems involving angles and distances in one plane are covered in this lesson. Florben G. Mendoza

Example 1: A navy aircraft is flying over a straight highway Example 1: A navy aircraft is flying over a straight highway. When the aircraft is in between the two cities that are 5 miles apart, he determines that the angle of depression to two cities to be 32° and 48° respectively. Find the distance of the aircraft from the two cities. 32° 48° C City A City B b = ? a = ? 32° 48° A 5 mi B Florben G. Mendoza

Given: A = 32° B = 48° c = 5 mi Find: a = ? b = ? Step 3: Step 1: sin C = a sin A Step 3: Step 1: ASA – Law of Sine 5 sin 100° = a sin 32° Step 2: C = 180° - (A+B) C = 180° - (32° + 48°) a (sin 100°) = 5 (sin 32°) C = 180° - 80° sin 100° sin 100° C = 100° a = 2.69 mi Florben G. Mendoza

Given: A = 32° B = 32° c = 5 mi Find: a = ? b = ? Step 4: a = ? b = ? 48° A B 5 mi c sin C = b sin B Step 4: 5 sin 100° = b sin 48° b (sin 100°) = 5 (sin 48°) b = 3.77 mi sin 100° sin 100° Florben G. Mendoza

Example 2: Three circles of radii 100, 140, & 210 cm respectively are tangent to each other externally. Find the angles of the triangle formed by joining their centers. 210 C C ? 310 350 100 A 140 B ? ? A 240 B Florben G. Mendoza

C 310 350 A 240 B Given: Find: a = 350 A = ? b = 310 B = ? c = 240 Step 1: SSS – Law of Cosine Step 2: cos A = b2 + c2 - a2 2bc cos A = (310)2 + (240)2 – (350)2 2(310)(240) Given: Find: cos A = 31 200 148 800 a = 350 A = ? b = 310 B = ? cos A = 0.21 c = 240 C = ? A = cos-1 0.21 A = 77.88° Florben G. Mendoza

C 310 350 A 240 B Step 3: cos B = a2 + c2 - b2 2ac 2(350)(240) cos B = 161 000 217 000 Step 4: C = 180° - (A+B) cos B = 0.74 C = 180° - (77.88° + 42.27°) B = cos-1 0.74 C = 180° - 120.15° B = 42.27° C = 59.85° Florben G. Mendoza

Example 3: A pole casts a shadow of 15 meters long when the angle of elevation of the sun is 61°. If the pole has leaned 15° from the vertical directly towards the sun, find the length of the pole. B 15° ? ? 61° 61° 105° A 15 m 15 m C Florben G. Mendoza

Step 1: Step 2: Step 3: Given: Find: A = 61° a = ? C = 105° b = 15 m B ASA – Law of Sine Step 2: B = 180° - (A + C) B = 180° - (61° + 105°) B = 180° - 166° B = 14° b sin B = a sin A Step 3: Given: Find: A = 61° a = ? 15 sin 14° = a sin 61° C = 105° a (sin 14°) = 15 (sin 61°) b = 15 m sin 14° sin 14° a = 54.23 m Florben G. Mendoza

Example 4: A tree on a hillside casts a shadow 215 ft down the hill Example 4: A tree on a hillside casts a shadow 215 ft down the hill. If the angle of inclination of the hillside is 22° to the horizontal and the angle of elevation of the sun is 52°, find the height of the tree. Florben G. Mendoza

C ? Step 1: B A Given: Find: A = 30° a = ? B = 112° C = 38° c = 215 ft sin C = a sin A Step 1: 215 sin 38° = a sin 30° a (sin 38°) = 215 (sin 30°) Given: Find: sin 38° sin 38° A = 30° a = ? B = 112° a = 174.61 ft C = 38° c = 215 ft Florben G. Mendoza

Example 5: A pilot sets out from an airport and heads in the direction N 20° E, flying at 200 mi/h. After one hour, he makes a course correction and heads in the direction N 50° E. Half an hour after that, engine trouble forces him to make an emergency landing. Find the distance between the airport & his final landing point. 100 mi 50° 200 mi 100 mi 150° ? A B C 40° 20° 200 mi ? 20° Florben G. Mendoza

100 mi C B 150° Step 1: 200 mi Step 2: b2 = a2 + c2 – 2ac cos B ? SAS – Law of Cosine Step 2: b2 = a2 + c2 – 2ac cos B b2 = (100)2 + (200)2 – 2(100)(200) (cos 150°) b2 = 50 000 – (- 34 641.02) Given: Find: b2 = 84 641.02 B = 150° b = ? b = 290.93 a = 100 mi c = 200 mi Florben G. Mendoza