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The Sine Law In the previous trigonometry units we have only been able to work with right-angle triangles. Although right angle triangles are really useful triangles, there are lots of other types of triangles besides those with right angles. We need a way to deal with those. If we're working with a non-right angle triangle, that means there's no hypotenuse, so we can't use Pythagorean theorem. Instead, we have two mathematical tools called the Sine Law and the Cosine Law. Let's take a look at the Sine Law first. An oblique triangle is a triangle that does not have a right angle. The Sine Law (or Law of Sines) can be used to solve oblique triangles when the following information is given: Two angles and any side (AAS or ASA) Two sides and an angle opposite one of the given sides (SSA) The Sine Law If A, B, C are the angle measures of a triangle and a, b, c are the lengths of the corresponding sides opposite these angles, then a b c sin A sin B sin C Or sin A sin B sin C a b c Proof First Construct triangle ABC to be a non right triangle. Draw a perpendicular from C to meet AB at E C b A h E Now writing h in terms of sin(A) and sin(B). a B h sin A b b sin A h h sin B a a sin B h Setting the h equations equal to each other. b sin A a sin B b a sin B sin A Now repeat by drawing a perpendicular from A to CB, and writing h in terms of sin(C) and sin(B). h sin C b b sin C h h sin B c c sin B h Setting the h equations equal to each other. b sin C c sin B b c sin B sin C Therefore putting these together a b c . sin A sin B sin C Example Solve ABC , if A 44.3 , a 11.5m, and b=77m. Round the side length to the nearest tenth of a metre and the angles to the tenth of a degree. Solution: This is Sine Law problem (ASS) First draw a diagram: C a=11.5 b=7.7 44.3 A B c Pick the two terms that have 3 of the 4 pieces of given information, and solve for the remaining term. sin B sin A b a sin B sin 44.3 7.7 11.5 sin B 0.4676345824 B 27.9 Solve for the remaining angle C 180 44.3 27.9 107.8 Use sine law to find c c a sin C sin A c 11.5 sin 107.8 sin 44.3 c 15.7 Summarizing: A 44.3 B 27.9 C 107.8 a b c 11.5m 77m 15.7m Example If P= 20, Q= 31 , p=210m, solve PQR . Q 31 r 210m q R 20 P Solution: First let’s find R R 180 20 31 129 Sum of angles in triangle Now apply sine law to find r r p sin R sin P r 210 sin 129 sin 20 210sin 129 r sin 20 477.2 Now apply the Sine Law to find q q p sin Q sin P q 210 sin 31 sin 20 210sin 31 q sin 20 316.2 Summarizing: P 20 Q 31 R 129 p q r 210m 316.2m 477.2 m Example An 8 metre radio antenna is located on top of an office building. At a distance d from the bottom of the building, the angle of elevation to the top of the antenna is 25, and the angle of elevation to the bottom of the antenna is 15. Determine the height of the office building. B C A 1525 h D Solution: From the diagram, determine angles and B B 902565 251510 Apply the Sine Law to determine AC BC AC sin sin B 8 AC sin 10 sin 65 8sin 65 AC sin 10 41.75375 Now apply primary trigonometric ratio to determine h h sin 15 41.75375 h 41.75375 sin 15 10.80667 Therefore the height of the building is about 10.8m Example A pilot, flying over a straight road, measures the angles of depression from km markers on opposite sides of the airplane: the angle of depression to one km marker is 85 , while the angle of depression to a km marker 4 km from the first marker is 6 . Determine the altitude of the plane to the nearest 100 metres. A 85 6 b c B h 85 6 D 4 km C Solution: We need to determine either c for ABD or side b for ADC . This way we have a right angle triangle to work with to determine h. Let’s choose c, with ABD . We will use the Sine Law with ABC . First we need A : A 180 856 89 Now, a c Sin A sin C 4 c sin 89 sin 6 4 sin 6 c sin 89 0.41818 km 418.18m Now we can use right triangle 1000 m = 1 km ABD to determine h h sin B c h sin 85 418.18 h 418.18 sin 85 416.59 To the nearest 100 m, the height is 400 m. Example: An observatory is 30 metres tall, and is situated on the top of a hill. From the top of the observatory, the angle of depression to a restaurant at the base of the hill is 46. From the base of the observatory the angle of depression to the restaurant is 35. Determine the distance, to the nearest metre, from the restaurant to the base of the observatory. 46 30 m 35 Solution: This is a job for the sine law, with x representing the required distance. Always try to first find as many angles as you can in the triangle. The angle opposite the longest side is 3590 125 The angle between the top of the observatory and the restaurant is 90 4644 The angle at the restaurant is 180 44125 11 Applying the sine law 30 x sin 11 sin 44 30 sin 44 x sin 11 109.218 The distance from the restaurant to the observatory is about 109 m. Example A hot air balloon is kept at a constant altitude by two ropes anchored at two points on the ground. One rope is 150 feet long and makes an angle of 68° with the ground. How long is the second rope, to the nearest ten feet, if it makes an angle of 75° with the ground? 150 ft x ft 75° 68° Solution: From the diagram, this is a job for the Sine Law x 150 sin 68 sin 75 150 sin 68 x sin 75 143.98 The second rope is about 144 feet long.