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The Law of Sines

The Law of Sines. If A , B , and C are the measures of the angles of a triangle, and a , b , and c are the lengths of the sides opposite these angles, then

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The Law of Sines

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  1. The Law of Sines

  2. If A, B, and C are the measures of the angles of a triangle, and a, b, and c are the lengths of the sides opposite these angles, then The ratio of the length of the side of any triangle to the sine of the angle opposite that side is the same for all three sides of the triangle. Law of Sines

  3. C The sum of the measurements of a triangle’s interior angles is 180º. A+B+C= 180º 50º +B+ 33.5º = 180º A = 50º and C = 33.5º. 33.5º b =76 a 83.5º +B= 180º Add. B= 96.5º Subtract 83.5º from both sides. 50º A B c Text Example • Solve triangle ABC if A= 50º, C= 33.5º, and b= 76. Solution We begin by drawing a picture of triangle ABC and labeling it with the given information. The figure shows the triangle that we must solve. We begin by finding B.

  4. C This is the known ratio. 33.5º b =76 a 50º A B c Text Example cont. Solve triangle ABC if A= 50º, C= 33.5º, and b= 76. Solution Keep in mind that we must be given one of the three ratios to apply the Law of Sines. In this example, we are given that b= 76 and we found that B= 96.5º. Thus, we use the ratio b/sin B, or 76/sin96.5º, to find the other two sides. Use the Law of Sines to find a and c. Find a: Find c: The solution is B= 96.5º, a 59, and c 42.

  5. Example • Solve the triangle shown with A=36º, B=88º and c=29 feet. Solution: A=36º, B=88º so 180-88-36=56º C=56º

  6. a=h andisjust theright length to form a right triangle. a is less than h and not long enough to form a triangle. a b b h=bsinA h=bsinA a A A a is greater than h and a is less than b. Two distinct triangles are formed. a is greater than h and a is greater than b. One triangle is formed. b a b h=bsinA a a A A The Ambiguous Case (SSA) Consider a triangle in which a, b, and A are given. This information may result in: No Triangle One Right Triangle Two Triangles One Triangle

  7. One Solution • Solve the triangle shown with A=43º, a=81 and b=62. 149 + 43 > 180. Thus there is only 1 solution.

  8. Two Solutions • Solve the triangle shown with X=40º, x=54 and z=62. 132 + 40 < 180, so we have 2 solutions!

  9. No Solution • Solve the triangle shown with A=75º, a=51 and b=71. No Solution!

  10. Area of An Oblique Triangle • The area of a triangle equals one-half the product of the lengths of two sides times the sine of their included angle. In the following figure, this wording can be expressed by the formulas

  11. Solution The triangle is shown in the following figure. Its area is half the product of the lengths of the two sides times the sine of the included angle. Area =1/2(24)(10)(sin 62º)  106 the area of the triangle is approximately 106 square meters. C b= 10 meters 62º A B c= 24 meters Text Example • Find the area of a triangle having two sides of lengths 24 meters and 10 meters and an included angle of 62º.

  12. Example • Find the area of a triangle having two sides of lengths 12 ft. and 20 ft. and an included angle of 57º. Solution:

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