7.1 The Law of Sines Congruence Axioms Side-Angle-Side (SAS) If two sides and the included angle of one triangle are equal, respectively, to two sides and the included angle of a second triangle, then the triangles are congruent. Angle-Side-Angle (ASA) If two angles and the included side of one triangle are equal, respectively, to two angles and the included side of a second triangle, then the triangles are congruent. Side-Side-Side (SSS) If three sides of one triangle are equal to three sides of a second triangle, the triangles are congruent.

7.1 Data Required for Solving Oblique Triangles Case 1 One side and two angles known: SAA or ASA Case 2 Two sides and one angle not included between the sides known: SSA This case may lead to more than one solution. Case 3 Two sides and one angle included between the sides known: SAS Case 4 Three sides are known: SSS

7.1 Derivation of the Law of Sines Start with an acute or obtuse triangle and construct the perpendicular from B to side AC. Let h be the height of this perpendicular. Then c and a are the hypotenuses of right triangle ADB and BDC, respectively.

7.1 The Law of Sines In a similar way, by constructing perpendiculars from other vertices, the following theorem can be proven. Alternative forms are sometimes convenient to use: Law of Sines In any triangle ABC, with sides a, b, and c,

7.1 Using the Law of Sines to Solve a Triangle Example Solve triangle ABC if A = 32.0°, B = 81.8°, and a = 42.9 centimeters. Solution Draw the triangle and label the known values. Because A, B, and a are known, we can apply the law of sines involving these variables.

7.1 Using the Law of Sines to Solve a Triangle To find C, use the fact that there are 180° in a triangle. Now we can find c.

7.1 Using the Law of Sines in an Application (ASA) Example Two stations are on an east-west line 110 miles apart. A forest fire is located on a bearing of N 42° E from the western station at A and a bearing of N 15° E from the eastern station at B. How far is the fire from the western station? Solution Angle BAC = 90° – 42° = 48° Angle B = 90° + 15° = 105° Angle C = 180° – 105° – 48° = 27° Using the law of sines to find b gives

7.1 Ambiguous Case If given the lengths of two sides and the angle opposite one of them, it is possible that 0, 1, or 2 such triangles exist. Some basic facts that should be kept in mind: For any angle , –1  sin   1, if sin  = 1, then  = 90° and the triangle is a right triangle. sin  = sin(180° –  ). The smallest angle is opposite the shortest side, the largest angle is opposite the longest side, and the middle-value angle is opposite the intermediate side (assuming unequal sides).

7.1 Ambiguous Case

7.1 Ambiguous Case for Obtuse Angle A

7.1 Solving the Ambiguous Case: No Such Triangle Example Solve the triangle ABC if B = 55°40´, b = 8.94 meters, and a = 25.1 meters. Solution Use the law of sines to find A. Since sin A cannot be greater than 1, the triangle does not exist.

7.1 Solving the Ambiguous Case: Two Triangles Example Solve the triangle ABC if A = 55.3°, a = 22.8 feet, and b = 24.9 feet. Solution

7.1 Solving the Ambiguous Case: Two Triangles To see if B2 = 116.1° is a valid possibility, add 116.1° to the measure of A: 116.1° + 55.3° = 171.4°. Since this sum is less than 180°, it is a valid triangle. Now separate the triangles into two: AB1C1 and AB2C2.

7.1 Solving the Ambiguous Case: Two Triangles Now solve for triangle AB2C2.

7.1 Number of Triangles Satisfying the Ambiguous Case Let sides a and b and angle A be given in triangle ABC. (The law of sines can be used to calculate sin B.) If sin B > 1, then no triangle satisfies the given conditions. If sin B = 1, then one triangle satisfies the given conditions and B = 90°. If 0 < sin B < 1, then either one or two triangles satisfy the given conditions If sin B = k, then let B1 = sin-1 k and use B1 for B in the first triangle. Let B2 = 180° – B1. If A + B2 < 180°, then a second triangle exists. In this case, use B2 for B in the second triangle.

7.1 Solving the Ambiguous Case: One Triangle Example Solve the triangle ABC, given A = 43.5°, a = 10.7 inches, and c = 7.2 inches. Solution The other possible value for C: C = 180° – 27.6° = 152.4°. Add this to A: 152.4° + 43.5° = 195.9° > 180° Therefore, there can be only one triangle.

7.1 Solving the Ambiguous Case: One Triangle