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Law of Sines 10.4 Holt Geometry Calculator Review 1. What is the third angle measure in a triangle with angles measuring 65° and 43°? 72° Find each value. Round trigonometric ratios to the nearest hundredth and angle measures to the nearest degree. 2. sin 73° 0.96 3. cos 18° 0.95 4. tan 82° 7.12 5. sin-1 (0.34) 6. cos-1 (0.63) 20° Holt Geometry 51° 7. tan-1 (2.75) 70° Objectives Use the Law of Sines and the Law of Cosines to solve triangles that are not right triangles. Solve a triangle by finding the measures of all sides and angles. Find the area of oblique (no right angles) triangles. Holt Geometry Finding Trigonometric Ratios for Obtuse Angles Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. A. tan 103° B. cos 165° C. sin 93° D. tan 175° E. cos 92° F. sin 160° Holt Geometry You can use the Law of Sines to solve a triangle if you are given • two angle measures and any side length (ASA or AAS) or • two side lengths and a non-included angle measure (SSA); called the ambiguous case and requires extra care. Holt Geometry Holt Geometry Using the Law of Sines AAS Find the length of FG. Round lengths to the nearest tenth and angles to the nearest degree. AAS, not the ambiguous case, no extra care required. Law of Sines Substitution. FG sin 39° = 40 sin 32° Cross Multiply. Divide both sides by sin 39. Holt Geometry Using the Law of Sines AAS Find the length of NP. Round lengths to the nearest tenth and angles to the nearest degree. AAS, not the ambiguous case, no extra care required. Law of Sines Substitution. NP sin 39° = 22 sin 88° Cross Multiply. Divide both sides by sin 39°. Holt Geometry Using the Law of Sines ASA Find the length of AC. Round lengths to the nearest tenth and angles to 69 the nearest degree. ASA, not the ambiguous case, no extra care required. mA + 67° + 44° = 180° mA = 69° Prop of ∆. Subtraction. Law of Sines Substitution. AC sin 69° = 18 sin 67° Holt Geometry Cross Multiply Divide both sides by sin 69°. Assignment Geometry: 10-4 Law of Sines Holt Geometry Using the Law of Sines SSA Find mL. K = 51°, k = 10, and l = 6. SSA, the ambiguous case, extra care required. Since the angle is obtuse and the opposite side is greater than then the adjacent side, there is one triangle. Law of Sines Substitute the given values. 10 sin L = 6 sin 125° Holt Geometry Cross Products Property Use the inverse sine function to find mL. Using the Law of Sines SSA Find mB with A = 51°, a = 3.5, and b = 5. SSA, the ambiguous case, extra care required. Law of Sines sin B sin 51 5 3.5 3.5 sin B = 5 sin 51° 5 sin 51 mB sin 3.5 1 Holt Geometry Substitute the given values. Cross Products Property Using the inverse function on the calculator indicates no such triangle exists. EXAMPLE 3 Using the Law of Sines SSA Find mB with A = 40°, a = 13, and b = 16. SSA, the ambiguous case, extra care required. Law of Sines sin B sin 40 16 13 Substitute the given values. Cross Products Property Use the inverse function on 16 sin 40 1 mB sin 52.3 the calculator, then check to 13 see if another triangle exists. 180 - 52.3 = 127.7 See if 127.7+ 40<180 If so, then 2 triangles exist, as in this case. Holt Geometry 13 sin B = 16 sin 40° Using the Law of Sines SSA Find mQ with P = 51°, p = 8, and q = 6. SSA, the ambiguous case, extra care required. Law of Sines Substitute the given values. Multiply both sides by 6. 180 - 36 = 144 Holt Geometry Use the inverse function on the calculator, then check to see if another triangle exists. See if sin 144 + 51 < 180. If not, then only one triangle exists, as in this case. Using the Law of Sines Mixed Review Find mB a. A = 105°, b = 13, a = 6 Since the angle is obtuse and the Sketch and classify C 6 13 c SSA obtuse B b. A = 110°, b = 100, a = 125 Sketch and classify C c SSA obtuse B B c Since the angle is obtuse and the opposite side is greater than the adjacent side, one triangle can be formed. C b = 100 110 Holt Geometry A 125 100 A a=6 105 b = 13 105 A opposite side is less than the adjacent side, a triangle cannot be formed. C a = 125 110 A c B sin B sin 110 mB 48.7 100 125 Using the Law of Sines Mixed Review Find mB sin B sin 70 c. A = 70°, a = 3.2, b = 3 mB 61.8 Sketch and classify 3 3.2 C 180 - 61.8 = 118.2 C 3.2 3 A 70 c SSA acute B B d. A = 76, a = 18, b = 20 Sketch and classify C A 76 c SSA acute Holt Geometry sin B sin 76 20 18 No such triangle exists. 18 20 Since 118.2 + 70 > 180 a = 3.2 only one triangle exists. B b=3 70 c A C b = 20 a = 18 76 B A Using the Law of Sines Mixed Review Find mB sin B sin 58 12.8 11.4 e. A = 58, a = 11.4, b = 12.8 Sketch and classify C 12.8 A 180 - 72.2 = 107.8 Since 107.8 + 58 < 180 two triangles exist. 11.4 58 c SSA acute B C C b = 12.8 a = 11.4 58 B c b = 12.8 a = 11.4 58 A mB 72.2 Holt Geometry mB 72.2 B c mB 107.8 A Using the Law of Sines Mixed Review Find the length of AB f. B = 60°, C = 10, a = 4.5 ASA just use the Law of Sines Sketch and classify The missing angle is 110 C C sin 110 sin 10 10 4.5 10 60 A B 4.5 AB .8 AB a = 4.5 b ASA acute A g. A = 65, B = 80, a = 17 Sketch and classify C 17 A 65 80 c AAS acute Holt Geometry 60 B AAS just use the Law of Sines The missing angle is 35 sin 65 sin 35 17 AB AB 10.8 c B Assignment Geometry: Ambiguous Case 1 Law of Sines Holt Geometry Area of an Oblique Triangle 1 bc 1 ab 1 bc 1Area 1abac sin sin CB Area sin A 1 Area Area 1 bc sinsin AA 1 ab sin Area sin CC ac ac sinsin A BB 11bcab sinAC 11 bc sin 22 22 2 2 22 22 22 C Example: Find the area of the triangle. A = 74, b = 103 inches, c = 58 inches Area = 1 bc sin A 2 = 1 (103)(58) sin 74 2 2871 square inches Holt Geometry 103 in a b A 74 c 58 in B Assignment Geometry: 10-4 Area Holt Geometry