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The Law of Sines and the Law of Cosines Lesson 8.5 Objective Use the Law of Sines to solve triangles. 1 Ex1: Finding Trigonometric Ratios for Obtuse Angles In this lesson, you will learn to solve any triangle. To do so, you will need to calculate trigonometric ratios for angle measures up to 180°. You can use a calculator to find these values. Use your calculator to find each trigonometric ratio. Round to the nearest hundredth. A. tan 103° B. cos 165° tan 103° –4.33 cos 165° –0.97 2 Holt McDougal Geometry C. sin 93° sin 93° 1.00 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). 3 Holt McDougal Geometry Example 2: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. FG Law of Sines FG sin 39° = 40 sin 32° 4 Holt McDougal Geometry Check It Out! Example 3 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. AC mA + mB + mC = 180° mA + 67° + 44° = 180° mA = 69° 5 Holt McDougal Geometry Prop of ∆. Check It Out! Example 3 Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. 69° Law of Sines AC sin 69° = 18 sin 67° Divide both sides by sin 69°. 6 Holt McDougal Geometry Check It Out! Example 4 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mX Law of Sines 7.6 sin X = 4.3 sin 50° Use the inverse sine function to find mX. 7 Holt McDougal Geometry Example 5: Using the Law of Sines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mQ Law of Sines 8 Holt McDougal Geometry Lesson Quiz: Part I Use a calculator to find each trigonometric ratio. Round to the nearest hundredth. 1. tan 154° –0.49 2. cos 124° –0.56 3. sin 162° 0.31 9 Holt McDougal Geometry Lesson Quiz Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree. 4. mB = 20°, mC = 31° and b = 210. Find a. 477.2 5. a = 16, b = 10, and mC = 110°. Find c. 21.6 10 Holt McDougal Geometry 8.5 Day Two Law of Sines Objective Use the Law of Cosines to solve triangles. 11 The Law of Sines cannot be used to solve every ∆. Instead, you can apply the Law of Cosines. You can use the Law of Cosines to solve a ∆ if you are given: • two side lengths and the included angle measure (SAS) or • three side lengths (SSS). 12 Holt McDougal Geometry Example 1: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. XZ XZ2 = XY2 + YZ2 – 2(XY)(YZ)cos Y Law of Cosines XZ2 = 352 + 302 – 2(35)(30)cos 110° Substitute XZ2 2843.2423 XZ 53.3 13 Holt McDougal Geometry Example 2 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. DE DE2 = EF2 + DF2 – 2(EF)(DF)cos F Law of Cosines = 182 + 162 – 2(18)(16)cos 21° DE2 42.2577 DE 6.5 14 Holt McDougal Geometry Example 3: Using the Law of Cosines Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mT RS2 = RT2 + ST2 – 2(RT)(ST)cos T 72 = 132 + 112 – 2(13)(11)cos T 49 = 290 – 286 cosT –241 = –286 cosT 15 Holt McDougal Geometry Example 4 Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK JL2 = LK2 + KJ2 – 2(LK)(KJ)cos K 82 = 152 + 102 – 2(15)(10)cos K 64 = 325 – 300 cosK –261 = –300 cosK 16 Holt McDougal Geometry Law of Cosines Substitute the given values. Example 4 Continued Find the measure. Round lengths to the nearest tenth and angle measures to the nearest degree. mK –261 = –300 cosK Solve for cosK. Use the inverse cosine function to find mK. 17 Holt McDougal Geometry Helpful Hint Do not round your answer until the final step of the computation. If a problem has multiple steps, store the calculated answers to each part in your calculator. 18 Holt McDougal Geometry Example 5 What if…? Another engineer suggested using a cable attached from the top of the tower to a point 31 m from the base. How long would this cable be, and what angle would it make with the ground? Round the length to the nearest tenth and the angle measure to the nearest degree. Step 1 Find the length of the cable. AC2 = AB2 + BC2 – 2(AB)(BC)cos B AC2= 312 + 562 – 2(31)(56)cos 100° AC2 4699.9065 19 AC 68.6 m Holt McDougal Geometry 31 m Example 5 Continued Step 2 Find the measure of the angle the cable would make with the ground. Law of Sines 20 Holt McDougal Geometry 31 m Lesson Quiz Use ΔABC for Items 4–6. Round lengths to the nearest tenth and angle measures to the nearest degree. 1. a = 20, b = 15, and c = 8.3. Find mA. 115° 21 Holt McDougal Geometry Lesson Quiz 2. An observer in tower A sees a fire 1554 ft away at an angle of depression of 28°. To the nearest foot, how far is the fire from an observer in tower B? To the nearest degree, what is the angle of depression to the fire from tower B? 1212 ft; 37° 22 Holt McDougal Geometry