Law of Cosines Skill 43

Objective HSG-SRT.7/8: Students are responsible for using Law of Cosines to solve problems.

Law of Cosines B c 𝒂 A C b 𝒂 𝟐 = 𝒃 𝟐 + 𝒄 𝟐 −𝟐𝒃𝒄 𝒄𝒐𝒔 𝑨 𝒃 𝟐 = 𝒂 𝟐 + 𝒄 𝟐 −𝟐𝒂𝒄 𝒄𝒐𝒔 𝑩 𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 −𝟐𝒂𝒃 𝒄𝒐𝒔 𝑪

Example 1; Using Law of Cosines (SAS) a) In ∆𝐴𝐵𝐶, 𝑚∠𝐵=44, 𝐴𝐵=10, and 𝐵𝐶=22, find 𝐴𝐶. A C B 𝒃 𝟏𝟎 𝟒𝟒° 𝟐𝟐 𝒃 𝟐 = 𝒂 𝟐 + 𝒄 𝟐 −𝟐𝒂𝒄 𝒄𝒐𝒔 𝑩 𝒃 𝟐 = 𝟐𝟐 𝟐 + 𝟏𝟎 𝟐 −𝟐 𝟐𝟐 𝟏𝟎 𝒄𝒐𝒔 𝟒𝟒 𝒃 𝟐 =𝟒𝟖𝟒+𝟏𝟎𝟎−𝟒𝟒𝟎 𝒄𝒐𝒔 𝟒𝟒 𝒃 𝟐 =𝟐𝟔𝟕.𝟒𝟗 𝒃= 𝟐𝟔𝟕.𝟒𝟗 ≈𝟏𝟔.𝟒

Example 1; Using Law of Cosines (SAS) b) In ∆𝑀𝑁𝐿, 𝑚∠𝐿=104, 𝑀𝐿=48, and 𝐿𝑁=29, find 𝑀𝑁. M N L 𝒍 𝟐𝟗 𝟒𝟖 𝟏𝟎𝟒° 𝒍 𝟐 = 𝒎 𝟐 + 𝒏 𝟐 −𝟐𝒎𝒏 𝒄𝒐𝒔 𝑳 𝒍 𝟐 = 𝟒𝟖 𝟐 + 𝟐𝟗 𝟐 −𝟐 𝟒𝟖 𝟐𝟗 𝒄𝒐𝒔 𝟏𝟎𝟒 𝒍 𝟐 =𝟑𝟏𝟒𝟓−𝟐𝟕𝟖𝟒 𝒄𝒐𝒔 𝟏𝟎𝟒 𝒍 𝟐 =𝟑𝟖𝟏𝟖.𝟓𝟏 𝒍= 𝟑𝟖𝟏𝟖.𝟓𝟏 ≈𝟔𝟏.𝟖

Example 2; Using Law of Cosines (SSS) a) In ∆𝑇𝑈𝑉, 𝑇𝑈=4.4, 𝑈𝑉=7.1, and 𝑇𝑉=6.7, find 𝑚∠𝑉. U V T 𝟒.𝟒 𝟔.𝟕 𝑽 𝒗 𝟐 = 𝒕 𝟐 + 𝒖 𝟐 −𝟐𝒕𝒖 𝒄𝒐𝒔 𝑽 𝟕.𝟏 𝟒.𝟒 𝟐 = 𝟕.𝟏 𝟐 + 𝟔.𝟕 𝟐 −𝟐 𝟕.𝟏 𝟔.𝟕 𝒄𝒐𝒔 𝑽 𝟏𝟗.𝟑𝟔=𝟓𝟎.𝟒𝟏+𝟒𝟒.𝟖𝟗−𝟗𝟓.𝟏𝟒 𝒄𝒐𝒔 𝑽 −𝟕𝟓.𝟗𝟒=−𝟗𝟓.𝟏𝟒 𝒄𝒐𝒔 𝑽 𝒄𝒐𝒔 𝑽 =𝟎.𝟕𝟗𝟖𝟏𝟗… 𝑽≈𝟑𝟕.𝟎°

Example 2; Using Law of Cosines (SSS) b) In ∆𝑇𝑈𝑉, 𝑇𝑈=4.4, 𝑈𝑉=7.1, and 𝑇𝑉=6.7, find 𝑚∠𝑇. U V T 𝑻 𝟒.𝟒 𝟔.𝟕 𝒕 𝟐 = 𝒖 𝟐 + 𝒗 𝟐 −𝟐𝒖𝒗 𝒄𝒐𝒔 𝑻 𝟕.𝟏 𝟕.𝟏 𝟐 = 𝟔.𝟕 𝟐 + 𝟒.𝟒 𝟐 −𝟐 𝟔.𝟕 𝟒.𝟒 𝒄𝒐𝒔 𝑻 𝟓𝟎.𝟒𝟏=𝟒𝟒.𝟖𝟗+𝟏𝟗.𝟑𝟔−𝟓𝟖.𝟗𝟔 𝒄𝒐𝒔 𝑻 −𝟏𝟑.𝟖𝟒=−𝟓𝟖.𝟗𝟔 𝒄𝒐𝒔 𝑻 𝒄𝒐𝒔 𝑻 =𝟎.𝟐𝟑𝟒𝟕… 𝑻≈𝟕𝟔.𝟒°

Example 3; Using Law of Cosines (SAS) a) An air traffic controller is tracking a plane 2.1 kilometers due south of the radar tower. A second plane is located 3.5 kilometers from the tower at a heading of N 75ᵒ E (75ᵒ east of north). To the nearest tenth of a kilometer, how far apart are the two planes? 𝒕 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 −𝟐𝒂𝒃 𝒄𝒐𝒔 𝑻 𝒕 𝟐 = 𝟑.𝟓 𝟐 + 𝟐.𝟏 𝟐 −𝟐 𝟑.𝟓 𝟐.𝟏 𝒄𝒐𝒔 𝟏𝟎𝟓 𝒕 𝟐 =𝟏𝟐.𝟐𝟓+𝟒.𝟒𝟏−𝟏𝟒.𝟕 𝒄𝒐𝒔 𝟏𝟎𝟓 𝒕 𝟐 =𝟏𝟔.𝟔𝟔−𝟏𝟒.𝟕 𝒄𝒐𝒔 𝟏𝟎𝟓 Tower Plane A Plane B North 𝒕 𝟐 =𝟐𝟎.𝟒𝟔𝟒… 𝟑.𝟓 𝒌𝒎 𝟕𝟓° 𝒕= 𝑨𝑵𝑺 𝟏𝟎𝟓° ≈𝟒.𝟓 𝒌𝒎 𝒕 𝟐.𝟏 𝒌𝒎 𝑻=𝟏𝟖𝟎−𝟕𝟓=𝟏𝟎𝟓°

Example 3; Using Law of Cosines (SAS) b) You and a friend hike 1.4 miles due west from a campsite. At the same time, two other friends hike 1.9 miles at a heading of S 11ᵒ W (11ᵒ west of south) from the campsite. To the nearest tenth of a mile, how far apart are the two groups? 𝒄 𝟐 = 𝒘 𝟐 + 𝒔 𝟐 −𝟐𝒘𝒔 𝒄𝒐𝒔 𝑪 𝒄 𝟐 = 𝟏.𝟗 𝟐 + 𝟏.𝟒 𝟐 −𝟐 𝟏.𝟗 𝟏.𝟒 𝒄𝒐𝒔 𝟕𝟗 𝒄 𝟐 =𝟑.𝟔𝟏+𝟏.𝟗𝟔−𝟓.𝟑𝟐 𝒄𝒐𝒔 𝟕𝟗 S E W N 𝒄 𝟐 =𝟓.𝟓𝟕−𝟓.𝟑𝟐 𝒄𝒐𝒔 𝟕𝟗 𝒄 𝟐 =𝟒.𝟓𝟓𝟒𝟖… 𝟏.𝟒 𝒎𝒊 𝑪 𝒄= 𝑨𝑵𝑺 𝒄 𝟏.𝟗 𝒎𝒊 𝟏𝟏° ≈𝟒.𝟓 𝒌𝒎 𝑻=𝟗𝟎−𝟏𝟏=𝟕𝟗°

#43: Law of Cosines Questions? Summarize Notes Homework Quiz