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Law of Cosines Skill 43.

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Presentation on theme: "Law of Cosines Skill 43."β€” Presentation transcript:

1 Law of Cosines Skill 43

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

3 Law of Cosines B c 𝒂 A C b 𝒂 𝟐 = 𝒃 𝟐 + 𝒄 𝟐 βˆ’πŸπ’ƒπ’„ 𝒄𝒐𝒔 𝑨
𝒃 𝟐 = 𝒂 𝟐 + 𝒄 𝟐 βˆ’πŸπ’‚π’„ 𝒄𝒐𝒔 𝑩 𝒄 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 βˆ’πŸπ’‚π’ƒ 𝒄𝒐𝒔 π‘ͺ

4 Example 1; Using Law of Cosines (SAS)
a) In βˆ†π΄π΅πΆ, π‘šβˆ π΅=44, 𝐴𝐡=10, and 𝐡𝐢=22, find 𝐴𝐢. A C B 𝒃 𝟏𝟎 πŸ’πŸ’Β° 𝟐𝟐 𝒃 𝟐 = 𝒂 𝟐 + 𝒄 𝟐 βˆ’πŸπ’‚π’„ 𝒄𝒐𝒔 𝑩 𝒃 𝟐 = 𝟐𝟐 𝟐 + 𝟏𝟎 𝟐 βˆ’πŸ 𝟐𝟐 𝟏𝟎 𝒄𝒐𝒔 πŸ’πŸ’ 𝒃 𝟐 =πŸ’πŸ–πŸ’+πŸπŸŽπŸŽβˆ’πŸ’πŸ’πŸŽ 𝒄𝒐𝒔 πŸ’πŸ’ 𝒃 𝟐 =πŸπŸ”πŸ•.πŸ’πŸ— 𝒃= πŸπŸ”πŸ•.πŸ’πŸ— β‰ˆπŸπŸ”.πŸ’

5 Example 1; Using Law of Cosines (SAS)
b) In βˆ†π‘€π‘πΏ, π‘šβˆ πΏ=104, 𝑀𝐿=48, and 𝐿𝑁=29, find 𝑀𝑁. M N L 𝒍 πŸπŸ— πŸ’πŸ– πŸπŸŽπŸ’Β° 𝒍 𝟐 = π’Ž 𝟐 + 𝒏 𝟐 βˆ’πŸπ’Žπ’ 𝒄𝒐𝒔 𝑳 𝒍 𝟐 = πŸ’πŸ– 𝟐 + πŸπŸ— 𝟐 βˆ’πŸ πŸ’πŸ– πŸπŸ— 𝒄𝒐𝒔 πŸπŸŽπŸ’ 𝒍 𝟐 =πŸ‘πŸπŸ’πŸ“βˆ’πŸπŸ•πŸ–πŸ’ 𝒄𝒐𝒔 πŸπŸŽπŸ’ 𝒍 𝟐 =πŸ‘πŸ–πŸπŸ–.πŸ“πŸ 𝒍= πŸ‘πŸ–πŸπŸ–.πŸ“πŸ β‰ˆπŸ”πŸ.πŸ–

6 Example 2; Using Law of Cosines (SSS)
a) In βˆ†π‘‡π‘ˆπ‘‰, π‘‡π‘ˆ=4.4, π‘ˆπ‘‰=7.1, and 𝑇𝑉=6.7, find π‘šβˆ π‘‰. U V T πŸ’.πŸ’ πŸ”.πŸ• 𝑽 𝒗 𝟐 = 𝒕 𝟐 + 𝒖 𝟐 βˆ’πŸπ’•π’– 𝒄𝒐𝒔 𝑽 πŸ•.𝟏 πŸ’.πŸ’ 𝟐 = πŸ•.𝟏 𝟐 + πŸ”.πŸ• 𝟐 βˆ’πŸ πŸ•.𝟏 πŸ”.πŸ• 𝒄𝒐𝒔 𝑽 πŸπŸ—.πŸ‘πŸ”=πŸ“πŸŽ.πŸ’πŸ+πŸ’πŸ’.πŸ–πŸ—βˆ’πŸ—πŸ“.πŸπŸ’ 𝒄𝒐𝒔 𝑽 βˆ’πŸ•πŸ“.πŸ—πŸ’=βˆ’πŸ—πŸ“.πŸπŸ’ 𝒄𝒐𝒔 𝑽 𝒄𝒐𝒔 𝑽 =𝟎.πŸ•πŸ—πŸ–πŸπŸ—β€¦ π‘½β‰ˆπŸ‘πŸ•.𝟎°

7 Example 2; Using Law of Cosines (SSS)
b) In βˆ†π‘‡π‘ˆπ‘‰, π‘‡π‘ˆ=4.4, π‘ˆπ‘‰=7.1, and 𝑇𝑉=6.7, find π‘šβˆ π‘‡. U V T 𝑻 πŸ’.πŸ’ πŸ”.πŸ• 𝒕 𝟐 = 𝒖 𝟐 + 𝒗 𝟐 βˆ’πŸπ’–π’— 𝒄𝒐𝒔 𝑻 πŸ•.𝟏 πŸ•.𝟏 𝟐 = πŸ”.πŸ• 𝟐 + πŸ’.πŸ’ 𝟐 βˆ’πŸ πŸ”.πŸ• πŸ’.πŸ’ 𝒄𝒐𝒔 𝑻 πŸ“πŸŽ.πŸ’πŸ=πŸ’πŸ’.πŸ–πŸ—+πŸπŸ—.πŸ‘πŸ”βˆ’πŸ“πŸ–.πŸ—πŸ” 𝒄𝒐𝒔 𝑻 βˆ’πŸπŸ‘.πŸ–πŸ’=βˆ’πŸ“πŸ–.πŸ—πŸ” 𝒄𝒐𝒔 𝑻 𝒄𝒐𝒔 𝑻 =𝟎.πŸπŸ‘πŸ’πŸ•β€¦ π‘»β‰ˆπŸ•πŸ”.πŸ’Β°

8 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 75α΅’ E (75α΅’ east of north). To the nearest tenth of a kilometer, how far apart are the two planes? 𝒕 𝟐 = 𝒂 𝟐 + 𝒃 𝟐 βˆ’πŸπ’‚π’ƒ 𝒄𝒐𝒔 𝑻 𝒕 𝟐 = πŸ‘.πŸ“ 𝟐 + 𝟐.𝟏 𝟐 βˆ’πŸ πŸ‘.πŸ“ 𝟐.𝟏 𝒄𝒐𝒔 πŸπŸŽπŸ“ 𝒕 𝟐 =𝟏𝟐.πŸπŸ“+πŸ’.πŸ’πŸβˆ’πŸπŸ’.πŸ• 𝒄𝒐𝒔 πŸπŸŽπŸ“ 𝒕 𝟐 =πŸπŸ”.πŸ”πŸ”βˆ’πŸπŸ’.πŸ• 𝒄𝒐𝒔 πŸπŸŽπŸ“ Tower Plane A Plane B North 𝒕 𝟐 =𝟐𝟎.πŸ’πŸ”πŸ’β€¦ πŸ‘.πŸ“ π’Œπ’Ž πŸ•πŸ“Β° 𝒕= 𝑨𝑡𝑺 πŸπŸŽπŸ“Β° β‰ˆπŸ’.πŸ“ π’Œπ’Ž 𝒕 𝟐.𝟏 π’Œπ’Ž 𝑻=πŸπŸ–πŸŽβˆ’πŸ•πŸ“=πŸπŸŽπŸ“Β°

9 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 11α΅’ W (11α΅’ west of south) from the campsite. To the nearest tenth of a mile, how far apart are the two groups? 𝒄 𝟐 = π’˜ 𝟐 + 𝒔 𝟐 βˆ’πŸπ’˜π’” 𝒄𝒐𝒔 π‘ͺ 𝒄 𝟐 = 𝟏.πŸ— 𝟐 + 𝟏.πŸ’ 𝟐 βˆ’πŸ 𝟏.πŸ— 𝟏.πŸ’ 𝒄𝒐𝒔 πŸ•πŸ— 𝒄 𝟐 =πŸ‘.πŸ”πŸ+𝟏.πŸ—πŸ”βˆ’πŸ“.πŸ‘πŸ 𝒄𝒐𝒔 πŸ•πŸ— S E W N 𝒄 𝟐 =πŸ“.πŸ“πŸ•βˆ’πŸ“.πŸ‘πŸ 𝒄𝒐𝒔 πŸ•πŸ— 𝒄 𝟐 =πŸ’.πŸ“πŸ“πŸ’πŸ–β€¦ 𝟏.πŸ’ π’Žπ’Š π‘ͺ 𝒄= 𝑨𝑡𝑺 𝒄 𝟏.πŸ— π’Žπ’Š 𝟏𝟏° β‰ˆπŸ’.πŸ“ π’Œπ’Ž 𝑻=πŸ—πŸŽβˆ’πŸπŸ=πŸ•πŸ—Β°

10 #43: Law of Cosines Questions? Summarize Notes Homework Quiz

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