Law of Sines Skill 42.

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Presentation transcript:

Law of Sines Skill 42

Objective HSG-SRT.7/8: Students are responsible for using the law of Sines in problems.

Law of Sines b B 𝒂 c A C 𝒔𝒊𝒏 𝑨 𝒂 = 𝒔𝒊𝒏 𝑩 𝒃 = 𝒔𝒊𝒏 𝑪 𝒄

Example 1; Using the Law of Sines (AAS) a) In ∆𝐴𝐵𝐶, 𝑚∠𝐴=48, 𝑚∠𝐵=93, and 𝐴𝐶=15, find 𝐵𝐶. A B C 𝒔𝒊𝒏 𝑩 𝒃 = 𝒔𝒊𝒏 𝑨 𝒂 𝟒𝟖° 𝒔𝒊𝒏 𝟗𝟑 𝟏𝟓 = 𝒔𝒊𝒏 𝟒𝟖 𝒂 𝟏𝟓 𝒂∗𝒔𝒊𝒏 𝟗𝟑 =𝟏𝟓∗𝒔𝒊𝒏 𝟒𝟖 𝟗𝟑° 𝒂= 𝟏𝟓∗𝒔𝒊𝒏 𝟒𝟖 𝒔𝒊𝒏 𝟗𝟑 𝒂 𝒂=𝟏𝟏.𝟐

Example 1; Using the Law of Sines (AAS) b) In ∆𝐴𝐵𝐶, 𝑚∠𝐴=48, 𝑚∠𝐵=93, and 𝐴𝐶=15, find 𝐴𝐵. A B C 𝒔𝒊𝒏 𝑩 𝒃 = 𝒔𝒊𝒏 𝑪 𝒄 𝟒𝟖° 𝒔𝒊𝒏 𝟗𝟑 𝟏𝟓 = 𝒔𝒊𝒏 𝟑𝟗 𝒄 𝒃 𝟏𝟓 𝒄∗𝒔𝒊𝒏 𝟗𝟑 =𝟏𝟓∗𝒔𝒊𝒏 𝟑𝟗 𝟑𝟗° 𝟗𝟑° 𝒄= 𝟏𝟓∗𝒔𝒊𝒏 𝟑𝟗 𝒔𝒊𝒏 𝟗𝟑 𝑪=𝟏𝟖𝟎− 𝟗𝟑+𝟒𝟖 𝑪=𝟑𝟗° 𝒄=𝟗.𝟓

Example 2; Using the Law of Sines (SSA) b) In ∆𝐾𝐿𝑀, 𝐿𝑀=9, 𝐾𝑀=14, and 𝑚∠𝐿=105, find 𝑚∠𝐾. 𝒔𝒊𝒏 𝑲 𝒌 = 𝒔𝒊𝒏 𝑳 𝒍 K L M 𝒔𝒊𝒏 𝑲 𝟗 = 𝒔𝒊𝒏 𝟏𝟎𝟓 𝟏𝟒 𝟏𝟒 𝟏𝟒∗𝒔𝒊𝒏 𝑲 =𝟗∗𝒔𝒊𝒏 𝟏𝟎𝟓 𝒔𝒊𝒏 𝑲 = 𝟗∗𝒔𝒊𝒏 𝟏𝟎𝟓 𝟏𝟒 𝟏𝟎𝟓° 𝟗 𝑲= sin −𝟏 𝟗∗𝒔𝒊𝒏 𝟏𝟎𝟓 𝟏𝟒 𝑲=𝟑𝟖.𝟒°

Example 2; Using the Law of Sines (SSA) a) A ship has been at sea longer than expected and has only enough fuel to safely sail another 42 miles. Port City Lighthouse and Cove Town Lighthouse are located 40 miles apart along the coast. At sea, the captain cannot determine distances by observation. The triangle formed by lighthouses and the ship is shown. Can the ship sail safely to either lighthouse? P S C 𝒔𝒊𝒏 𝑺 𝒔 = 𝒔𝒊𝒏 𝑷 𝒑 𝟕𝟎° 𝒔𝒊𝒏 𝟓𝟖 𝟒𝟎 = 𝒔𝒊𝒏 𝟕𝟎 𝒑 𝟒𝟎 𝒎𝒊 𝒑∗𝒔𝒊𝒏 𝟓𝟖 =𝟒𝟎∗𝒔𝒊𝒏 𝟕𝟎 𝟓𝟐° 𝟓𝟖° 𝒑= 𝟒𝟎∗𝒔𝒊𝒏 𝟕𝟎 𝒔𝒊𝒏 𝟓𝟖 𝑺=𝟏𝟖𝟎− 𝟗𝟑+𝟒𝟖 𝒑=𝟒𝟒.𝟑 𝒎𝒊. 𝑺=𝟓𝟖° Can’t go to Cove Town

Example 2; Using the Law of Sines (SSA) a) A ship has been at sea longer than expected and has only enough fuel to safely sail another 42 miles. Port City Lighthouse and Cove Town Lighthouse are located 40 miles apart along the coast. At sea, the captain cannot determine distances by observation. The triangle formed by lighthouses and the ship is shown. Can the ship sail safely to either lighthouse? P S C 𝒔𝒊𝒏 𝑺 𝒔 = 𝒔𝒊𝒏 𝑪 𝒄 𝟕𝟎° 𝒔𝒊𝒏 𝟓𝟖 𝟒𝟎 = 𝒔𝒊𝒏 𝟓𝟐 𝒄 𝟒𝟎 𝒎𝒊 𝒄∗𝒔𝒊𝒏 𝟓𝟖 =𝟒𝟎∗𝒔𝒊𝒏 𝟓𝟐 𝟓𝟐° 𝟓𝟖° 𝒄= 𝟒𝟎∗𝒔𝒊𝒏 𝟓𝟐 𝒔𝒊𝒏 𝟓𝟖 𝒄=𝟑𝟕.𝟐 𝒎𝒊. Can go to Port City

Example 2; Using the Law of Sines (SSA) b) The right-fielder fields a softball between first base and second base as shown. If the right-fielder throws the ball to second base, how far does she throw the ball? RF 2nd 1st 𝒔𝒊𝒏 𝑹𝑭 𝟔𝟎 = 𝒔𝒊𝒏 𝟏𝒔𝒕 𝒙 𝟕𝟐° 𝒔𝒊𝒏 𝟕𝟐 𝟔𝟎 = 𝒔𝒊𝒏 𝟒𝟎 𝒙 𝟔𝟖° 𝟔𝟎 𝒇𝒕 𝒙∗𝒔𝒊𝒏 𝟕𝟐 =𝟔𝟎∗𝒔𝒊𝒏 𝟒𝟎 𝟒𝟎° 𝒙= 𝟔𝟎∗𝒔𝒊𝒏 𝟒𝟎 𝒔𝒊𝒏 𝟕𝟐 𝑹𝑭=𝟏𝟖𝟎− 𝟔𝟖+𝟒𝟎 𝒙=𝟒𝟎.𝟔 𝒇𝒕. 𝑹𝑭=𝟕𝟐°

#42: Law of Sines Questions? Summarize Notes Homework Quiz