急! F5 Trigonometry 11q5

a) tan∠RBC＝　ＲＣ／ＢＣ　＝　１８／６０∴∠RBC　＝　１６。７°
the angle of elevation ofthe top R of the rod from B is 16.7°ｂ）　∠ＮˊＢＣ＝　３０ﾟ　（Ｇｉｖｅｎ）∠ＮˊＢＡ　＝　７０°（Ｇｉｖｅｎ）∴ＡＢＣ　＝　１００°ｂｙ　Ｓｉｎｅ　Ｒｕｌｅ，　ｉｎ　△ＡＢＣＡＣ／ｓｉｎ∠ＡＢＣ　＝　ＢＣ／ｓｉｎ∠ＢＡＣＡＣ　＝　６０ｓｉｎ１００°／ｓｉｎ２０°　　＝　１７３ｍｃ）∠ＡＣＢ　＝　１８０°　－　１００°　－２０°　＝　６０°　ｔｈｅ　ａｒｅａ　ｏｆ　△ＡＢＣ＝　１／２　（ＡＣ）（ＢＣ）　ｓｉｎ６０°
　＝　１／２　（１７２。７６）（６０）ｓｉｎ６０°　＝　４４８８。５≒　４４９０　（ｓｑ）ｍ ａ）　ａｔ　４ｐｍ，　ｂｅａｒｉｎｇ　ｏｆ　ｓｈｉｐ　Ａ　ｆｒｏｍ　ｔｈｅ　ｐｏｒｔ　Ｐ　ｉｓ　Ｎ３５°Ｗ，　ａｎｄ　ＡＰ　＝　８０Ｋｍｂｅａｒｉｎｇ　ｏｆ　ｓｈｉｐ　Ｂ　ｆｒｏｍ　ｔｈｅ　ｐｏｒｔ　Ｐ　ｉｓ　Ｎ２５°Ｅ，　ａｎｄ　ＡＰ　＝　４５Ｋｍ∴　∠　ＡＰＢ　＝　６０° AB^2 = AP^2 + PB^2 -2(AP)(PB)cos ∠APB= 80^2 +45^2 - 2(80)(45)cos６０° = 4825∴ AB = 69.5 Km b) Cos ∠PAB = [AP^2 +AB^2 -PB^2]/2(AP)(AB) = [80^2 + 69.46^2 - 45^2]/2(80)(69.45) = 0.827788∠PAB = 34.1° the bearing of B from A is S 69.1° E
for 34.1 + 35 = 69.1