How do you solve the triangle ABC, given a=5, A=42, b=7?

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How do you solve the triangle ABC, given a=5, A=42, b=7?

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Answer 1 · 2017-01-08T12:17:55

Solution: $a = 5; b = 7; c = 6.95, ∠ A = 42^{0} : ∠ B = {69.52}^{0} : ∠ C = {68.48}^{0}$ $a=5 ; b=7 ; c= 6.95 , /_A=42^0 : /_B=69.52^0 : /_C=68.48^0$

Explanation:

We know by sine law $a/sinA=b/sinB or 5/sin42=7/sinB or sin B =(7 sin 42)/5= 0.93678 :./_B=sin^-1(0.93678)=69.52^0(2dp) :. /_C=180-(42+69.52)=68.48^0$

Similarly , $a/sinA=c/sinC or 5/sin42=c/sin68.48 or c=5*sin68.48/sin42 =6.95$

Solution $a=5 ; b=7 ; c= 6.95 , /_A=42^0 : /_B=69.52^0 : /_C=68.48^0$ [Ans]

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How do you solve the triangle ABC, given a=5, A=42, b=7?

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