Solve the triangle using the Law of Sines Assume a 20 b 35

Solve the triangle using the Law of Sines. Assume a = 20, b = 35, and.A = 28degree. (Round the length to one decimal place and the angles to the nearest whole number. If either set of answers is not necessary, cross out those boxes.) A pilot flies in a straight path for 1.5 hours. She then makes a course correction, heading 10degree to the right of her original course, and flies 2 hours in the new direction. If she maintains a constant speed of 500 mi/h, how far is she from her starting position?

Solution

Solution : Given a = 20, b = 35 and A = 35.

We have      b sin A = 35 sin 28^o = 16.43,    and   35 sin 28^o < 20 < 35.

Since    b sin A < a < b, therefore two triangles are possible.

We know sine rule

         a/sin A= b/ sinB

or 20/sin28 = 35/ sin B

or       sin B = 35 sin 28^o/ 20

                 =35*0.27/ 20

                 = 0.82157523

Therefore angle   B₁ = 55^o   and B₂ = 180^o - 55^o= 125.

                                                        Ans.

We also know that angle C = 180-(28+55.28), and C = 180-(28+125),   

             Angle C₁ = 97^o and C₂ = 27^o                 Ans.

Again use sine rule,

    a/sin A = c₁/sin C₁

             c₁ = a sin C₁/ sinA

               = 20*sin 97/ sin 28

           c₁= 46.3.                                                                 Ans

Similarly, by the sine rule,

    a/sin A = c₂/sin C₂

             c₂ = a sin C₂/ sinA

               = 20*sin 27^o/ sin 28^o

           c₂ = 19.3                                                              Ans