Trigonometry

Question

(i) Given that cos A =3/5 and cos(A+B) =-5/13, where A and B are acute, evaluate sin B.

 

(ii) Another acute angle C is such that sin C=12/13. Find sin(A+B+C).

 

(iii) Determine whether A, B and C are angles of a triangle.

 

 

 

Ans: (i) 56/65

       (ii) 0

       (iii) Yes

Answer 1

i.

 

sin(A) = √(1-cos²A) = ± 4/5

since A is acute then sin(A) = +4/5

 

sin(A+B) = √(1-cos²(A+B))  = ± 12/13

since both A and B are acute , then A+B is either in first or second quadrants ,  therefore Sin(A+B) = + 12/13

 

We have two equations for cos(A+B) and Sin(A+B)

Cos(A+B) = Cos A Cos B - Sin A Sin B    =>  -5/13 = 3/5 Cos B  - 4/5 Sin B  ... eqn I

Sin(A+B)  = Sin A Cos B + Cos A Sin B   => 12/13 = 4/5 Cos B + 3/5 Sin B  ... eqn II

 

multiply eqn I by 3/4 and add it to eqn II

-5*3/(4*13)+12/13 = (9/20 + 4/5) Cos B => Cos B = 33/65 => Sin B = √(1-cos^₂B) = 56/65

 

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ii . 

Cos(C) = √(1-sin²C) = 5/13  (positive as C is acute)

 

Sin (A+B+C) = Sin (A+B) Cos C + Cos (A+B) Sin C = 12/13*5/13 + (-5/13)*12/13 = 0

 

iii.

Since Sin(A+B+C) = 0  => A+B+C = 180 or 360

but since A, B and C are acute => their sum is less than 90+90+90 ( i.e. 270)

Therefore A+B+C = 180

Therefore they can form angles of a triangle