Search 74,111 tutors
0 0

# Trigonometry

(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

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-cos2B) = 56/65

------------------------------------
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