^{2}A) = ± 4/5

^{2}(A+B)) = ± 12/13

^{2}B) = 56/65

^{2}C) = 5/13 (positive as C is acute)

(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

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Ahmed F. | Smart and Patient Math and Physics Tutor Smart and Patient Math and Physics Tutor...

i.

sin(A) = √(1-cos^{2}A) = ± 4/5

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

sin(A+B) = √(1-cos^{2}(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^{2}B) = 56/65

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

ii .

Cos(C) = √(1-sin^{2}C) = 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

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