Question

You are given that cos(A)=12/13, with A in Quadrant I, and cos(B)=5/13, with B in Quadrant I. Find cos(A−B). Give your answer as a fraction.

Answer 1

Ans. Cos(A-B) = 120/169

Ars Given Cos (A)= 12 Cos (B) = s 13 M Also, it is given that A & B dies in quachent 1. Therefore, the trignomenteie cagles will be positive, because in birst quachant all the teignemeteic ratios positive, Now, Cos A に 12 Bose Hypotenuse 13 Using Pythagores theorem, (hypotenuse)2 = (Perpendicular) + (Base)? (13)? - 144 (Perpendicular)" + (121² 169 (Perpendicular) + (Perpendicular) ?- 169-144 - (Pexpendicular) = 25 Perpendicula a 125 = (Resecting the negative value because angle A lies in First quadeant) - Perpendicula = 5 † 5 Therefore, Sin A Perpendicular Hypotenuse مام s 13 Sin A= S 13

Also, Cos B = 5 13 Base Hypotenuse → Similarly, using bythagoras theoren- - (My petenuse)" = (Perpendicular)² + (Base)? (13)? (Perpendicelor) + (5) 169 (Perpendicular)? - -) (Perpendicular) = 169-25 - (Peependicula) - Al Perpendicular - 5144 = + 25 + 12 Derpendicula = 12 (Rejecting the negative value) Sin B = Perpendicular hypotenuse 12 13 =) Sin B 12 13 Now, using identity Cas (A-B) = COSA. (OSB & Sina. SinB Putting the velves in dave bornola- Cas (A-B) = ( 12 + 5) =) (as (A-B) + 169 60 169 =) (as (A-B) Got Go 16 | Cos (A-B) 120 169 . Ans.

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