Question

if 3cotA =2,find the value of (4sinA - 3cosA)/2sinA + 6cosA

Answer 1

Answer:

3cotA=2

cotA=2/3. (cot=base/perpendicular)

so, base=2, perpendicular =3

sinA=? (sin=per../hypotenuse)

by P. G. T:-

(2)sq. +(3)sq. =(hypotenuse)sq.

4+9= (hypotenuse)sq.

root 13 =hypotenuse

sinA=3/root 13

cosA=? (base/hypotenuse)

cosA=2/root 13

Now,,,(4sinA - 3cosA)/2sinA + 6cosA

(4*3/root13 -3*2/root13)/2*3/root 13 + 6*2/root 13

(12/root 13 - 6/root13)/ 6/root 13 + 12/root13

(6/root13) / 18/root13

=6/18

=1/3