If sin (A+2B)-and cos (A+4B)-0, A B and A +4B < 90°, then find the va

1.	If sin (A+2B)-and cos (A+4B)-0, A B and A +4B < 90°, then find the value of A and B.
Answer» sin(A+2B)=sin60°.=>A+ 2B=60°. cos(A+4B)=cos90° =>A+4B=90°. Now we have Subtract both equation. - 2B=-30° =>B=15°. Hence A+2.15°=60°=>A=60°-30°=30°. A=30°,B=15° As we have sin(A+2B)=sin60°.=>A+ 2B=60°. cos(A+4B)=cos90° =>A+4B=90°. Now we havenow

If sin (A+2B)-and cos (A+4B)-0, A B and A +4B < 90°, then find the value of A and B.

Answer»

sin(A+2B)=sin60°.=>A+ 2B=60°. cos(A+4B)=cos90° =>A+4B=90°. Now we have

Subtract both equation. - 2B=-30° =>B=15°. Hence A+2.15°=60°=>A=60°-30°=30°. A=30°,B=15°

As we have sin(A+2B)=sin60°.=>A+ 2B=60°. cos(A+4B)=cos90° =>A+4B=90°. Now we havenow