ifsin(A+2B)=√3/2,cos(A+4B)=0,A>B and A+4B<90°,then find A and B
Answers
Answered by
3
Here is ur answer. Hope it helps.!!!!
Attachments:
himanshuawasthi:
bhai division galat kya hain 15*4 =60 hota hain ok
15×4=60 hota h
15×5=75 hota h
ok
Answered by
4
sin(A+2B)=√3/2
sin(A+2B)=sin60°
then
A+2B=60°-----------(1)
cos(A+4B)=0
cos(A+4B)=cos90°
A+4B=90°-----------(2)
eq2 is subtracted to eq1
A+4B-A-2B=90°-60°
2B=30°
B=15°
Putting in eqn(1)
A+2×15°=60°
A+30°=60°
A=30°
sin(A+2B)=sin60°
then
A+2B=60°-----------(1)
cos(A+4B)=0
cos(A+4B)=cos90°
A+4B=90°-----------(2)
eq2 is subtracted to eq1
A+4B-A-2B=90°-60°
2B=30°
B=15°
Putting in eqn(1)
A+2×15°=60°
A+30°=60°
A=30°
Similar questions