if sin(A+B)=1 and cos(A-B)=1 0 degree is lesser than a+b and is lesser than 90 degrees and a is greater than b find a b
Answers
Answered by
3
Answer:
Given sin(A+B)=1
⇒sin(A+B)=sin90
o
(∵sin90
o
=1)
⇒A+B=90
o
…………(1)
Again, cos(A−B)=
3
/2
⇒cos(A−B)=cos30
o
(∵cos30
o
=
3
/2)
⇒A−B=30
o
…………(2)
Adding (1)+(2)
A+B+A−B=90
o
+30
o
⇒2A=120
o
⇒A=120/2
⇒A=60
o
Putting A=60
o
in equation-(2) we get
A−B=30
o
⇒60
o
−B=30
o
⇒60
o
−30
o
=B
⇒B=30
o
∴A=60
o
;B=30
o
.
Step-by-step explanation:
PLZ MARK ME BRAINLIST
Answered by
5
Answer:
A = 45° B = 45°
Step-by-step explanation:
sin(A+B) =1
sin(A+B) = sin90°
A+B = 90° -(i)
cos(A-B) = 1
cos(A-B) = cos0°
A-B = 0° -(ii)
On adding both (i) and (ii)
2A = 90°
A = 45°
Putting the value of A in (i)
45° + B = 90°
B = 90° - 45°
B = 45°
PLEASE MARK IT AS BRAINLIEST
Similar questions