Question

if sin a+b =1 and cos A-B root 3/2 0° <A+B _<90°,A+B, then find a and b​

Answer 1

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:

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