Question

if sin(a+b)=1 and cos(a-b)=root3/2 find a and b.​

Answer 1

Answer: step by step explaination:

sin(A+B)=1

cos(A-B)=√3/2

We know sin-¹ =90

A+B=9 _ _ _ _(1)

cos (A-B)=root 3/2

A-B=cos-¹ root 3/2

We know cos-¹ root 3/2=30

A-B=30 _ _ _ _ _ (2)

Solve (1) & (2)

A+B=90

A-B=30 add

2A=120

A=60 B=30 (Ans)

Answer 2

Step-by-step explanation:

hiiiii.......

Here is your answer------

sin (a+ b) = 1

We can replace 1 by sin 90° as (sin 90° = 1)

sin ( a+ b ) = sin 90°

Now, a+ b= 90° ..... 1st equation

Also, cos (a- b ) = root 3/2

which means,

cos (a-b ) = cos30°

a - b = 30° ....... 2nd equation

solving 1st and 2nd equation by using elimination method..

a + b = 90°

a - b = 30°

2a = 120°

a= 120/2 °

a = 60°

Now, a + b = 90°

60° + b = 90°

b = 90° - 60° = 30 °

Hence, a = 60° and b= 30°

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Hope it helps ✌️✌️

Thanks ..