Question

If cos (a+b) =0 and sin (a-b) 0=3,then find the value of a and b where a and b are acute angles

Answer 1

If cos (a+b) =0 and sin (a-b) = 0,then find the value of a and b where a and b are acute angles. [ correct question]

Given:

cos (a+b) =0 and sin (a-b) = 0

To find:

The value of a and b

Solution:

As per the given data of the question:

cos (a+b) =0 and sin (a-b) = 0

so,

cos (a+b) = cos 90° and sin (a-b) = sin 0°

As per the rule of the trigonometric value

cos 90° = 0 and sin 0° = 0

So

a+b = 90°

a-b = 0°

adding the equation we get the values as:

a = 45° and b = 45°

The value of a and b is 45°

Answer 2

Answer:

A=60°. ,B=30°

A=60°. ,B=30°Step-by-step explanation:

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i)

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i) 2b=60

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i) 2b=60b=30

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i) 2b=60b=30Putting the value of B in (i), we get

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i) 2b=60b=30Putting the value of B in (i), we get⇒ A = 30° + 30° = 60°

A=60°. ,B=30°Step-by-step explanation:cos (A+B)=0. =cos 90. ,sin(a-b)=1/2=sin30a+b=90. ,a-b=30 ,a=30+b ....(i)30+b+b=90. from (i) 2b=60b=30Putting the value of B in (i), we get⇒ A = 30° + 30° = 60°∴ A = 60°, B = 30°