Question

If cot (A-B)=√3 2cos(A+B)=1 find the acute angles A and B ​

Answer 1

Step-by-step explanation:

Given, sin(A+2B)=√3/2

cos(A+4B)=0

A>B,

We know that, sin60= √3/2

and cos90 =0

Consider,

sin(A+2B)= √3/2

and sin60 = √3/2

⟹(A+2B)=60 ---------------(i)

Consider,

cos(A+4B)=0 and cos90 =0

⟹(A+4B)=90 ---------------(ii)

Solve (i) and (ii) :

(A+2B)=60

(A+4B)=90

Subtracting (i) from (ii),

2B=30

B= 30/2 =15

From (ii)

(A+4B)=90

Also, B=15

A=90 −60

A=30

∴A=30,B=15

Answer 2

Step-by-step explanation:

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