Question

If SinA/SinB=√2 and TanA/TanB=√3, find A and B

Answer 1

SinA/sinB=√2

or, sinA=√2sinB

or, sin²A=2sin²B

∴, cos²A=(1-sin²A) [∵, sin²A+cos²A=1]

or, cos²A=1-2sin²B

∴, tanA/tanB=√3

or, tan²A/tan²B=3

or, (sin²A/cos²A)/(sin²B/cos²B)=3

or, 2sin²B/1-2sin²B×cos²B/sin²B=3

or, 2cos²B/(1-2sin²B)=3

or, 2cos²B=3-6sin²B

or, 2(1-sin²B)=3-6sin²B

or, -2sin²B+6sin²B=3-2

or, 4sin²B=1

or, sin²B=1/4

or, sinB=1/2 [∵, B is an acute angle]

∴, sinB=sin30°

or, B=30° and

sinA=√2sinB

or, sinA=√2sin30°

or, sinA=√2×1/2

or, sinA=1/√2

or, sinA=sin45°

or, A=45°

∴, A=45°, B=30°

Answer 2

Answer:

A =45° ; B =30°

Step-by-step explanation:

sinA /sinB. =√2

sinA = √2sinB

by assumption ,

sin 45° = √2 sin 30°

1/√2 =√2 . 1/2

1/√2 = √2/2

1/√2 =1/√2

hence our assumption is correct