Question

SinA / sinB = √2 and tanA / tanB =√3 find A and B if A and B acute angles hai

Answer 1

A = 45°

B= 30°

sinA/sinB = √2

tanA/tanB = √3

( tanA= sinA/cosA)

⇒tanA/tanB = sinA/cosA÷sinB/cosB = √3

sinA/cosA × cosB/sinB

(sinA/sinB = √2)

⇒√2cosB/cosA =√3

⇒cosB/cosA=√3/√2

⇒√1-sin²B/√1-sin²A = √3/√2

⇒√1-sin²B/√1-2sin²B = √3/√2

⇒2(1-sin²B) =3(1-2sin²B)

⇒2-2sin²B =3-6sin²B

4sin²B=1

sin²B =1/4

sinB=±1/2

for acute angles

sinB =1/2 =sin30°

B=30°

sinA =sinB√2

sinA =√2/2

sinA=1/√2

A=45°