In ΔABC, m∠A = 90 and if AB : BC = 1:2 find sinB, cosc, tanC.
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In ∆ABC ,
m<A = 90°
and AB : BC = 1 : 2
Let AB = x , BC = 2x
By Phythogarian theorem ,
BC² = BA² + AC²
=> AC² = BC² - BA²
=> AC² = ( 2x )² - x²
= 4x² - x²
= 3x²
Therefore ,
AC = √3 x
Now ,
i ) SinB= CA/BC
= (√3 x)/2x
= √3/2
Sin B = √3/2
ii ) Cos C = CA/BC
= ( √3 x )/2x
= √3/2
Cos C = √3/2
iii ) tan C = BA/CA
= 1x/(√3 x)
= 1/√3
tan C = 1/√3
••••
m<A = 90°
and AB : BC = 1 : 2
Let AB = x , BC = 2x
By Phythogarian theorem ,
BC² = BA² + AC²
=> AC² = BC² - BA²
=> AC² = ( 2x )² - x²
= 4x² - x²
= 3x²
Therefore ,
AC = √3 x
Now ,
i ) SinB= CA/BC
= (√3 x)/2x
= √3/2
Sin B = √3/2
ii ) Cos C = CA/BC
= ( √3 x )/2x
= √3/2
Cos C = √3/2
iii ) tan C = BA/CA
= 1x/(√3 x)
= 1/√3
tan C = 1/√3
••••
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