In a right angle triangle ABC, right angle is at B, if tan A = 3 then find the value of
(i) sin A cos C + cos A sin C (ii) cos A cos C − sin A sin C
Answers
Answered by
18
It is given that ,
In ∆ABC ,
<B = 90°
tan A = 3/1 = BC/AB
BC = 3 , AB = 1
AC² = BC² + AB²
[ By Phythogarian theorem ]
AC² = 3² + 1²
AC² = 9 + 1
AC = √10
i ) SinAcosA + CosAsinC
= ( 3/√10 )( 3/√10 ) + (1/√10)(1/√10 )
= 9/10 + 1/10
= 10/10
= 1
ii ) cosAcosC - sinAsinC
= (1/√10)(3/√10) - ( 3/√10 )(1/√10)
= 3/√10 - 3/√10
= 0
I hope this helps you.
: )
In ∆ABC ,
<B = 90°
tan A = 3/1 = BC/AB
BC = 3 , AB = 1
AC² = BC² + AB²
[ By Phythogarian theorem ]
AC² = 3² + 1²
AC² = 9 + 1
AC = √10
i ) SinAcosA + CosAsinC
= ( 3/√10 )( 3/√10 ) + (1/√10)(1/√10 )
= 9/10 + 1/10
= 10/10
= 1
ii ) cosAcosC - sinAsinC
= (1/√10)(3/√10) - ( 3/√10 )(1/√10)
= 3/√10 - 3/√10
= 0
I hope this helps you.
: )
Attachments:
Similar questions