Question

If in a ∆ ABC right angled at B, AB= 6 units BC =6 units, find the value of Sin A. cos C + cos A. sin C.
(Class 10 Maths Sample Question Paper)

Answer 1

SOLUTION:
Given :
In ∆ ABC , AB = 6 , BC = 8

AC² = AB² + BC²
AC² = 6² + 8²
AC² = 36 + 64
AC² = 100
AC = √ 100
AC = 10

sin A = P / H = 8/10
cos A = B / H= 6/10

For ∠C , Base = BC, Perpendicular= AB, hypotenuse = AC
sin C = P /H =  AB / AC = 6/10
cos C = B / H  = BC /AC= 8/ 10

sin A. cos C + cos A. sin C

= (8/10)(8/10) + (6/10)(6/10)
= (64/100) + (36/100)
= (64 +36 )/100
= 100 / 100
= 1

HOPE THIS WILL HELP YOU....

Answer 2

............. Dont know