if in the right angled triangle ABC, <b=90° ,bc= 7cm , AC -AB =1cm ,then find the values of sin A and Cos A
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AC-AB=1
AC=1+AB
so by Pythagoras theorem
AC^2=AB^2+BC^2
(1+AB)^2=AB^2+7^2
1+2AB+AB^2=AB^2+49
1+2AB=49
2AB=49-1
2AB=48
AB=24
now AC=25
so sinA=7/25
and CosA=24/25
AC=1+AB
so by Pythagoras theorem
AC^2=AB^2+BC^2
(1+AB)^2=AB^2+7^2
1+2AB+AB^2=AB^2+49
1+2AB=49
2AB=49-1
2AB=48
AB=24
now AC=25
so sinA=7/25
and CosA=24/25
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