if sinA=3/4 calculate cosA and tanA
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kevin20:
same answer
what
i have also solved this question by same method
kk
Answered by
25
let Δ ABC be right angled at ∠B
sinA=3/4=BC/AC
let BC=3k,AC=4k
inΔABC,∠B=90°
∴By Pythagoras theorem,
(AC)²=(AB)²+(BC)²
⇒(4k)²=(AB)²+(3k)²
⇒(AB)²=16k²-9k²
⇒(AB)²=7k²
⇒AB=√7k
∵cos A=base/hypotenuse
∴cos A=AB/AC=√7k/4k=√7/4
∵tan A=perpendicular/base
∴tanA=BC/AB=3k/√7k=3/√7
sinA=3/4=BC/AC
let BC=3k,AC=4k
inΔABC,∠B=90°
∴By Pythagoras theorem,
(AC)²=(AB)²+(BC)²
⇒(4k)²=(AB)²+(3k)²
⇒(AB)²=16k²-9k²
⇒(AB)²=7k²
⇒AB=√7k
∵cos A=base/hypotenuse
∴cos A=AB/AC=√7k/4k=√7/4
∵tan A=perpendicular/base
∴tanA=BC/AB=3k/√7k=3/√7
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