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If tan A=3/4, then sinAcosA = 12/25....

If `tan A=3/4`, then `sinAcosA = 12/25`.

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Given, `tanA= 3/4=P/B= ("Perpendicular")/("Base"0`
Let P=3k and B=4k
By pythagoras theorem,
`H^(2) = P^(2) + B^(2)= (3K)^(2)+(4k)^(2)`
`=9k^(2) + 16k^(2)= 25k^(2)`
`rArr H=5k` (since, side cannot be negative)
`therefore sin A = P/H=(3k)/(5k) = 3/5` and `cosA = B/H= (4k)/(5k)=4/5`
Now, sinAcosA=`3/5.4/5=12/25`
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