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Heya!!
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Hope it helps u :)
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Hope it helps u :)
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Sanaya0:
thanks ^_^
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tanФ = perpedicular/basetanA = AC/BC
if , tanA = 3/4 then,
AC = 3
BC = 4
In ΔABC ∠C = 90°,
AB² = BC² + AC²
AB² = (4)² + (3)²
AB² = 16+9
AB² = 25
AB = √25
AB = 5
LHS =sinA*cosB + cosA*sinB
LHS = (3/5)(3/5) + (4/5)(4/5)
LHS = 9/25 + 16/25
LHS = 9+16/25
LHS = 25/25
LHS = 1
if , tanA = 3/4 then,
AC = 3
BC = 4
In ΔABC ∠C = 90°,
AB² = BC² + AC²
AB² = (4)² + (3)²
AB² = 16+9
AB² = 25
AB = √25
AB = 5
LHS =sinA*cosB + cosA*sinB
LHS = (3/5)(3/5) + (4/5)(4/5)
LHS = 9/25 + 16/25
LHS = 9+16/25
LHS = 25/25
LHS = 1
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