Question

If 3 tan A = 4, then find sin A and cos A ?

Answer 1

given

3tanA =4


tan A = 4/3


tan A = perpendicular /base

perpendicular = 4

base = 3


by Pythagoras theorem

(hypotenuse)²= (perpendicular)²+(base)²

(hypotenuse)² = (4)²+(3)² = 16+9 = 25

hypotenuse =± 5

thus sinA = perpendicular/hypotenuse

= ±3/5

cosA = base /hypotenuse

= ±4/5

Answer 2

→3tan A = 4.

→3tan A = 4.→tan A 4/3

→3tan A = 4.→tan A 4/3→P/ b = tan A

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,→sin A = P/h

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,→sin A = P/h→sin A = 3/5

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,→sin A = P/h→sin A = 3/5→ cos A = b/h

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,→sin A = P/h→sin A = 3/5→ cos A = b/h→cos A = 4/5

→3tan A = 4.→tan A 4/3→P/ b = tan A→perpendicular(p) = 4, Base(b)= 3 BY PYTHAGORES THEOREME :→h² = p² + b² [h = hypotenous]→h² = {4}² + {3}²→ h² = 16 + 9→h² = 25→h=√25→(h = 5)NOW,,,→sin A = P/h→sin A = 3/5→ cos A = b/h→cos A = 4/5→HOPE IT HELPS ...

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