Question

if sin O =3/5 find , tan O​

Answer 1

Answer:

sin O=3/5

sinO=P/H

so base =√H²+P²

B=√(5)²+(3)²

B=√25+9

B=√34

so tan O=P/B

=3/√34

is correct answer

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Answer 2

Given :-

${sin\theta} = \dfrac{3}{5}$

To find :-

${tan\theta}$

To know :-

${sin\theta} = \dfrac{opposite}{Hypotenuse}$

${tan\theta} = \dfrac{opposite}{adjacent}$

Solution:-

We know only opposite side and Hypotenuse

From Pythagoras theorem,

AB² + BC² = AC²

AB² +(3)² = (5)²

AB² + 9 = 25

AB² = 25-9

AB² = 16

AB² =(4)²

AB = 4

Now,

${tan\theta} = \dfrac{opposite}{adjacent}$

adjacent side = 4

opposite side = 3

${tan\theta} = \dfrac{opposite}{adjacent}$

${tan\theta} = \dfrac{3}{4}$

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Know more:-

Trigon metric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

csc²θ - cot²θ = 1

Trigometric relations

sinθ = 1/cscθ

cosθ = 1 /secθ

tanθ = 1/cotθ

tanθ = sinθ/cosθ

cotθ = cosθ/sinθ

Trigonmetric ratios

sinθ = opp/hyp

cosθ = adj/hyp

tanθ = opp/adj

cotθ = adj/opp

cscθ = hyp/opp

secθ = hyp/adj