Question

if sina=3/4 calculate cosA and secA​

Answer 1

Given:-

• Sin A = 3/4

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To find:-

• Cos A and Sec A.

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$\huge \tt \: Solution$

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sin θ = $\frac{Perpendicular}{Hypotenuse}$

$\implies \tt \: sin A \: = \frac{3}{4}$

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• We need AB beacuse AB is base.

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By using Pythagoras theorem that is

☛ (Base)² + (Perpendicular)² = (Hypotenuse)²

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➢Base = AB = ?

➢Perpendicular = BC = 3

➢Hypotenuse = AC = 4

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Put the values,

⇒(AB)² + (3)² = (4)²

⇒(AB)² = (4)² - (3)²

⇒(AB)² = 16 - 9

⇒(AB)² = 7

⇒AB = √7

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So,

$\large\bf\underline\red{Cos A}$

☘ Cos θ = $\frac{Base}{Hypotenuse}$

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➢Base = AB = √7

➢Hypotenuse = AC = 4

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☆ cos A = √7/4

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$\large\bf\underline\purple{SecA}$

☘ sec θ = $\frac{Hypotenuse}{Base}$

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➢Base = AB = √7

➢Hypotenuse = AC = 4

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☆sec A = 4/√7

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More ratio's:-

☘ tan θ = $\frac{Perpendicular}{Base}$

☘cot θ = $\frac{Base}{Perpendicular}$

☘ cosec θ = $\frac{Hypotenuse}{Perpendicular}$