Question

3
3. If sin A= 3/4
calculate cos A and tan A
​

Answer 1

1. cosA= underroot 7/4
2. tanA=3/underroot 7

Step-by-step explanation:

sin^2A+cos^2A=1

(3/4)^2+cos^2A=1

9/16+cos^2A=1

cos^2A=1-9/16

cos^2A=16-9/16

cos^2A=7/16

cosA=under root 7/4

tanA=sinA/cosA

tanA=3/4/under root 7/4

tanA=3/under root 7

Answer 2

$\huge \sf Answer :$

$\sf cosA = \dfrac{\sqrt{7}}{4}$

$\sf tanA = \dfrac{3}{\sqrt{7}}$

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$\huge \sf Solution :$

We know that,

$\sf sin \theta = \dfrac{Perpendicular}{Hypotenuse}$

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According to Statement,

$\sf sin \theta = \dfrac{3}{4}$

$\sf \implies \dfrac{Perpendicular}{Hypotenuse} = \dfrac{3}{4}$

∴ Perpendicular for θ = 3x

and the hypotenuse for θ = 4x

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By Pythagoras' Theorem,

Hypotenuse² = Perpendicular² + Base²

$\sf⇒ AC² = AB² + BC²$

$\sf⇒ (4x)² = (3x)² + BC²$

$\sf⇒ 16x² = 9x² + BC²$

$\sf⇒ 16x² - 9x² = BC²$

$\sf⇒ 7x² = BC²$

$\sf⇒ \sqrt{7x^{2}} = BC$

$\sf⇒ BC = \sqrt{7}x$

⠀ ⠀

Now, We have cosA and tanA will be,

$\sf cosA = \dfrac{Base}{Hypotenuse}$

$\sf⇒ cosA = \dfrac{\sqrt{7}x}{4x}$

$\sf⇒ cosA = \dfrac{\sqrt{7}}{4}$

⠀ ⠀

$\sf tanA = \dfrac{Perpendicular}{Base}$

$\sf⇒ tanA = \dfrac{3x}{\sqrt{7}x}$

$\sf⇒ tanA = \dfrac{3}{\sqrt{7}}$