Question

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Qᴜᴇsᴛɪᴏɴ-:
Iғ sɪɴ A=3/4, ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴄᴏs A ᴀɴᴅ ᴛᴀɴ A?

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Answer 1

Sin∅ = perpendicular/Hypotenuse.....

In right angle triangle

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

(4)²=(3)²+(base)²

16 = 9 + base²

16-9 = base²

base = √7

SinA = 3/4

CosA = base/hypotenuse = √7/4

TanA = SinA/CosA = (3/4)/(√7/4) = 3/√7

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Answer 2

Hii there !

Given:-

• SinA = $\sf\dfrac{3}{4}$

To find:-

• TanA
• CosA

formulas to be used:-

$\underline{\boxed{\sf SinØ = Perpendicular/hypotenuse}}$

$\underline{\boxed{\sf hypotenuse ² = perpendicular ²+base²}}$

Supposition:-

• let the perpendicular be 3x and hypotenuse be 4x.

Solution:-

according to Pythagoras theorem

$\implies$ H² = P²+B²

$\implies$ (4x)² = (3x)²+B²

$\implies$ 16x²=9x²+B²

$\implies$ B² = 7x²

$\implies$ B = √7x.

Putting the value of B in the formula of TanØ

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

TanA = $\sf\dfrac{3x}{√7x}$ = $\sf\dfrac{3}{√7}$

Putting the value of B in the formula of CosØ

$\implies$ CosA = $\sf\dfrac{Base}{Hypotenuse}$

$\implies$ CosA = $\sf\dfrac{√7x}{4x}$ = $\sf\dfrac{√7}{4}$