Question

Tell me this please tell me correctly​

Answer 1

$\huge{\underline{\underline{Question\::-}}}$

>> If value of Sin∅ = $\dfrac{\sqrt{3}}{2}$, Find out all ratio of ∅ .

$\huge{\underline{\underline{Answer\::-}}}$

$\underline{\underline{Given\::-}}$

Sin∅ = $\dfrac{\sqrt{3}}{2}$

• Then As we know that

Sin∅ = $\dfrac{Perpendicular }{Hypotenuse}$

>>Then from above's information we can get the ratio of base with respect to the Perpendicular and Hypotenuse

$\underline{\underline{Solving\::-}}$

• Let

>>the common divisor be m

>> base be x

• Then

>> Perpendicular = $\sqrt{3}$ m

>> Hypotenuse = 2 m

• By using Pythagorean theorem

$Hypotenuse^2 = Perpendicular^2 + Base^2$

=> $(2m)^2 = (\sqrt{3}m)^2 + x^2$

=> $4m^2 = 3m^2 + x^2$

=> $x^2 = 4m^2 - 3m^2$

=> $x^2 = m^2$

=> $x = \sqrt{m^2}$

=> x = m

• Now we will find out other ratio

>> Cos∅ = $\dfrac{Base}{Hypotenuse}$

= $\dfrac{m}{2m}$

=$\dfrac{1}{2}$

>> Tan∅ = $\dfrac{Perpendicular}{Base }$

= $\dfrac{\sqrt{3}m}{m}$

= $\dfrac{\sqrt{3}}{1}$

= $\sqrt{3}$

>> Cot∅ = $\dfrac{Base}{Perpendicular}$

= $\dfrac{m}{\sqrt{3}m}$

= $\dfrac{1}{\sqrt{3}}$

>> Cose∅ = $\dfrac{Hypotenuse}{Perpendicular}$

= $\dfrac{2m}{\sqrt{3}m}$

= $\dfrac{2}{\sqrt{3}}$

>> Sec∅ = $\dfrac{Hypotenuse}{Base}$

= $\dfrac{2m}{m}$

= $\dfrac{2}{1}$

= 2