Question

4. If sin =1/2 then find cos ​

Answer 1

Given : $\sf{\sin \theta = \dfrac{1}{2}}$

Need To Find : The value of cos $\theta$ .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

❍ Basic Formulas of Trigonometry is given by :

$\boxed { \begin{array}{c c} \\ \dag \qquad \large {\underline {\bf{ Some \:Basic\:Formulas \:For\:Trigonometry \::}}}\\\\ \sf{ In \:a \:Right \:Angled \: Triangle-:} \\\\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star  \tan \theta = \dfrac{Perpendicular}{Base}} \end{array}}$

Then,

• $\sf{\sin \theta = \dfrac{1}{2}= \dfrac{Perpendicular} { Hypotenuse} }$

Therefore,

• Perpendicular of Right Angled Triangle is 1 cm .

• Hypotenuse of Right angled triangle is 2 cm .

⠀⠀⠀⠀⠀Finding Base of Right angled triangle :

$\sf{\underline {\dag As, \:We \:Know\:that \::}}\\ \\ \bf{ By \:Pythagoras\:Theorem\::}$

$\underline {\boxed {\sf{\star (Perpendicular)^{2} + Base^{2}  = ( Hypotenuse)^{2} }}}$

❍ Let's Consider Base of Triangle be x .

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

⠀⠀⠀⠀⠀$:\implies \tt{ 1^{2} + x^{2} = 2^{2}}$

⠀⠀⠀⠀⠀$:\implies \tt{ 1 + x^{2} = 4}$

⠀⠀⠀⠀⠀$:\implies \tt{  x^{2} = 4- 1}$

⠀⠀⠀⠀⠀$:\implies \tt{  x^{2} = 3}$

⠀⠀⠀⠀⠀$:\implies \tt{  x = \sqrt{3}}$

⠀⠀⠀⠀⠀$\underline {\boxed{\pink{ \mathrm {  x = \sqrt {3}\: }}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\underline {\therefore\:{ \mathrm {  Base  \:of\:Right \:Angled \:triangle \:is\:\sqrt {3}cm\: }}}$

❒ Finding value of cos $\theta$ by using found values :

$\sf{\underline {\dag As, \:We \:Know\:that \::}}$

• $\sf{\cos \theta =  \dfrac{Base} { Hypotenuse} }$

Where ,

• Base of Right Angled Triangle is $\sqrt {3} cm$.

• Hypotenuse of Right angled triangle is 2 cm .

Therefore,

⠀⠀⠀⠀⠀$\sf{\cos \theta = \dfrac{\sqrt{3}}{2}= \dfrac{Base} { Hypotenuse} }$

$\underline {\boxed{\pink{ \mathrm { Hence,\:The\:Value \:of\:\cos \theta  = \dfrac{\sqrt{3}}{2}\: }}}}\:\bf{\bigstar}$

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Answer 2

Given :-

• sin ø = $\large{\rm{\frac{1}{2}}}$

To find :-

• The value of cos ø

❍ Basic Formulae of Trigonometry is

Given by,

$\begin{gathered}\boxed { \begin{array}{c c} \\ \dag \qquad \large {\underline {\bf{ Some \:Basic\:Formulas \:For\:Trigonometry \::}}}\\\\ \sf{ In \:a \:Right \:Angled \: Triangle-:} \\\\ \sf {\star Sin \theta = \dfrac{Perpendicular}{Hypotenuse}} \\\\ \sf{ \star \cos \theta = \dfrac{ Base }{Hypotenuse}}\\\\ \sf{\star \tan \theta = \dfrac{Perpendicular}{Base}} \end{array}}\\\end{gathered}$

Now,

• $\boxed{\rm{sin \: ø = \frac{1}{2}} = {\frac{Base} {Hypotenuse}}}$$\large\star$

Therefore,

• Perpendicular of Right Angled Triangle is 1 cm.
• Hypotenuse of Right Angled Triangle is 2 cm.

★ Finding Base of Right Angled Triangle

$\large\underline{\frak{We \: know \: that,}}$

$\large\underline{\rm{By \: Pythagoras \: Theorem}}$

• ${\underline{\boxed{\green{\frak{(Perpendicular)²~+~Base²~=~Hypotenuse)² }}}}}$

❍ Let's consider Base of Triangle be x

★ Substituting the value,

:$\implies$1² + x² = 2²

$~~~~~$:$\implies$1 + x² = 4

$~~~~~~~~~~$:$\implies$x² = 4 - 1

$~~~~~~~~~~~~~~~$:$\implies$x² = 3

$~~~~~~~~~~~~~~~~~~~~$:$\implies$${\sf{x = \sqrt{3} }}$

$~~~~~~~~~~~~~~~~~~~~~~~~~$:$\implies$$\large{\underline{\boxed{\pink{\frak{x~= \sqrt{3}}}}}}$★

$\large\dag$ Hence,

• Base of the Right Angled Triangle is $\large{\rm{\sqrt{3}}}$

• Finding value of cose ø
• Using find value :

$\large\underline{\frak{We \: know \: that,}}$

• $\large\boxed{\rm{sin \: ø = \frac{Base}{Hypotenuse}}}$$\large\star$

Therefore,

• Base of Right Angled Triangle is $\large{\rm{\sqrt{3 \: cm}}}$
• Hypotenuse of Right Angled Triangle is 2 cm.

Henceforth,

• $\large\boxed {\rm {cos \: ø = \frac {\sqrt3}{2}} = {\frac{Base}{Hypotenuse} }}$$\large\star$

$\large\dag$ Hence Proved,

• The value of the cos ø = $\large{\rm{\sqrt{3}{2}}}$