Question

$sin \alpha = \frac{3}{4} \cos \alpha and \tan \alpha$
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Answer 1

GIVEN :-

• sina = 3/4.

TO FIND :-

• The value of cosa and tana.

SOLUTION :-

As we know that,

★ cos∅ = √(1 - sin²∅) ★

★ tan∅ = sin∅/cos∅ ★

Firstly we will find the value of cosa,

$\implies \displaystyle \sf \: \cos( \alpha ) = \sqrt{1 - \sin ^{2} ( \alpha ) }$

$\implies \displaystyle \sf \: \cos( \alpha ) = \sqrt{ 1 - \bigg( \frac{3}{4} \bigg) ^{2} }$

$\implies \displaystyle \sf \: \cos( \alpha ) = \sqrt{1 - \frac{9}{16} }$

$\implies \displaystyle \sf \: \cos( \alpha ) = \sqrt{ \frac{16 - 9}{16} }$

$\implies \displaystyle \sf \: \cos( \alpha ) = \sqrt{ \frac{7}{16} }$

$\implies \underline{ \boxed{ \displaystyle \sf \: \cos( \alpha ) = \frac{ \sqrt{7} }{4} }}$

Now we will Put the value of tana,

$\implies \displaystyle \sf \: \tan( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) }$

$\implies \displaystyle \sf \: \tan( \alpha ) = \frac{ \bigg( \dfrac{3}{4} \bigg) }{ \bigg( \dfrac{ \sqrt{7} }{4} \bigg)}$

$\implies \displaystyle \sf \: \tan( \alpha ) = \frac{3}{4} \times \frac{4}{ \sqrt{7} }$

$\implies \underline{ \boxed{ \displaystyle \sf \: \tan( \alpha ) = \frac{3}{ \sqrt{7} } }}$

Answer 2

Question ⤵

$sin \alpha = \frac{3}{4} \cos \alpha and \tan \alpha$

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Solution ⤵

Given $\rightarrow$ $\frac{3}{4}$

$\Rightarrow$ $\frac{BC}{AC}$ = $\frac{3}{4}$

BC = 3K and AC = 4k

$\because$ k is constant of Proportionality.

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Using Pythagoras Theorem ⤵

⇛${AC}^{2} = {AB}^{2} + {BC}^{2}$

⇛${AB}^{2} = {AC}^{2} - {BC}^{2}$

⇛${AB}^{2} = {4k}^{2} - {3k}^{2}$

⇛${AB}^{2} = {16k}^{2} - {9k}^{2}$

⇛AB = $\sqrt{7}$

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So,

⇛Cos A $\frac{AB}{AC}$

⇛$\frac{ \sqrt{7k} }{4k}$

⇛Cos A$\frac{ \sqrt{7} }{4}$

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⇛Tan A $\frac{BC}{AB}$

⇛$\frac{3k}{ \sqrt{7k} }$

⇛ Tan A$\frac{3}{ \sqrt{7} }$

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