Question

Find CosA, if sinA=3/4​

Answer 1

Solution:-

Given

$\rm \implies \: \sin A = \dfrac{3}{4}$

To find

$\rm \implies \: the \: value \: of \: \cos A$

Now take

$\rm \implies \: \sin A = \dfrac{3}{4} = \dfrac{p}{h}$

Using pythagoras theorem

Where

$\implies \rm p = 3 \: \: , \: h = 4 \: and \: b = x$

$\rm \implies {h}^{2} = {b}^{2} + {p}^{2}$

$\rm \implies \: {4}^{2} = {x}^{2} + {3}^{2}$

$\rm \implies \: 16 = {x}^{2} + 9$

$\rm \implies {x}^{2} = 16 - 9 = 7$

$\rm \implies \: x = b = \sqrt{7}$

So

$\rm \cos \: A \: = \dfrac{ \sqrt{7} }{4} = \dfrac{b}{h}$

More solution:-

$\rm \implies \tan A = \dfrac{3}{ \sqrt{7} } = \dfrac{p}{b}$

$\rm \implies \csc A = \dfrac{4}{3} = \dfrac{h}{p}$

$\rm \implies \: \sec A = \dfrac{ 4 }{ \sqrt{7} } = \dfrac{h}{b}$

$\rm \implies \cot A = \dfrac{ \sqrt{7} }{ 3} = \dfrac{b}{p}$

Answer 2

Answer:

Given :-

• sinA = $\dfrac{3}{4}$

To Find :-

• What is the value of cosA

Solution :-

Given : sinA = $\dfrac{3}{4}$

we know that,

$\mapsto$ sinA = $\dfrac{perpendicular}{hypotenuse}$ = $\dfrac{3}{4}$

Let, perpendicular (BC) = 3x

And, hypotenuse (AC) = 4x

Now, by using Phythagorus Theorem we know that,

✦ $\boxed{\bold{\large{AB\: =\: \sqrt{{AC}^{2} - {BC}^{2}}}}}$

⇒ AB = $\sqrt{({4x})^{2} - ({3x})^{2}}$

⇒ AB = $\sqrt{{16x}^{2} - {9x}^{2}}$

⇒ AB = $\sqrt{{7x}^{2}}$

➠ AB = $\sqrt{7}$x

Hence, the value of cosA is,

↦ cosA = $\dfrac{base}{hypotenuse}$

↦ cosA = $\dfrac{AB}{AC}$

↦ cosA = $\dfrac{\sqrt{7}x}{4x}$

➙ cosA = $\dfrac{\sqrt{7}}{4}$

$\therefore$ The value of cosA is $\dfrac{\sqrt{7}}{4}$

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