Question

if sin A=3/4 calculate cos A and tan A.​

Answer 1

Answer:

cosA = √7/4

tanA = 3/√7

Step-by-step explanation:

We know that;

$\sf \Longrightarrow sin \theta = \dfrac{Opp \ Side}{Hypotenuse}$

ATQ,

$\sf \Longrightarrow sin \theta = \dfrac{3}{4}$

Therefore, the side opposite to θ has the value "3x" and the hypotenuse has the value of "4x".

(We suffix the values with x since the fraction has been reduced to it's lowest form)

By using Pythagoras Theorem we get;

⇒ Hypotenuse² = Altitude² + Base²

⇒ AC² = AB² + BC²

⇒ (4x)² = (3x)² + BC²

⇒ 16x² = 9x² + BC²

⇒ 16x² - 9x² = BC²

⇒ 7x² = BC²

⇒ $\sf \sqrt{7x^{2}}$ = BC

⇒ BC = √7x

Now, We'll find cosA and tanA.

CosA

⇒ cosA = adjacent side/hypotenuse.

⇒ cosA = (√7x)/4x

⇒ cosA = √7/4

TanA

⇒ tanA = opposite side/adjacent side.

⇒ tanA = (3x)/(√7x)

⇒ tanA = 3/√7

Answer 2

Given that,

• $\rm { sin \: A = \dfrac{3}{4} }$

We have to find,

• $\rm { cos \: A \: and \: tan \: A }$

Solution,

Here, we know that

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

Hence,

• $\rm { Perpendicular = 3}$
• $\rm { Hypotenuse = 4}$

_____________

Applying pythagoras property-

$\rm\red { {H}^{2} = {B}^{2} + {P}^{2} }$

$\rm { \leadsto {B}^{2} = {H}^{2} - {P}^{2} }$

$\rm { \leadsto {B}^{2} = {4}^{2} - {3}^{2} }$

$\rm { \leadsto {B}^{2} = 16 - 9 = 7}$

$\rm\blue { \leadsto B = \sqrt{7} }$

________________

• $\rm { Hypotenuse = 4 }$
• $\rm { Base = \sqrt{7} }$
• $\rm { Perpendicular = 3 }$

$\leadsto \rm\red{ cos \: A = \dfrac{ Base}{ Hypotenuse} = \dfrac{ \sqrt{7} }{4} }$

$\:$

$\leadsto \rm\red{ tan \: A = \dfrac{ Perpendicular}{ Base} = \dfrac{ 3 }{\sqrt{7}} }$