Question

If $sin A=\frac{9}{41}$, compute cos A and tan A.

Answer 1

Firstly, express sin A in terms of perpendicular and hypotenuse i.e. sin A = perpendicular / hypotenuse  

and then by using Pythagoras theorem in right ∆ ABC (AC² = AB² + BC²) , find the third  side of the triangle (Base) . Further determine the value of cos A and tan A by using the formula :

cos A = base /hypotenuse

tan A = perpendicular/ base

SOLUTION :  

Given:  sin A = 9/ 41

sin A = Perpendicular /Hypotenuse = 9/41

Perpendicular side = 9 and

Hypotenuse = 41

We draw a ∆ ABC right angled at B.  

In ΔABC,  

Let BC = Perpendicular = 9  ,  Hypotenuse (AC) = 41, base (AB)

AC² = AB² + BC

[by using Pythagoras theorem]

41² = AB² + 9²

AB² = 41² - 9²

AB² = 1681 – 81

AB= √1600

AB = 40

Hence, length of side AB = 40

Now

cos A= Base / Hypotenuse

cos A = AB/AC

cosA = 40/41

tan A= Perpendicular/ Base

tan A= BC/AB

tan A = 9/40

Hence, cos A = 40/41 , tan A=9/40

HOPE THIS ANSWER WILL HELP YOU...

Answer 2

Sin A = 9/41 which is = p/h
using pythagoras theorem
${9}^{2} = ( {41)}^{2} + {b}^{2} \\ 81 = 1681 + {b}^{2} \\ b = 40$

cos A = 40/41
tan A = 9/40
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