Question

if sin a =3/5 find cos A, tan A, sec A, and cosec A?​

Answer 1

sin A =3/5

perpendiular=3 , hypotenuse=5

5^2 = 3^2+ base^2

25 =9 + b^2

b^2=25-9

b=√16

b=4

cosA = b/h

=4/5

tanA = p/b

=3/4

secA = h/p

= 5/3

cosec A = h/b

= 5/4

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Answer 2

Answer:

Given : sin A=$\frac{3}{5}$

To find : cos A, tan A , sec A and cosec A

Solution :

Let ∆BAC be the required rt. triangle.

sin A = $\frac{3}{5}$

$\frac{Perpendicular}{Hypotenuse}=\frac{3x}{5x}$

Using Pythagoras's Theorem;

$(Hypotenuse)^{2} = (Base)^2 + (Perpendicular) ^2\\\\\implies (Base)^2 = (Hypotenuse) ^2 - (Perpendicular)^2$

Putting values in the above equation;

(Base)² = (5x)² - (3x)²

$\implies (Base)^2= (2x)^2$

Base = 4x

Hence,

$cos A=\frac{Base}{Hypotenuse}=\frac{4x}{5x} =\frac{4}{5}\\\\\\tanA=\frac{Perpendicular}{Base} = \frac{3x}{4x}=\frac{3}{4} \\\\\\cosecA=\frac{1}{sinA} =\frac{5}{3} \\\\\\secA=\frac{1}{cosA}=\frac{5}{4}\\\\\\cotA=\frac{1}{tanA}=\frac{4}{3}$