Question

If sin=3/5'then find cos and tan​

Answer 1

We have sin a = 3/5

therefore Sin θ = Perpendicular / Hypotenuse = AB/AC

therefore AB = 3

AC = 5

therefore as per the Pythagoras theorem

we have AC2 = BC2 + AB2

52 = BC2 + 32

25 = BC2 + 9

25-9 = BC2

16 = BC2

BC = 4

So now we have AB = 3

AC = 5

BC = 4

Now to find cos a and tan a

Cosine θ = Base / Hypotenuse = BC / AC

so cos a = 4/5

and Tangent θ = Perpendicular / Base = AB / BC

so tan a = 3/4

Hope it helps

Answer 2

Step-by-step explanation:

The question uses the concept of Trigonometrical ratios in trigonometry.

Let us assume that the angle is 'A',

Given:

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

To be Found: $cosA\hspace{1 mm} and \hspace{1 mm}tanA$  

Formula used:

$sin A = \frac{height}{hypotenuse} \\cos A = \frac{base}{hypotenuse} \\tan A= \frac{height}{base}$

Also, if we know Pythagoras theorem, then we know the relation between the 3 sides of a right angled triangle.

$hypotenuse^{2}=base^{2}+height^{2}$

where,

• hypotenuse= angle opposite to right angle
• base and height= the other two sides

In this question, we have:

$sinA= \frac{height}{hypotenuse} = \frac{3}{5}$

Since ratio between height and hypotenuse is $3:5$, we can assume hypotenuse to be $5x$ and height to be $3x$.

Now, $base^{2}=hypotenuse^{2}-height^{2} = (5x)^{2}-(3x)^{2}= 25x^2-9x^2= 16x^2$

Taking square root both sides, $base=4x$

So, $cosA=\frac{base}{hypotenuse}=\frac{4x}{5x} = \frac{4}{5} \\tanA=\frac{height}{base}=\frac{3x}{4x} = \frac{3}{4}$

Therefore, if $sinA=3/5$ then $cosA=4/5$ and $tanA=3/4$