Question

In a triangle ABC, COS A =3/5
then the sin²A + cos² A

plzzz help me with full explanation of answer I need
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Answer 1

Answer:

cosA= 3/5and

cos A= base/hypotenuse

so on comparing ,

base =3

and hypotenuse=5

now using Pythagoras theorem ,

perpendicular length = 4

so,

sin A= 4/5

on substitution values

we get,

sin^2A + cos^2A=1

Answer 2

Given :

• In a Triangle ABC ,
• Cos A = 3/5

Find :

• sin²A + cos²A

Solution :

As we know that Cos A = B/H

Where,

B = Base

H = Hypotenuse

Therefore, by using Pythagoras Theorem

(Hypotenuse)² = (Base )² + (Perpendicular)²

Substituting the values in it, we get

↬ (5)² = (3)² + (Per.)²

↬25 = 9 + ( Per )²

↬ 25 - 9 = ( per )²

↬ 16 = ( Per )²

↬ √16 = Perpendicular

↬ 4 = Perpendicular

Now,

As we know Sin = P /H

And P = 4 , whereas H = 5

So, Sin = 4/5$.$

Then, we got the appropriate values of :

• Sin A = 4/5
• Cos A = 3/5

→ Sin²A + Cos² A

Putting the value in the above equation :

→ (4/5)² + (3/5)²

→ (16/25) + (9/25)

→ (16 +9)/25

→(25/25)

→ 1

Therefore, the value of Sin²A + Cos² A is 1