Question

1f sin A + COSA 7/5 and sin A cos A= 12/25, find the values of sin A and cos A.​

Answer 1

Answer:

Step-by-step explanation:

Given,

sinA + cosA = 7/5

sinA.cosA = 12/25

To Find :-

Values of 'sinA' and 'cosA'

Formula Required :-

1) sin2A = 2sinAcosA

2) sin2A = 2tanA/1+tan^2A

3) Pythagoras theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

4) tanA = opposite side/adjacent side

5) sinA = opposite side/hypotenuse side

6) cosA = adjacent side/hypotenuse side

Solution :-

sinA.cosA = 12/25

[Let it be equation - 1]

Multiplying with '2' for equation - 1

2 (sinA.cosA) = 2(12/25)

2sinAcosA = 24/25

sin2A = 24/25

[ ∴ sin2A = 2sinAcosA ]

$\dfrac{2tanA}{1+tan^2A}=\dfrac{24}{25}$

[ ∴ sin2A = 2tanA/1 + tan²A ]

Cross multiplication :-

25(2tanA) = 24(1 + tan²A)

50tanA = 24 + 24tan²A

24tan²A - 50tanA + 24 = 0

2(12tan²A - 25tanA + 12) = 0

12tan²A - 25tanA + 12 = 0

Solving by using splitting the middle term process :-

12tan²A - 9tanA - 16tanA + 12 = 0

3tanA(4tanA - 3) - 4(4tanA - 3) = 0

Taking common :-

(4tanA - 3)(3tanA - 4) = 0

Equating both terms to '0' :-

4tanA - 3 = 0 , 3tanA - 4 = 0

4tanA = 3 , 3tanA = 4

tanA = 3/4 , tanA = 4/3

∴ tanA = 3/4 , 4/3

Finding value of 'sinA and cosA' if 'tanA = 3/4' :-

tanA = 3/4

opposite side/adjacent side = 3/4

→ opposite side = 3 , adjacent side = 4

Let, hypotenuse side = x

Applying Pythagoras theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

(x)² = (3)² + (4)²

x^2 = 9 + 16

x^2 = 25

x = √25

x = 5

∴ Hypotenuse side = x = 5

sinA = opposite side/hypotenuse side

= 3/5

cosA = adjacent side/hypotenuse side

= 4/5

∴sinA = 3/5 , cosA = 4/5 [If and only if 'tanA = 3/4' ]

Finding value of 'sinA and cosA' if 'tanA = 4/3' :-

opposite side/adjacent side = 4/3

→ opposite side = 4 , adjacent side = 3

Let, hypotenuse side = x

Applying Pythagoras theorem :-

(hypotenuse side)² = (opposite side)² + (adjacent side)²

(x)^2 = (4)^2 + (3)^2

x^2 = 16 + 9

x^2 = 25

x = √25

x = 5

∴ Hypotenuse side = x = 5

sinA = opposite side/hypotenuse side

= 4/5

cosA = adjacent side/hypotenuse side

= 3/5

∴sinA = 4/5 , cosA = 3/5 [If and only if 'tanA = 4/3']

Answer 2

Given :-

Sin A + Cos A = 7/5 and Sin A Cos A = 12/25

To Find :-

Values of Sin A and Cos A

Used Concepts :-

• At First , according to the situation assume the equation/equations .
• All variables in the both equation are same . So find the value of one variable in terms of variable/variables .
• Substitute the value in 2nd equation .
• Find the value .
• Put the value in 1st equation and find the value of other variables .
• Done !

Solution :-

Let us assume that ,

Sin A + Cos A = 7/5 -------( i )

Sin A . Cos A = 12/25 -------( ii )

By ( ii ) ,

=> Sin A . Cos A = 12/25

=> Sin A = 12/25 Cos A

Put Sin A = 12/25 Cos A in ( i ) ,

=> 12/25 Cos A + Cos A = 12/25

=> 12 + 25 Cos²A/25 Cos A = 12/25

=> Doing Cross Multiplication ,

=> 5 ( 12 + 25 Cos²A ) = 7 ( 25 Cos A )

=> 60 + 125 Cos²A = 175 Cos A

=> 125 Cos²A - 175 Cos A + 60 = 0

=> You can see that it is of the form of a quadratic equation . Where ,

a = 125 , b = - 175 and c = 60

=> D = b² - 4ac = ( -175 )² - 4 × 125 × 60 = 30625 - 30000 = 625

=> √D = √625 = 25

Now Values of Cos A are given by the quadratic formula ,

=> Cos A = -b + √D/2a , -b - √D/2a

=> Cos A = - ( -175 ) + 25/2 × 125 , - ( -175 ) - 25/2 × 125

=> Cos A = 175 + 25/250 , 175 - 25/250

=> Cos A = 200/250 , 150/250

=> Cos A = 4/5 , 3/5

Now By ( i ) ,

=> When Cos A = 4/5

=> Sin A + Cos A = 7/5

=> Sin A + 4/5 = 7/5

=> Sin A = 7/5 - 4/5

=> Sin A = 3/5

=> When Cos A = 3/5

=> Sin A + Cos A = 7/5

=> Sin A + 3/5 = 7/5

=> Sin A = 7/5 - 3/5

=> Sin A = 4/5

=> Sin A = 4/5 , 3/5

Finally We get ,

Sin A = 3/5 , 4/5

Cos A = 4/5 , 3/5

Henceforth , Value of Sin A is 4/5 when value of Cos A is 3/5 , and value of Sin A is 3/5 when value of Cos A is 4/5 .