Question

E 9. Given 3 cos A - 4 sin A = 0; evaluate without using tables
sin A + 2 cos A
----------------------
3 cos A-sin A​

Answer 1

Answer:

11/9

Step-by-step explanation:

Given,

3cosA - 4sinA = 0

To Find :-

Value of :-

(sinA + 2cosA)/(3cosA - sinA)

How  To Do :-

By Solving the given equation we will get the value of 'tanA' and by applying pythagoras theorem we will get the values of both 'sinA' and 'cosA' and we need to substitute these values to get the required answer.

Solution :-

3cosA - 4sinA = 0

3cosA = 4sinA

3/4 = sinA/cosA

3/4 = tanA [ ∴ sinA/cosA = tanA]

tanA = opposite side/adjacent side = 3/4

By comparing both we will get :-

opposite side = 3 , adjacent side = 4

Applying pythagoras theorem :-

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

(hypotenuse)² = (3)² + (4)²

(hypotenuse)² = 9 + 16

(hypotenuse)² = 25

hypotenuse = √25

∴Hypotenuse side = 5

sinA = opposite side /hypotenuse = 3/5

cosA = adjacent side/hypotenuse = 4/5

As we need to find ' (sinA + 2cosA)/(3cosA - sinA) ' :-

1) sinA + 2cosA

= 3/5 + 2(4/5)

= 3/5 + 8/5

= (3 + 8)/5

= 11/5

∴ sinA + 2cosA = 11/5.

2) 3cosA - sinA

= 3(4/5) - 3/5

= 12/5 - 3/5

= 12 - 3/5

= 9/5

∴  3cosA - sinA = 9/5.

(sinA + 2cosA)/(3cosA - sinA) :-

$=\dfrac{\dfrac{11}{5}}{\dfrac{9}{5}}$

$=\dfrac{11}{5}\times \dfrac{5}{9}$

= 11/9

(sinA + 2cosA)/(3cosA - sinA)  = 11/9