Question

If 3 cot θ = 2, find the value of $\frac{4sin\Theta-3cos\Theta}{2sin\Theta+6cos\Theta}$.

Answer 1

SOLUTION IS IN THE ATTACHMENT.

** Trigonometry is the study of the relationship between the sides and angles of a triangle.

The ratio of the sides of a right angled triangle with respect to its acute angles are called trigonometric ratios.

** For any acute angle in a right angle triangle the side opposite to the acute angle is called a perpendicular(P),  the side adjacent to this acute angle is called the base(B) and side opposite to the right angle is called the hypotenuse(H).

** Find the third  side of the right ∆ ABC by using Pythagoras theorem (AC² = AB² + BC²).

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Answer 2

HELLO DEAR,




GIVEN:- 3cot∅ = 2

cot∅ = 2/3

cos∅/sin∅ = 2x/3x

cos∅ = 2x , sin∅ = 3x


now,

$\bold{\frac{4sin\Theta - 3cos\Theta}{2sin\Theta + 6cos\Theta}}$

$\bold{\implies \frac{4*3x - 3*2x}{2*3x + 6*2x}}$

$\bold{\implies \frac{12x - 6x}{6x + 12x}}$

$\bold{\implies \frac{6x}{18x} = 1/3}$

hence, $\bold{\frac{4sin\Theta - 3cos\Theta}{2sin\Theta + 6cos\Theta} = 1/3}$




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