Question

find the value of x, if 2x tan^2 60° + 3x sin^2 30°= 27cos^2 45°/4sin^2 60°.​

Answer 1

Given :

2 x tan²60° + 3 x sin²30° = ( 27 cos²45° ) / 4 sin²60° )

In the attachment , I have given the trigonometric table .

From there we have :

tan 60° = √3

sin 30° = 1/2

cos 45° = 1 /√2

sin 60° = √3 / 2

Hence we can write the above equation as :

2 x ( √3 )² + 3 x ( 1/2 )² = ( 27 × ( 1/√2 )² ) / ( 4 × ( √3/2 )² )

⇒ 2 x × 3 + 3 x × 1/4 = ( 27/2 / 3 )

⇒ 6 x + 3 x /4 = 9/2

⇒ 2 x + x/4 = 3/2

⇒ x ( 2 + 1/4 ) = 3/2

⇒ x ( 8 + 1 )/4 = 3/2

⇒ x × 9/4 = 3/2

⇒ x = 3/2 × 4/9

⇒ x = 2/3

The value of x is 2/3

↪ sin A = x , then sin²A is x² . sin²A is spelled as sin squared A .

↪ The ratios sin , cos , cot , tan , sec and cosec are all positive in the first quadrant .

↪ Thereby the ratios of acute angles are positive .

↪ It means that ratios of trigonometry involving 0° - 90° are always positive .

Answer 2

$\huge\sf{Answer:}$

$\sf\frac{2}{3}$

Step-by-step explanation:

Given,

$\sf 2x (tan^260^{\circ}) + 3x(sin^230^{\circ}) = \frac{27cos^245^{\circ}}{4sin^260^{\circ}}$

Substitue the values of the trigonometric ratios. We know that :-

$\sf tan60^{\circ} = \sqrt{3}\\\\\implies tan^2 60^{\circ} = 3 \\\\sin30^{\circ} = \frac{1}{2} \\\\\implies sin^230^{\circ} = \frac{1}{4}\\\\ \\\sf cos45^{\circ} = \frac{1}{\sqrt{2}}\\\\\implies cos^245^{\circ} = \frac{1}{2}\\\\sin60^{\circ} = \frac{\sqrt{3}}{2}\\\\\implies sin^260^{\circ} = \frac{3}{4}$

So, $\sf 2x (tan^260^{\circ}) + 3x(sin^230^{\circ}) = \frac{27cos^245^{\circ}}{4sin^260^{\circ}}$

$\sf = 2x(3) + 3x(\frac{1}{4}) = \frac{27(\frac{1}{2})}{4(\frac{3}{4})}\\\\\implies 6x + \frac{3x}{4}= \frac{\frac{27}{2}}{3}\\\\\implies \frac{24x + 3x}{4}= \frac{27}{2} \div 3\\\\\implies \frac{27x}{4} = \frac{9}{2}\\\\\implies x = \frac{9}{2}\times\frac{4}{27}\\\\\implies x = \frac{2}{3}$