Question

If x tan30° + y cot 60° = 0 and 2x-y tan45° = 1 then find the value of x and y.​

Answer 1

$\: \: \: \: \: \: x \tan(30) + y \cot(60) = 0 \\ = > x. \frac{1}{ \sqrt{3} } + y. \frac{1}{ \sqrt{3} } = 0 \\ = > \frac{1}{ \sqrt{3} } (x + y) = 0 \\ = > (x + y) = 0.......(i)$

Now,

$\: \: \: \: \: \: \: 2x - y \tan(45) = 1 \\ = > 2x - y \times 1 = 1 \\ = > 2x - y = 1......(ii)$

Adding equ. (i) and (ii)

$\: \: \: \: \: \: \: \: 2x - y + x + y = 1 \\ = > 3x = 1 \\ = > x = \frac{1}{3}$

Putting value of x in eq (i)

$\: \: \: \: \: \: \: \: \: x + y = 0 \\ = > y = 0 - x \\ = > y = 0 - \frac{1}{3} \\ = > y = \frac{ - 1}{3}$

Answer 2

$\huge\underline{\boxed{\mathtt{\red{Given:}}}}$

$\sf{x \tan {30}^{0} + y \cot {60}^{0} = 0}$

$\sf{2x - y \tan {45}^{0} = 1}$

$\huge\underline{\boxed{\mathtt{\red{ToFind:}}}}$

• The value of X and y.

$\rule{300}{2}$

$\sf{x \tan {30}^{0} + y \cot {60}^{0} = 0}$

$\sf{\implies x. \frac{1}{ \sqrt{3} } + y . \frac{1}{ \sqrt{3} } = 0}$

$\sf{\implies \frac{1}{ \sqrt{3} } (x + y) = 0}$

$\sf{\implies x + y = 0}$--------(1)

$\rule{300}{2}$

$\mathtt{Now,}$

$\sf{\implies 2x - y \tan {45}^{0} = 1}$

$\sf{\implies 2x - y.1 = 1}$

$\sf{\implies 2 x - y = 1}$------------(2)

$\rule{300}{2}$

$\mathtt{Adding \: eq(1) \: and \: eq(2) \: we \: get,}$

$\sf{x + y + 2x - y = 0 + 1}$

$\sf{\implies 3x = 1}$

$\sf{\implies x = \frac{1}{3}}$

$\rule{300}{2}$

$\mathtt{Putting \: the \: value \: of \: x \: in \: eq(1)we \: get}$

$\sf{x + y = 0}$

$\sf{\implies \frac{1}{3} + y = 0}$

$\sf{\implies y = - \frac{1}{3} }$

$\mathtt{Hence, \: the \: required \: value \: of \: x } \\ \mathtt{is \: \frac{1}{3} \: and \: y \: is \: - \frac{1}{3}. }$

$\rule{300}{2}$

$\huge\underline{\boxed{\mathtt{\red{ThankYou}}}}$