Question

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

Answer 1

Step-by-step explanation:

x tan30° + y cot 60° = 0

TAN 30°=1/√3

COT 60°=1/√3

x tan30° + y cot 60° = 0

=>x×1/√3 + y × 1/√3 =0

=>x+y=0.......(i)

2x-y tan45°=1

TAN45°=1

=>2x-y×1=1

=>2x-y=1.........(ii)

solving both eqn

=>x+y=0.......(i)

=>2x-y=1.........(ii)

adding both eqn

2x+x+y-y=0+1

=>3x=1

x=1/3

y=0-x=(-1/3)

x=1/3

y=(-1/3)

Answer 2

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

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

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

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

• 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)

$\rule{300}{2}$

$\mathtt{Now,}$

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

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

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

$\rule{300}{2}$

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

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

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

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

$\rule{300}{2}$

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

$\sf{x + y = 0}$

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

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

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

$\rule{300}{2}$

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