Question

) If sin ( x + y) = 1 and cos (x-y) = √3 2 , x> y fid the value of x and y

Answer 1

Required Answer:-

Given:

• sin(x + y) = 1
• cos(x - y) = √3/2
• ∠x > ∠y

To find:

• ∠x and ∠y.

Solution:

Given that,

➡ sin(x + y) = 1

➡ cos(x - y) = √3/2

From Trigonometry Ratio Table,

➡ sin(x + y) = sin(90°)

➡ x + y = 90° ......(i)

Also,

➡ cos(x - y) = cos(30°)

➡ x - y = 30° ........(ii)

Adding both equations (i) and (ii), we get,

➡ 2x = 120°

➡ ∠x = 60°

From (i), ∠y = 90° - ∠x

➡ ∠y = 90° - 60°

➡ ∠y = 30°

Hence,

➡ ∠x = 60°

➡ ∠y = 30°

Answer:

• ∠x = 60°
• ∠y = 30°

Trigonometry Ratio Table:

$\sf Trigonometry\: Value \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle x & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin(x) & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos(x)& 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan(x) & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec(x) & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec(x)& 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot(x)& \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$

Answer 2

Answer:

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