Question

If√3sinA-cosA=0and.0<A<90findthevalueofA.​

Answer 1

Required Answer:-

Given:

• √3 sin(x) - cos(x) = 0

To find:

• The value of x.

Solution:

Given that,

➡ √3 sin(x) - cos(x) = 0

➡ √3 sin(x) = cos(x)

➡ √3 = cos(x)/sin(x)

➡ √3 = 1/tan(x)

➡ tan(x) = 1/√3

From Trigonometry Ratio Table,

➡ tan(x) = tan(30°)

➡ x = 30°

Hence, the value of x is 30°

Trigonometry Ratio Table:

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

Relationship Between Trigo Functions:

• sin(x) = 1/cosec(x)
• cos(x) = 1/sec(x)
• tan(x) = 1/cot(x)
• sin(x)/cos(x) = tan(x)

Answer 2

Answer:

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