Question

If 3 sin - cos0 = 0 and o" << 90°, find the value of tita​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

If √3 sin θ - cos θ = 0 and 0° < θ < 90°, find the value of θ.

Given,

$\sqrt{3}sin\theta-cos\theta=0$

To Find :-

Value of 'theta'

Solution :-

$\sqrt{3}sin\theta-cos\theta=0$

$\sqrt{3}sin\theta=cos\theta$

$\sqrt{3}\times \dfrac{sin\theta}{cos\theta}=1$

$\sqrt{3}\times tan\theta=1$   [ sinA/cosA = tanA]

$tan\theta=\dfrac{1}{\sqrt{3}}$

$tan\theta=tan30^{\circ}$

Cancelling 'tan' on both sides :-

$\theta=30^{\circ}$

Know More :-

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

Answer 2

$\begin{gathered} \sf\pink{2sin30° + cos0° + 3sin90°} \\ \sf\pink{ = > 2 \times \frac{1}{2} + 1 + 3 \times 1 } \\ \sf\pink{ = > 1 + 1 + 3} \\ \sf\pink{ = > 5} \\ \sf\purple{sin0 =0 \: \: cos0° = 1 \: tan0° = 0 } \\ \sf\purple{sin30° \: = \frac{1}{2 \: } cos30° = \frac{ \sqrt{3} }{2} \: tan = \frac{1}{ \sqrt{3} } } \\ \sf\purple{ sin45° = \frac{1}{ \sqrt{2} } \: cos45° = \frac{1}{ \sqrt{2} } \: tan45° = 1} \\ \sf\purple{sin60° = \frac{ \sqrt{3} }{2} \: cos60° = \frac{1}{2} \: tan60° = \sqrt{3} } \\ \sf\purple{sin90° = 1 \: cos0° = 0 \: tan0° = \infty }\end{gathered}$

Know More :-

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