Answer
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Hint:
we are give a trigonometric equation and with the known values $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$and the identity $\sin (A + B) = \sin A\cos B + \cos A\sin B$ we can find the value of $\theta $
Complete step by step solution:
We are given that $\cos \theta + \sqrt 3 \sin \theta = 2$
Diving by 2 on both sides we get,
$ \Rightarrow \dfrac{1}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta = 1$
We know that $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Substituting in the above equation we get
$ \Rightarrow \;\sin {30^ \circ }\cos \theta + \sin \theta \cos {30^ \circ } = 1$
By the identity
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
We get,
$
\Rightarrow \sin (30 + \theta ) = 1 \\
\Rightarrow {30^ \circ } + \theta = {\sin ^{ - 1}}1 \\
\Rightarrow {30^ \circ } + \theta = {90^ \circ } \\
\Rightarrow \theta = {60^ \circ } \\
$
Hence, the value of $\theta = {60^ \circ } $
Note:
There are six trigonometric functions: sine, cosine, tangent and their reciprocal functions, secant, cosecant and cotangent. These functions are found by the ratios of a triangle's sides
The trigonometric ratios for the angles 30°, 45° and 60° can be found using two special triangles.
An equilateral triangle with side lengths of 2 cm can be used to find exact values for the trigonometric ratios of 30° and 60°.
we are give a trigonometric equation and with the known values $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$and the identity $\sin (A + B) = \sin A\cos B + \cos A\sin B$ we can find the value of $\theta $
Complete step by step solution:
We are given that $\cos \theta + \sqrt 3 \sin \theta = 2$
Diving by 2 on both sides we get,
$ \Rightarrow \dfrac{1}{2}\cos \theta + \dfrac{{\sqrt 3 }}{2}\sin \theta = 1$
We know that $\sin {30^ \circ } = \dfrac{1}{2}$ and $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Substituting in the above equation we get
$ \Rightarrow \;\sin {30^ \circ }\cos \theta + \sin \theta \cos {30^ \circ } = 1$
By the identity
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
We get,
$
\Rightarrow \sin (30 + \theta ) = 1 \\
\Rightarrow {30^ \circ } + \theta = {\sin ^{ - 1}}1 \\
\Rightarrow {30^ \circ } + \theta = {90^ \circ } \\
\Rightarrow \theta = {60^ \circ } \\
$
Hence, the value of $\theta = {60^ \circ } $
Note:
There are six trigonometric functions: sine, cosine, tangent and their reciprocal functions, secant, cosecant and cotangent. These functions are found by the ratios of a triangle's sides
The trigonometric ratios for the angles 30°, 45° and 60° can be found using two special triangles.
An equilateral triangle with side lengths of 2 cm can be used to find exact values for the trigonometric ratios of 30° and 60°.
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