
In a\[\Delta ABC\], if \[A = {30^\circ }b = 2,c = \sqrt 3 + 1\], then \[\frac{{C - B}}{2} = \]
A. \[{15^9}\]
B. \[{30^\circ }\]
C. \[{45^\circ }\]
D. None of these
Answer
162.6k+ views
Hint:
Here in this case, it is given that in \[\Delta ABC\] one of the angles is \[A = {30^\circ }\] . So it is confirmed that the triangle is an isosceles triangle because the one of the angle’s measurements is \[{30^\circ }\] in isosceles and the other two are of equal length. So, we can solve accordingly to get the desired result.
Formula used:
To calculate the value of the given angle we can use the formula as below,
\[\tan \theta = \frac{{opposite}}{{Adjacent}} = \frac{a}{b}\]
Complete Step-by Step Solution:
We have been provided in the given data that in a triangle ABC,
In, \[{\rm{A}} = {30^\circ }\]
\[ \Rightarrow {\rm{B}} + {\rm{C}} = {150^\circ },{\rm{b}} = 2,{\rm{c}} = \sqrt 3 + 1\]
Now, we have to rewrite the equation in terms of trigonometry ratio as below,
\[\tan \left( {\frac{{{\rm{C}} - {\rm{B}}}}{2}} \right) = \frac{{{\rm{c}} - {\rm{b}}}}{{{\rm{c}} + {\rm{b}}}}\cot \frac{{\rm{A}}}{2}\]
Now, we have to substitute the value from the given data in the equation above, we get
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 3}}\cot {15^0}\]
Now, we have to solve the above equation, we obtain
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 (\sqrt 3 + 1)}}\left[ {\frac{{\cot {{30}^0} + 1}}{{\cot {{30}^0} - 1}}} \right]\]
On simplifying the above trigonometry terms from the obtained equation, we get
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 (\sqrt 3 + 1)}}\left[ {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right]\]
On simplification the above equation, we get
\[ = \frac{1}{{\sqrt 3 }}\]
The obtained term implies,
\[ \Rightarrow \frac{{{\rm{C}} - {\rm{B}}}}{2}\]
Now, we have to write in terms of trigonometry ratio of the above expression, we obtain
\[ = \frac{\pi }{6}\]
On further simplification of the above trigonometry value in terms of angles, we get
\[ = {30^0}\]
Therefore, if\[A = {30^\circ }b = 2,c = \sqrt 3 + 1\], then \[\frac{{C - B}}{2} = \] \[{30^\circ }\]
Hence, the option B is correct.
Note:
Always remember that the total of a triangle's interior angles is always 180 degrees. Before you can calculate the angle, you must first check the cosine function values. This is because the cosine function is negative in the second quadrant, and if it is negative, the triangle must have an obtuse angle. We can't use the sine function to figure it out because it's positive in the second quadrant as well. Students are likely to make mistakes in these types of problems because it includes trigonometry values, angles and radians. So, one should be very cautious in solving these types of problems.
Here in this case, it is given that in \[\Delta ABC\] one of the angles is \[A = {30^\circ }\] . So it is confirmed that the triangle is an isosceles triangle because the one of the angle’s measurements is \[{30^\circ }\] in isosceles and the other two are of equal length. So, we can solve accordingly to get the desired result.
Formula used:
To calculate the value of the given angle we can use the formula as below,
\[\tan \theta = \frac{{opposite}}{{Adjacent}} = \frac{a}{b}\]
Complete Step-by Step Solution:
We have been provided in the given data that in a triangle ABC,
In, \[{\rm{A}} = {30^\circ }\]
\[ \Rightarrow {\rm{B}} + {\rm{C}} = {150^\circ },{\rm{b}} = 2,{\rm{c}} = \sqrt 3 + 1\]
Now, we have to rewrite the equation in terms of trigonometry ratio as below,
\[\tan \left( {\frac{{{\rm{C}} - {\rm{B}}}}{2}} \right) = \frac{{{\rm{c}} - {\rm{b}}}}{{{\rm{c}} + {\rm{b}}}}\cot \frac{{\rm{A}}}{2}\]
Now, we have to substitute the value from the given data in the equation above, we get
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 + 3}}\cot {15^0}\]
Now, we have to solve the above equation, we obtain
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 (\sqrt 3 + 1)}}\left[ {\frac{{\cot {{30}^0} + 1}}{{\cot {{30}^0} - 1}}} \right]\]
On simplifying the above trigonometry terms from the obtained equation, we get
\[ = \frac{{\sqrt 3 - 1}}{{\sqrt 3 (\sqrt 3 + 1)}}\left[ {\frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}}} \right]\]
On simplification the above equation, we get
\[ = \frac{1}{{\sqrt 3 }}\]
The obtained term implies,
\[ \Rightarrow \frac{{{\rm{C}} - {\rm{B}}}}{2}\]
Now, we have to write in terms of trigonometry ratio of the above expression, we obtain
\[ = \frac{\pi }{6}\]
On further simplification of the above trigonometry value in terms of angles, we get
\[ = {30^0}\]
Therefore, if\[A = {30^\circ }b = 2,c = \sqrt 3 + 1\], then \[\frac{{C - B}}{2} = \] \[{30^\circ }\]
Hence, the option B is correct.
Note:
Always remember that the total of a triangle's interior angles is always 180 degrees. Before you can calculate the angle, you must first check the cosine function values. This is because the cosine function is negative in the second quadrant, and if it is negative, the triangle must have an obtuse angle. We can't use the sine function to figure it out because it's positive in the second quadrant as well. Students are likely to make mistakes in these types of problems because it includes trigonometry values, angles and radians. So, one should be very cautious in solving these types of problems.
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