How do you solve \[\cot 2a = \dfrac{5}{{12}}\]?
Answer
557.1k+ views
Hint: Here we have to determine the value, and the given function is an inverse trigonometry. The cot is a cotangent trigonometry. We simplify the term which is present in RHS and by using the table of trigonometry ratios for the standard angles and by the scientific calculator or by the Clark’s table we determine the solution for the question.
Complete step by step solution:
The sine, cosine, tangent, cosecant, secant and cotangent are the trigonometry ratios of trigonometry. It is abbreviated as sin, cos, tan, cosec, sec and cot. Here in this question, we have \[\cot 2a = \dfrac{5}{{12}}\]
Taking the cot to the RHS, the trigonometry ratio will become the inverse. So we have
\[ \Rightarrow 2a = {\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\]
where \[{\cot ^{ - 1}}\] represents the inverse of a cotangent function. So we have to find \[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\].
We consider the table of trigonometry ratios for standard angles.
The table of cotangent function for standard angles is given as
Now consider the given function
\[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\]
When we consider the fraction term and if we simplify the fraction term we obtain 0.416667
So according to the table the value will be greater than 60 but less than 90.
When we find the value by using the scientific calculator or Clark’s table the value of \[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\] is \[67.3801351\]with 5 decimal places and we round off the number it is nearly equal to 67.38013
\[ \Rightarrow 2a = 67.38013\]
On dividing by 2 we get
\[ \Rightarrow a = 33.69006\]
Note:
The trigonometry and inverse trigonometry are inverse for each other. The inverse of a function is represented as an arc of the function or the function is raised by the power -1. For the trigonometry and the inverse trigonometry we need to know about the table of trigonometry ratios for the standard angles.
Complete step by step solution:
The sine, cosine, tangent, cosecant, secant and cotangent are the trigonometry ratios of trigonometry. It is abbreviated as sin, cos, tan, cosec, sec and cot. Here in this question, we have \[\cot 2a = \dfrac{5}{{12}}\]
Taking the cot to the RHS, the trigonometry ratio will become the inverse. So we have
\[ \Rightarrow 2a = {\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\]
where \[{\cot ^{ - 1}}\] represents the inverse of a cotangent function. So we have to find \[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\].
We consider the table of trigonometry ratios for standard angles.
The table of cotangent function for standard angles is given as
| Angle | 0 | 30 | 45 | 60 | 90 |
| cot | \[\infty \] | \[\sqrt 3 \] | \[1\] | \[\dfrac{1}{{\sqrt 3 }}\] | \[0\] |
Now consider the given function
\[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\]
When we consider the fraction term and if we simplify the fraction term we obtain 0.416667
So according to the table the value will be greater than 60 but less than 90.
When we find the value by using the scientific calculator or Clark’s table the value of \[{\cot ^{ - 1}}\left( {\dfrac{5}{{12}}} \right)\] is \[67.3801351\]with 5 decimal places and we round off the number it is nearly equal to 67.38013
\[ \Rightarrow 2a = 67.38013\]
On dividing by 2 we get
\[ \Rightarrow a = 33.69006\]
Note:
The trigonometry and inverse trigonometry are inverse for each other. The inverse of a function is represented as an arc of the function or the function is raised by the power -1. For the trigonometry and the inverse trigonometry we need to know about the table of trigonometry ratios for the standard angles.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Find the value of the expression given below sin 30circ class 11 maths CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

