
If $tanA=11$, find the value of the other trigonometric ratios.
Answer
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Hint: From the given value of $\tan A$, we can find other trigonometric ratios such a$\sin A,\tan A,\cot A,\sec A\And \cos ecA$ . We know that $tanA=\dfrac{P}{B}$ from this equation we can find the hypotenuse of the triangle corresponding to angle A by Pythagoras theorem. Now, we have all the sides so we can easily find the other trigonometric ratios corresponding to angle A.
Complete step-by-step answer:
The value of $\tan A$ given in the above question is:
$tanA=11$
The below figure is showing a right triangle ABC right angled at B.
In the above figure, “P” stands for perpendicular with respect to angle A, “B” stands for the base of a triangle with respect to angle A and “H” stands for the hypotenuse of the triangle with respect to angle A.
We know from the trigonometric ratio that:
$tanA=\dfrac{P}{B}$
Comparing this $\tan A$ with the $\tan A$ given in the question we found that $P=11\And B=1$. In the above equation, P stands for perpendicular and B stands for base of the triangle with respect to angle A so from the Pythagoras theorem we can find the hypotenuse of the triangle and we are representing the hypotenuse with a symbol “H”.
$\begin{align}
& {{H}^{2}}={{P}^{2}}+{{B}^{2}} \\
& \Rightarrow {{H}^{2}}=121+1 \\
& \Rightarrow {{H}^{2}}=122 \\
& \Rightarrow H=\sqrt{122} \\
\end{align}$
Now, we can easily find the other trigonometric ratios with respect to angle A.
We are going to find the trigonometric ratio sin A.
$\sin A=\dfrac{P}{H}$
Plugging $P=11$ and $H=\sqrt{122}$ we get,
$\sin A=\dfrac{11}{\sqrt{122}}$
We are going to find the trigonometric ratio of $\cos A$ .
$\cos A=\dfrac{B}{H}$
Plugging $B=1$ and $H=\sqrt{122}$in the above equation we get,
$\cos A=\dfrac{1}{\sqrt{122}}$
We know that reciprocal of $\cos A$ is $\sec A$ so,
$\sec A=\sqrt{122}$
We know that $\cos ecA$ is the reciprocal of $\sin A$ so,
$\cos ecA=\dfrac{\sqrt{122}}{11}$
We know that $\cot A$ is the reciprocal of $\tan A$ so,
$\cot A=\dfrac{1}{11}$
The value of $\tan A$ is already given in the question.
Hence, we have found all the trigonometric ratios corresponding to angle A.
Note: In the above problem, you might get confused about what are the trigonometric ratios and if you could understand the trigonometric ratios you might get confused like do I have to find the trigonometric ratios for all the angles of the given triangle.
The solution to all this confusion is that trigonometric ratios are $\sin ,\tan ,\cot ,\sec \And \cos ec$ of a particular angle and as $\tan A$ is given in the question so we have to find the trigonometric ratios corresponding to angle A.
Complete step-by-step answer:
The value of $\tan A$ given in the above question is:
$tanA=11$
The below figure is showing a right triangle ABC right angled at B.
In the above figure, “P” stands for perpendicular with respect to angle A, “B” stands for the base of a triangle with respect to angle A and “H” stands for the hypotenuse of the triangle with respect to angle A.
We know from the trigonometric ratio that:
$tanA=\dfrac{P}{B}$
Comparing this $\tan A$ with the $\tan A$ given in the question we found that $P=11\And B=1$. In the above equation, P stands for perpendicular and B stands for base of the triangle with respect to angle A so from the Pythagoras theorem we can find the hypotenuse of the triangle and we are representing the hypotenuse with a symbol “H”.
$\begin{align}
& {{H}^{2}}={{P}^{2}}+{{B}^{2}} \\
& \Rightarrow {{H}^{2}}=121+1 \\
& \Rightarrow {{H}^{2}}=122 \\
& \Rightarrow H=\sqrt{122} \\
\end{align}$
Now, we can easily find the other trigonometric ratios with respect to angle A.
We are going to find the trigonometric ratio sin A.
$\sin A=\dfrac{P}{H}$
Plugging $P=11$ and $H=\sqrt{122}$ we get,
$\sin A=\dfrac{11}{\sqrt{122}}$
We are going to find the trigonometric ratio of $\cos A$ .
$\cos A=\dfrac{B}{H}$
Plugging $B=1$ and $H=\sqrt{122}$in the above equation we get,
$\cos A=\dfrac{1}{\sqrt{122}}$
We know that reciprocal of $\cos A$ is $\sec A$ so,
$\sec A=\sqrt{122}$
We know that $\cos ecA$ is the reciprocal of $\sin A$ so,
$\cos ecA=\dfrac{\sqrt{122}}{11}$
We know that $\cot A$ is the reciprocal of $\tan A$ so,
$\cot A=\dfrac{1}{11}$
The value of $\tan A$ is already given in the question.
Hence, we have found all the trigonometric ratios corresponding to angle A.
Note: In the above problem, you might get confused about what are the trigonometric ratios and if you could understand the trigonometric ratios you might get confused like do I have to find the trigonometric ratios for all the angles of the given triangle.
The solution to all this confusion is that trigonometric ratios are $\sin ,\tan ,\cot ,\sec \And \cos ec$ of a particular angle and as $\tan A$ is given in the question so we have to find the trigonometric ratios corresponding to angle A.
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