Question

Suppose that in a right angled triangle, \[\cos t = \dfrac{3}{4}\]. How do you find cot t?

Answer 1

Hint: We will first mention the cosine in terms of base and hypotenuse and then find the value of perpendicular using the hypotenuse theorem and then we will find the value of cotangent from it.

Complete step-by-step solution:
We are given that in a right angled triangle, \[\cos t = \dfrac{3}{4}\] and we need to find the value of cot t.
Now, we already know that the ratio of base and the hypotenuse gives us the value of cosine of any angle.
Since, we are given that \[\cos t = \dfrac{3}{4}\] and $\cos \theta = \dfrac{B}{H}$.
Therefore, our base is 3x units and hypotenuse is 4x units.
Now, we know that the Pythagorean theorem states that: In a right angled triangle, ${H^2} = {P^2} + {B^2}$, where H is the hypotenuse, B is the base and P is the perpendicular.
Putting the values, we will get: ${(4x)^2} = {P^2} + {(3x)^2}$.
Simplifying the values, we will get: $16{x^2} = {P^2} + 9{x^2}$
Simplifying it further, we will get: $P = \sqrt 7 x$ units
And we know that cotangent of any angle is given by the ratio of base and perpendicular.
Since $\cot \theta = \dfrac{B}{P}$, on putting the values, we will get the following equation:
$ \Rightarrow \cot \theta = \dfrac{{3x}}{{\sqrt 7 x}}$
Cutting – off the x from both the numerator and the denominator, we will get the following equation:-
$ \Rightarrow \cot \theta = \dfrac{3}{{\sqrt 7 }}$

Therefore the value of $\cot t$ is equal to $\dfrac{3}{{\sqrt 7 }}$

Note: The students must know that the cotangent of any angle is the inverse of the tangent of any angle. And, since we know that in tangent, the numerator has perpendicular and the denominator is the base, therefore, the cotangent has base in the numerator and perpendicular in denominator.
Here, we assumed x with the 3 and 4’s because we are just given the ration, there may be multiples of them involved and thus we have that as cosine. Therefore, for a generalized answer, we assumed it as the coefficient of x and thus found the required ratio.