Question

If the value of 15cot A = 8, find sin A and sec A.

Answer 1

Hint: In this particular question first draw the pictorial representation of the given problem it will give us a clear picture of what we have to find out then later in the solution. Use the property of Pythagoras theorem and trigonometric ratios. Here use $cot\theta = \dfrac{base}{perpendicular}$ or $tan\theta = \dfrac{perpendicular}{base}$ . So use these concepts to reach the solution of the question.

Complete step-by-step answer:

Given trigonometric equation is
15 cot A = 8.
Therefore, cot A = (8/15)
Now as we know that cot x = (1/tan x) so use this property in the above equation we have,
Therefore, (1/tan A) = (8/15)
Therefore, tan A = (15/8)................ (1)
Now as we know that in a right angle triangle tan is the ratio of perpendicular to base.
So consider a right angle triangle ABC, which is right angled at B as shown in the above figure.
In this triangle, Hypotenuse = AC, Perpendicular = BC and Base = AB.
And $\angle BAC = A$
So in this triangle, tan A = (perpendicular/base) = (BC/AB)............ (2)
Now compare this equation from equation (1) we have,
BC = 15 and AB = 8 units.
So by Pythagoras theorem we have,
$ \Rightarrow {\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{perpendicular}}} \right)^2} + {\left( {{\text{base}}} \right)^2}$
Now substitute the variable we have,
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {{\text{BC}}} \right)^2} + {\left( {{\text{AB}}} \right)^2}$
Now substitute the values we have,
$ \Rightarrow {\left( {{\text{AC}}} \right)^2} = {\left( {15} \right)^2} + {\left( 8 \right)^2} = 225 + 64 = 289 = {\left( {17} \right)^2}$
So, AC = 17 units.
Now as we know that in a right angle triangle sine is the ratio of perpendicular to hypotenuse so we have,
$ \Rightarrow \sin A = \dfrac{{BC}}{{AC}}$
Now substitute the values we have,
$ \Rightarrow \sin A = \dfrac{{15}}{{17}}$
Now as we know that in a right angle triangle cosine is the ratio of base to hypotenuse so we have,
$ \Rightarrow \cos A = \dfrac{{AB}}{{AC}}$
Now substitute the values we have,
$ \Rightarrow \cos A = \dfrac{8}{{17}}$
Now as we know that, sec A = (1/cos A) so we have,
$ \Rightarrow \sec A = \dfrac{{17}}{8}$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that according to the Pythagoras theorem square of the hypotenuse is equal to the sum of the square of the two legs of the right angle triangle, so first calculate all the sides of the triangle as above then use that in a right angle triangle sine is the ratio of perpendicular to hypotenuse and in a right angle triangle cosine is the ratio of base to hypotenuse so just substitute the values we will get the required answer.