Chapter 4.3, Problem 1E

To determine

To match: The trigonometric function with its right triangle definition.

Answer to Problem 1E

Correct matching is $\left(a\right)\to \left(v\right),\left(b\right)\to \left(iv\right),\left(c\right)\to \left(vi\right),\left(d\right)\to \left(iii\right),\left(e\right)\to \left(i\right),\left(f\right)\to \left(ii\right)$

Explanation of Solution

Given information:

Trigonometric functions are:

  $\left(a\right)\text{ sine }\left(b\right)\text{ cosine }\left(c\right)\text{ tangent }\left(d\right)\text{ cosecant }\left(e\right)\text{ secant }\left(f\right)\text{ cotangent}$

Right triangle definition:

  $\left(\text{i}\right)\frac{\text{hypotenuse}}{\text{adjacent}}\left(\text{ii}\right)\frac{\text{adjacent}}{\text{opposite}}\left(\text{iii}\right)\frac{\text{hypotenuse}}{\text{opposite}}\left(iv\right)\frac{\text{adjacent}}{\text{hypotenuse}}\left(v\right)\frac{\text{opposite}}{\text{hypotenuse}}\left(vi\right)\frac{\text{opposite}}{\text{adjacent}}$

Consider, a right triangle,

  

Now,

  $\begin{array}{c}\sin \theta =\frac{\text{opposite}}{\text{hypotenuse}}\\ \cos \theta =\frac{\text{adjacent}}{\text{hypotenuse}}\\ \tan \theta =\frac{\text{opposite}}{\text{adjacent}}\\ \cot \theta =\frac{\text{adjacent}}{\text{opposite}}\\ \sec \theta =\frac{\text{hypotenuse}}{\text{adjacent}}\\ \csc \theta =\frac{\text{hypotenuse}}{\text{opposite}}\end{array}$

Therefore, correct matching is,

  $\begin{array}{c}\left(a\right)\text{ sine }\to \left(v\right)\frac{\text{opposite}}{\text{hypotenuse}}\\ \left(b\right)\text{ cosine }\to \left(iv\right)\frac{\text{adjacent}}{\text{hypotenuse}}\\ \left(c\right)\text{ tangent }\to \left(vi\right)\frac{\text{opposite}}{\text{adjacent}}\\ \left(d\right)\text{ cosecant }\to \left(iii\right)\frac{\text{hypotenuse}}{\text{opposite}}\\ \left(e\right)\text{ secant }\to \left(i\right)\frac{\text{hypotenuse}}{\text{adjacent}}\\ \left(f\right)\text{ cotangent}\to \left(ii\right)\frac{\text{adjacent}}{\text{opposite}}\end{array}$