Question

1. find the angle \\( \theta \\) in the given right triangle. leave your answer in terms of an inverse trigonometric function.

(a) triangle 1
\\( \sin \theta = \frac { 7 } { 10 } \\)
\\( \sin ^ { - 1 } ( \frac { 7 } { 10 } ) \\)
(b) triangle 2
\\( \tan \theta = \frac { 12 } { 19 } \\)
\\( \tan ^ { - 1 } ( \frac { 12 } { 19 } ) \\)

Explanation:

Step1: Recall trigonometric ratio definitions

In a right - triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\) and \(\tan\theta = \frac{\text{opposite}}{\text{adjacent}}\).
For Triangle 1:
The side opposite to \(\theta\) is \(7\) and the hypotenuse is \(10\). Using the definition of the sine function \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), we get \(\sin\theta=\frac{7}{10}\). To find \(\theta\), we use the inverse sine function. So, \(\theta=\sin^{- 1}(\frac{7}{10})\).
For Triangle 2:
The side opposite to \(\theta\) is \(12\) and the side adjacent to \(\theta\) is \(19\). Using the definition of the tangent function \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), we get \(\tan\theta=\frac{12}{19}\). To find \(\theta\), we use the inverse tangent function. So, \(\theta=\tan^{-1}(\frac{12}{19})\).

Answer:

(a) \(\theta=\sin^{-1}(\frac{7}{10})\)
(b) \(\theta=\tan^{-1}(\frac{12}{19})\)