Chapter 12, Problem 10CR

To determine

Find $\left(\theta \right)$ to nearest ten minutes if $\cot \theta =-0.7046\text{, }0°<\theta <180°$ .

Answer to Problem 10CR

The value of $\left(\theta \right)$ is $125°10\text{'}$ , so option (b) is correct.

Explanation of Solution

Given information: The given trigonometric function is,

  $\cot \theta =-0.7046\text{, }0°<\theta <180°\mathrm{....}\left(1\right)$

Calculation: From (1) the angle $\left(\theta \right)$ lies in between $0°$ and $180°$ , therefore evaluating (1) gives,

  $\begin{array}{c}\cot \theta =-0.7046\text{, }0°<\theta <180°\mathrm{....}\left(\text{From }\left(1\right)\right)\\ \theta ={\cot }^{-1}\left(-0.7046\right)\\ \theta =125.17°\\ \theta =125°+\left(0.17°\right)\\ \theta =125°+\left(0.17\right)\left(60\right)\text{'}\\ \theta =125°+10.2\text{'}\\ \theta =125°10\text{'}\end{array}$

Hence the value of $\left(\theta \right)$ is $125°10\text{'}$ , so option (b) is correct.