Chapter 67, Problem 1A

Determine the sizes of $\angle A$and $\angle B$to the nearest minute.

To determine

Find the sizes of $\angle A$ and $\angle B$ to the nearest minute.

Answer to Problem 1A

The angle:-

  $\begin{array}{l}\angle A={70}^{o}7\text{'}\\ \angle B={19}^{o}53\text{'}\end{array}$

Explanation of Solution

Given information:

The triangle shown with the problem.

Calculations:

From the triangle shown-

  $\begin{array}{l}\tan \angle A=\frac{98.4mm}{35.6mm}\\ \therefore \angle A={\tan }^{-1}\left(2.764\right)={70.11}^o\\ \text{since }{1}^o=60\text{'}\\ \text{simplifying the above}\text{value}\\ \angle A={70}^o\left({0.11}^o\times \frac{60\text{'}}{10}\right)\\ \angle A={70}^{o}6.6\text{'}\\ or\\ \Rightarrow \angle A={70}^{o}7\text{'}\end{array}$

  $\begin{array}{l}Again\text{ from the triangle given:}\\ \tan \angle B=\frac{35.6mm}{98.6mm}\\ \angle B={\tan }^{-1}\left(0.3617\right)\\ \angle B={19.89}^o\\ \text{simplifying the}\text{value to nearest minute}-\\ \angle B={19}^o\left({0.89}^o\times \frac{60\text{'}}{1^0}\right)\\ \angle B={19}^{o}53.4\text{'}\end{array}$

  $\begin{array}{l}or\\ \Rightarrow \angle B={19}^{o}53\text{'}\end{array}$

Conclusion: