Chapter 13.3, Problem 15MCQ

a

To determine

The value of the h.

Answer to Problem 15MCQ

The height(h) of the computer monitor is $\text{9}\text{in}$ .

Explanation of Solution

Given:

The front of the monitor is measured as shown below:

  

From diagram the height of computer monitor is evaluated as follows:

From Pythagoras theorem:

  ${\text{(height}\text{of}\text{computer}\text{monitor)}}^{\text{2}}\text{+}{\left(\text{base}\text{of}\text{computer}\text{monitor}\right)}^{\text{2}}\text{=}{\left(\text{diagonal}\text{of}\text{computer}\text{monitor}\right)}^{\text{2}}$

Therefore,

  $\begin{array}{l}{h}^2+{12}^2={15}^2\\ h^2={15}^2-{12}^2\\ h^2=9^2\\ h=9\end{array}$

Hence, the height of the computer monitor is $9in$ .

b

To determine

To show: $\cot \theta =\frac{\cos \theta }{\sin \theta }$ using the given diagram.

Explanation of Solution

Given:

The front of the monitor is measured as shown below:

  

From the diagram, evaluating the value of the $\cos \theta \text{and sin}\theta$ :

  $\begin{array}{c}\text{cosθ=}\frac{\text{base}\text{of}\text{computer}\text{monitor}}{\text{digonal}\text{of}\text{computer}\text{monitor}}\\ \text{=}\frac{\text{12}}{\text{15}}\\ \text{sinθ=}\frac{\text{height}\text{of}\text{computer}\text{monitor}}{\text{digonal}\text{of}\text{computer}\text{monitor}}\\ \text{=}\frac{\text{9}}{\text{15}}\end{array}$

Now evaluate the value of $\cot \theta$ :

  $\cot \theta =\frac{\cos \theta }{\sin \theta }$

Therefore,

  $\begin{array}{c}\cot \theta =\frac{\frac{12}{15}}{\frac{9}{15}}\\ =\frac{12}{9}\\ =\frac{4}{3}\end{array}$

Hence, the solution is $\cot \theta =\frac{4}{3}$ .

Since,

  $\begin{array}{c}\text{Cot}\theta \text{=}\frac{\text{Base}}{\text{Perpendicular}}\\ =\frac{12}{9}\\ =\frac{4}{3}\end{array}$

Therefore, the equation is verified $\text{cot}\theta \text{=}\frac{\text{cos}\theta }{\sin \theta }$ .