Chapter 9.9, Problem 3PSA

a.

To determine

To calculate:A ratio formed by sine of $45°$ .

Answer to Problem 3PSA

The ratio formed by sine of $45°$ is $\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information:

A triangle with angles as $45°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $45°$ .

Formula used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

Calculation:

Consider a right angled triangle ABC .

  

To find $\sin {45}^{\circ }$ value,

  $\sin {45}^{\circ }=\frac{oppositeside}{hypotenuseside}$

We know that,

  $\sin {45}^{\circ }=\frac{Perpendicular}{Hypotenuse}$

A simple method by means of which we can calculate the value of sine ratios for all the degrees is discussed here.

  $\begin{array}{l}\sin 0^{\circ }=\sqrt{\frac{0}{4}}=0\\ \sin {30}^{\circ }=\sqrt{\frac{1}{4}}=\frac{1}{2}\\ \sin {45}^{\circ }=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}\end{array}$

Therefore, $\sin {45}^{\circ }$ equals to fractional value of $\frac{1}{\sqrt{2}}$ .

b.

To determine

To find: A ratioformed by cosine of $45°$ .

Answer to Problem 3PSA

The ratioformed by cosine of $45°$ is $\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information:

A triangle with angles as $45°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $45°$ .

Formula used:

  $\cos \theta =\sin ({90}^{\circ }-\theta )$

Calculation:

Consider a right angled triangle ABC .

  

To find $\sin {45}^{\circ }$ value,

  $\sin {45}^{\circ }=\frac{oppositeside}{hypotenuseside}$

We know that,

  $\sin {45}^{\circ }=\frac{Perpendicular}{Hypotenuse}$

A simple method by means of which we can calculate the value of sine ratios for all the degrees is discussed here.

  $\begin{array}{l}\sin 0^{\circ }=\sqrt{\frac{0}{4}}=0\\ \sin {30}^{\circ }=\sqrt{\frac{1}{4}}=\frac{1}{2}\\ \sin {45}^{\circ }=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}\\ \sin {60}^{\circ }=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\\ \sin {90}^{\circ }=\sqrt{\frac{4}{4}}=1\end{array}$

From the values of sine, we can easily find the cosine function values. Now, to find the cos values, fill the opposite order the sine function values. It means that

  $\begin{array}{l}\cos 0^{\circ }=\sin {90}^{\circ }\\ \cos {30}^{\circ }=\sin {60}^{\circ }\\ \cos {45}^{\circ }=\sin {45}^{\circ }\\ \cos {60}^{\circ }=\sin {30}^{\circ }\\ \cos {90}^{\circ }=\sin 0^{\circ }\end{array}$

Therefore $\cos {45}^{\circ }$ equals to fractional value of $\frac{1}{\sqrt{2}}$ .

c.

To determine

To calculate: A ratio formed by tangent of $45°$ .

Answer to Problem 3PSA

The ratioformed by tangent of $45°$ is $1$ .

Explanation of Solution

Given information:

A triangle with angles as $45°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $45°$ .

Formula used:

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

Calculation:

Consider a right angled triangle ABC .

  

To find $\sin {45}^{\circ }$ value,

  $\sin {45}^{\circ }=\frac{oppositeside}{hypotenuseside}$

We know that,

  $\sin {45}^{\circ }=\frac{Perpendicular}{Hypotenuse}$

A simple method by means of which we can calculate the value of sine ratios for all the degrees is discussed here.

  $\begin{array}{l}\sin 0^{\circ }=\sqrt{\frac{0}{4}}=0\\ \sin {30}^{\circ }=\sqrt{\frac{1}{4}}=\frac{1}{2}\\ \sin {45}^{\circ }=\sqrt{\frac{2}{4}}=\frac{1}{\sqrt{2}}\\ \sin {60}^{\circ }=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}\\ \sin {90}^{\circ }=\sqrt{\frac{4}{4}}=1\end{array}$

From the values of sine, we can easily find the cosine function values. Now, to find the cos values, fill the opposite order the sine function values. It means that

  $\begin{array}{l}\cos 0^{\circ }=\sin {90}^{\circ }\\ \cos {30}^{\circ }=\sin {60}^{\circ }\\ \cos {45}^{\circ }=\sin {45}^{\circ }\\ \cos {60}^{\circ }=\sin {30}^{\circ }\\ \cos {90}^{\circ }=\sin 0^{\circ }\end{array}$

  $\cos {45}^{\circ }$ equals to fractional value of $\frac{1}{\sqrt{2}}$ .

  $\begin{array}{l}\tan \theta =\frac{\sin \theta }{\cos \theta }\\ \tan {45}^{\circ }=\frac{\sin {45}^{\circ }}{\cos {45}^{\circ }}=\frac{(\frac{1}{\sqrt{2}})}{(\frac{1}{\sqrt{2}})}=1\end{array}$

Therefore $\tan {45}^{\circ }$ equals to value of $1$ .