Chapter 9.9, Problem 2PSA

a.

To determine

To calculate: The ratio of sine of ${30}^{\circ }$ .

Answer to Problem 2PSA

The ratio of sine of ${30}^{\circ }$ is $\frac{1}{2}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

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

Calculation:

  

Consider an equilateral triangle ABC.

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

To find $\sin {30}^{\circ }$ value,

  $\sin {30}^{\circ }=\frac{oppositeside}{hypotenuseside}$

We know that,

  $\sin {30}^{\circ }=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$

Therefore, $\sin {30}^{\circ }$ equals to fractional value of $\frac{1}{2}$ .

b.

To determine

To find: A ratio formed by cosine of ${30}^{\circ }$ .

Answer to Problem 2PSA

The ratioformed by cosine of ${30}^{\circ }$ is $\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

  $\cos \theta =\frac{adjacentside}{hypotenuseside}$

Calculation:

  

Consider an equilateral triangle ABC .

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

By Pythagoras theorem,

  $AD=\sqrt{3}a$

To find $\cos {30}^{\circ }$ value,

  $\cos {30}^{\circ }=\frac{adjacentside}{hypotenuseside}$

We know that,

  $\cos {30}^{\circ }=\frac{AD}{AB}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$

Therefore, $\cos {30}^{\circ }$ equals to fractional value of $\frac{\sqrt{3}}{2}$ .

c.

To determine

To calculate: A ratio formed by tangent of ${30}^{\circ }$ .

Answer to Problem 2PSA

The ratio formed by tangent of ${30}^{\circ }$ is $\frac{1}{\sqrt{3}}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

  $\tan \theta =\frac{oppositeside}{adjacentside}$

Calculation:

  

Consider an equilateral triangle ABC .

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

By Pythagoras theorem,

  $AD=\sqrt{3}a$

To find $\tan {30}^{\circ }$ value,

  $\tan {30}^{\circ }=\frac{oppositeside}{adjacentside}$

We know that,

  $\tan {30}^{\circ }=\frac{BD}{AD}=\frac{a}{\sqrt{3}a}=\frac{1}{\sqrt{3}}$

Therefore, $\tan {30}^{\circ }$ equals to fractional value of $\frac{1}{\sqrt{3}}$ .

d.

To determine

To find: A ratio formed by sine of ${60}^{\circ }$ .

Answer to Problem 2PSA

The ratio formed by sine of ${60}^{\circ }$ is $\frac{\sqrt{3}}{2}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

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

Calculation:

  

Consider an equilateral triangle ABC .

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

By Pythagoras theorem,

  $AD=\sqrt{3}a$

To find $\sin {60}^{\circ }$ value,

  $\sin {60}^{\circ }=\frac{oppositeside}{hypotenuseside}$

We know that,

  $\sin {60}^{\circ }=\frac{AD}{AB}=\frac{\sqrt{3}a}{2a}=\frac{\sqrt{3}}{2}$

Therefore, $\sin {60}^{\circ }$ equals to fractional value of $\frac{\sqrt{3}}{2}$ .

e.

To determine

To calculate: A ratio formed by cosine of ${60}^{\circ }$ .

Answer to Problem 2PSA

The ratio formed by cosine of ${60}^{\circ }$ is $\frac{1}{2}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

  $\cos \theta =\frac{adjacentside}{hypotenuseside}$

Calculation:

  

Consider an equilateral triangle ABC .

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

By Pythagoras theorem,

  $AD=\sqrt{3}a$

To find $\cos {60}^{\circ }$ value,

  $\cos {60}^{\circ }=\frac{adjacentside}{hypotenuseside}$

We know that,

  $\cos {60}^{\circ }=\frac{BD}{AB}=\frac{a}{2a}=\frac{1}{2}$

Therefore, $\cos {60}^{\circ }$ equals to fractional value of $\frac{1}{2}$ .

f.

To determine

To calculate: A ratioformed by tangent of ${60}^{\circ }$ .

Answer to Problem 2PSA

The ratioformed by tangent of ${60}^{\circ }$ is $\sqrt{3}$ .

Explanation of Solution

Given information:

A triangle with angles as $30°$ and $90°$ .

As sum of angles in a triangle is $180°$ , the third angle is $60°$ .

Formula used:

  $\tan \theta =\frac{oppositeside}{adjacentside}$

Calculation:

  

Consider an equilateral triangle ABC .

  

Since each angle is an equilateral triangle is $60°$ , therefore

  $\angle A=\angle B=\angle C=60°$

Draw the perpendicular line AD from A to the side BC.

Now, $\Delta ABD\cong \Delta ACD$

Therefore BD=DC and also

  $\angle BAD\cong \angle CAD$

Now observe that the triangle ABD is a right triangle, right angled at D with

  $\angle BAD={30}^{\circ }$ and $\angle ABD={60}^{\circ }$ .

As you know, for finding the trigonometric ratios, we need to know the lengths of the sides of the triangle. So, let us suppose that

  $AB=2a$

  $BD=\frac{1}{2}BC=a$

By Pythagoras theorem,

  $AD=\sqrt{3}a$

To find $\tan {60}^{\circ }$ value,

  $\tan {60}^{\circ }=\frac{oppositeside}{adjacentside}$

We know that,

  $\tan {60}^{\circ }=\frac{AD}{BD}=\frac{\sqrt{3}a}{a}=\sqrt{3}$

Therefore $\tan {60}^{\circ }$ equals to fractional value of $\sqrt{3}$ .