Introduction to Trigonometry

 Introduction to Trigonometry




 

Trigonometry is a branch of mathematics dealing with relations involving lengths and angles of triangles. It can, in a simpler manner, be called the study of triangles. The angles are either measured in degrees or radians

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Opposite & Adjacent Sides in a Right Angled Triangle

In the Î”ABC right-angled at B, BC is the side opposite to ∠A, AC is the hypotenuse an










Trigonometric Ratios

For the right - angled triangle ABC , right-angled at ∠B , the trigonometric ratios of the ∠A are as follows: 

 · sin A=`\frac{opposite side}{hypoten use}`= `\frac{BC}{AC }`

·        ·    cos A=`\frac{adjacent side}{hypoten use}`=`\frac{AB}{AC }`

·       ·      tan A=`\frac{opposite side}{adjacent side}`=`\frac{BC}{AB }`

·        ·   cosec A=`\frac{opposite side}{adjacent side}`=`\frac{AC}{BC }`  

      ·   sec A=`\frac{hypoten use}{opposite side}`=`\frac{AC}{AB }` 

·        ·      cot A=`\frac{adjacent side}{opposite side}`=`\frac{AB}{BC }`

Relation between Trigonometric Ratios

cosec θ =`\frac{1}{sin θ}`

sec θ = `\frac{1}{cos θ}`

tan θ = `\frac{sin θ}{cos θ}`

cot θ = `\frac{cosθ}{sin θ}` = `\frac{1}{tan θ}`

Trigonometric Ratios of Specific Angles

Range of Trigonometric Ratios from 0 to 90 degrees

For 0∘≤θ≤90∘,

0≤ sinθ ≤1

0≤cosθ≤1

0≤tanθ<∞

1≤secθ<∞

0≤cotθ<∞

1≤cosecθ<∞

tanθ and secθ are not defined at  90∘.

cotθ and cosecθ are not defined at 0∘.

Variation of trigonometric ratios from 0 to 90 degrees

As θ increases from 0∘ to 90∘

sin θ increases from 0 to 1

cos θ decreases from 1 to 0

tan θ increases from 0 to ∞

cosec θ decreases from ∞ to 1

sec θ increases from 1 to ∞

cot θ decreases from ∞ to 0

Standard values of Trigonometric ratios

∠A

0o

30o

45o

60o

90o

sin A

0

 1/2

 1/√2

 √3/2

 1

cos A

1

 √3/2

1/√2

 1/2

0

tan A

0

 1/√3

 1

√3

 not defined

cosec A

not defined

 2

 √2

 2/√3

 1

sec A

1

2/√3

√2

 2

 not defined

cot A

not defined

 √3

 1

1/√3

0



Trigonometric Ratios of Complementary Angles

Complementary Trigonometric ratios

If θ is an acute angle, its complementary angle is 90∘−θ. The following relations hold true for trigonometric ratios of complementary angles.

sin (90∘− θ) = cos θ

cos(90∘− θ) = sin θ

tan (90∘− θ) = cot θ

cot (90∘− θ) = tan θ

cosec (90∘− θ) = sec θ

sec (90∘− θ) = cosec θ

 

Trigonometric Identities


sin^{2}θ+cos^{2}θ=1

1+cot^{2}θ=coesc^{2}θ

1+tan^{2}θ=sec^{2}θ

Example 1:

Find Sin A and Sec A, if 15 cot A = 8.

Solution:


Given that 15 cot A = 8

Therefore, cot A = 8/15.

We know that tan A = 1/ cot A

Hence, tan A = 1/(8/15) = 15/8.

Thus, Side opposite to ∠A/Side Adjacent to ∠A = 15/8

Let BC be the side opposite to ∠A and AB be the side adjacent to ∠A and AC be the hypotenuse of the right triangle ABC respectively.

Hence, BC = 15x and AB = 8x.

Hence, to find the hypotenuse side, we have to use the Pythagoras theorem.

(i.e) AC2 = AB2 + BC2

AC2 = (8x)2+(15x)2

AC2 = 64x2+225x2

AC2 = 289x2

AC = 17x.

Therefore, the hypotenuse AC = 17x.

Finding Sin A:

We know Sin A = Side Opposite to ∠A / Hypotenuse

Sin A = 15x/17x

Sin A = 15/17.

Finding Sec A:

To find Sec A, find cos A first.

Thus, cos A = Side adjacent to ∠A / Hypotenuse

Cos A = 8x/17x

We know that sec A = 1/cos A.

So, Sec A = 1/(8x/17x)

Sec A = 17x/8x

Sec A = 17/8.

Therefore, Sin A = 15/17 and sec A = 17/8.

Example 2:

If tan (A+ B) =√3, tan (A-B) = 1/√3, then find A and B. [Given that  0° <A+B ≤ 90°; A>B ]

Solution:

Given that 

Tan (A+B) = √3.

We know that tan 60 = √3.

Thus, tan (A+B) = tan 60° = √3.

Hence A+B= 60° …(1)

Similarly, given that,

Tan (A-B) = 1/√3.

We know that tan 30° = 1/√3.

Thus, tan (A-B) = tan 30° = 1/√3.

Hence, A-B = 30° …(2)

Now, adding the equations (1) and (2), we get

A+B+A-B = 60° + 30°

2A = 90°

A = 45°.

Now, substitute A = 45° in equation (1), we get

45° +B = 60°

B = 60°- 45°

B = 15°

Hence, A = 45 and B = 15°.






EXERCISE 1.1

1. A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°.


Solution:

Length of the rope is 20 m and angle made by the rope with the ground level is 30°.

Given: AC = 20 m and angle C = 30°

To Find: Height of the pole

Let AB be the vertical pole

In right ΔABC, using sine formula

sin 30° = AB/AC

Using value of sin 30 degrees is ½, we have

1/2 = AB/20

AB = 20/2

AB = 10

Therefore, the height of the pole is 10 m.

2. A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Solution:

Using given instructions, draw a figure. Let AC be the broken part of the tree. Angle C = 30°

BC = 8 m

To Find: Height of the tree, which is AB




From figure: Total height of the tree is the sum of AB and AC i.e. AB+AC

In right ΔABC,

Using Cosine and tangent angles,

cos 30° = BC/AC

We know that, cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

Therefore, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.







3. A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Solution:

As per contractor’s plan,












Let, ABC is the slide inclined at 30° with length AC and PQR is the slide inclined at

60° with length PR.

To Find: AC and PR

In right ΔABC,

sin 30° = AB/AC

1/2 = 1.5/AC

AC = 3

Also,

In right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3

Hence, length of the slide for below 5 = 3 m and

Length of the slide for elders children = 2√3 m

4. The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Solution:

Let AB be the height of the tower and C is the point elevation which is 30 m away from the foot of the tower.

To Find: AB (height of the tower)

In right ABC

tan 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

Thus, the height of the tower is 10√3 m.

5. A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Solution:

Draw a figure, based on given instruction,

Let BC = Height of the kite from the ground, BC = 60 m

AC = Inclined length of the string from the ground and

A is the point where string of the kite is tied.

To Find: Length of the string from the ground i.e. the value of AC

From the above figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

Thus, the length of the string from the ground is 40√3 m.

6. A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Solution:

Let the boy initially stand at point Y with inclination 30° and then he approaches the building to the point X with inclination 60°.

To Find: The distance boy walked towards the building i.e. XY

From figure,

XY = CD.

Height of the building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

Measure of AB is 28.5 m

In right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

Again,

In right ΔABC,

tan 60° = AB/BC

√3 = 28.5/BC

BC = 28.5/√3 = 28.5√3/3

Therefore, the length of BC is 28.5√3/3 m.

XY = CD = BD – BC = (28.5√3-28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 = 19√3  m.

Thus, the distance boy walked towards the building is 19√3 m.

7. From a point on the ground, the angles of elevation of the bottom and the top of a

transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Solution:

Let BC be the 20 m high building.

D is the point on the ground from where the elevation is taken.

Height of transmission tower = AB = AC – BC




















To Find: AB, Height of the tower

From figure, In right ΔBCD,

tan 45° = BC/CD

1 = 20/CD

CD = 20

Again,

In right ΔACD,

tan 60° = AC/CD

√3 = AC/20

AC = 20√3

Now, AB = AC – BC = (20√3-20) = 20(√3-1)

Height of transmission tower = 20(√3 – 1) m.

8. A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

Solution:

Let AB be the height of statue.

D is the point on the ground from where the elevation is taken.

To Find: Height of pedestal = BC = AC-AB






















In right triangle BCD,

tan 45° = BC/CD

1 = BC/CD

BC = CD …..(1)

 

Again,

In right ΔACD,

tan 60° = AC/CD

√3 = ( AB+BC)/CD

√3CD = 1.6 + BC

√3BC = 1.6 + BC (using equation (1)

√3BC – BC = 1.6

BC(√3-1) = 1.6

BC = [(1.6)(√3+1)]/[(√3-1)(√3+1)]

BC = [1.6(√3+1)]/(2) m

BC = 0.8(√3+1)

Thus, the height of the pedestal is 0.8(√3+1) m.

 

9. The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Solution:

Let CD be the height of the tower. AB be the height of the building. BC be the distance between the foot of the building and the tower. Elevation is 30 degree and 60 degree from the tower and the building respectively.




















In right ΔBCD,

tan 60° = CD/BC

√3 = 50/BC

BC = 50/√3 …(1)

Again,

In right ΔABC,

tan 30° = AB/BC

⇒ 1/√3 = AB/BC

Use result obtained in equation (1)

AB = 50/3

Thus, the height of the building is 50/3 m.