SOME APPLICATION OF TRIGNOMETRY

Made By: Shivansh J., Shivam, Benson

Psalm 9001:1001 – “Now who da Man !”

Introduction

Trigonometry is the branch of mathematics that deals with triangles particularly right triangles. For one thing trigonometry works with all angles and not just triangles. They are behind how sound and light move and are also involved in our perceptions of beauty and other facets on how our mind works. So trigonometry turns out to be the fundamental to pretty much everything!

Basic Fundamentals Angle of Elevation: In the picture below, an

observer is standing at the top of a building is looking straight ahead (horizontal line). The observer must raise his eyes to see the airplane (slanting line). This is known as the angle of elevation.

Angle of Depression: The angle below horizontal that an observer must look to see an object that is lower than the observer. Note: The angle of depression is congruent to the angle of elevation (this assumes the object is close enough to the observer so that the horizontals for the observer and the object are effectively parallel).

600

300

600

a

2a2a

If θ is an angleThe 90-θ is it’s complimentary angle

Angle of elevation

Angle of depressio

n

11

145

2

1 45

2

145

Tan

Cos

Sine

A trigonometric function is a ratio of certain parts of a triangle. The names of these ratios are: The sine, cosine, tangent, cosecant, secant, cotangent.

Let us look at this triangle…

ac

b

ө A

B

C

Given the assigned letters to the sides and angles, we can determine the following trigonometric functions.

The Cosecant is the inversion of the sine, the

secant is the inversion of the cosine, the cotangent is the

inversion of the tangent.With this, we can find the sine of the value of angle A by dividing side a by side c. In order to find the angle itself, we must take the sine of the angle and invert it (in other words, find the cosecant of the sine of the angle).

Sinθ=

Cos θ=

Tan θ=

Side Opposite

Side Adjacent

Side AdjacentSide Opposite

Hypothenuse

Hypothenuse

=

=

= a

bca

b

c

4560

12

h

d

d toequals also is 39.16732.0

12

732.0732.112

;732.112h

(2)(1)..in ngSubstituti

)2(732.1312

60

(1)-------hd ;..145

h

hhhh

d

hTan

ord

hTan

The angle of elevation of the top of a tower from a point At the foot of the tower is 300 . And after advancing 150mtrs Towards the foot of the tower, the angle of elevation becomes 600 .Find the height of the tower

150

h

d

30 60

mh

h

hh

hh

hh

dofvaluethengSubstituti

hd

From

hdFrom

d

hTan

d

hTan

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31502

31503

31503

)1503(3

..........

)150(3

)2(

3)1(

)2(150

360

)1(3

130

Questions based on trigonometry :- The angle of elevation of the top of a pole measures 48°

from a point on the ground 18 ft. away from its base. Find the height of the flagpole.

Solution Step 1: Let’s first visualize the situationStep 2: Let ‘x’ be the height of the flagpole.STEP 3: From triangle ABC,tan48=x/18Step 4: x = 18 × tan 48° = 18 × 1.11061… = 19.99102…» 20Step 5: So, the flagpole is about 20 ft. high.

(Simple)Questions on trigonometry:-

Question #1John wants to measure the height of a tree. He walks exactly 100 feet from the base of the tree and looks up. The angle from the ground to the top of the tree is 30º . How tall is the tree?

Question #2A building is 50 feet high. At a distance away from the building, an observer notices that the angle of elevation to the top of the building is 45º. How far is the observer from the base of the building?

45

50

B

A

C

D

A hoarding is fitted above a building. The height of the building is 12 m. When I look at the lights fitted on top of the hoarding, the angle of elevation is 500 and when I look at the top of the building from the same place, the angle is 450. If the height of the flat on each floor is equal to the height of the hoarding, the max floors on the building are? (Tan 500=1.1917)

ANSWER : Let AB denote the height of the building, Let AC denote the height of the hoarding on top of the building Thus, Tan500 = (12 + AC) ÷ 12 1.1917 = 1 + (AC ÷ 12) 1.1917 – 1 = AC ÷ 12 12 ÷ AC = 1 ÷ 0.1917 ~ 5

45

BA

CDE

60

I see a bird flying at a constant speed of 1.7568 kmph in the sky. The angle of elevation is 600. After ½ a minute, I see the bird again and the angle of elevation is 450. The perpendicular distance of the bird from me, now will be(horizontal distance) ?

ANSWER : Let A be the initial position and B be the final position of the bird, <AED= 600 , <BED = 450

Let E be my position. Time required to cover distance from A to B=30 sec.

Speed of bird= 1.7568 × m/s

Distance travelled by bird in 30 sec. = 1.7568 × × 30 = 14.64 m

In right angled = Tan 600 . Thus, ED =

In right angled As EC=ED+DC ,,, BC= +DC ,,, BC= + 14.64

185

185

ED

ADAED,

3

AD

ECBCBCE ,

3

AD

3

BC

64.143

11

BC

3

11

164.14BC 320

1732.1

364.14

m

THE END

Made By:-Shivansh J. ShivamPsalm 9001:1001 – “Now who da Man !”