# Height And Distance Formulas

## Height and Distances related Formulas

Trigonometry was invented because its need arose in astronomy. Since then the astronomers have used it, for instance, to calculate distances from the Earth to the planets and stars. In this page we are going to learn about various Formulas for Height and Distances which will include different types of trigonometric ratios and figures. For your knowledge ### Height and Distance Formulas

• There are basically two terms associated with heights and distances which are as follows :
• Angle of Elevation.
• Angle of Depression.

### Formulas for Angle of Elevation

• The Angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object

### Formulas for Angle of Depression

• The angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed. ### Related Banners

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### Height and Distance Formulas for Trigonometric ratio

• ΔABC is a right angled triangle where is AB is the perpendicular, AC is the hypotenuse, and BC is the base.

Then,

• Sin θ  = $\frac{AB}{AC}$
• Cos θ = $\frac{BC}{AC}$
• Tan θ  = $\frac{AB}{BC}$
• Cosec θ  = $\frac{AC}{AB}$
• Sec θ  = $\frac{AC}{BC}$
• Cot θ  = $\frac{BC}{AB}$
• Trigonometrical Identities:
• sin2θ + cos2 θ = 1
• 1 + tan2θ = sec2 θ
• 1 + cot2θ = cosec2 θ ### Definition of Harmonic Progression (H.P)

Harmonic progression is the series when the reciprocal of the terms are in AP.

For example, $\frac{1}{a}, \frac{1}{ (a + d)}, \frac{1}{(a + 2d)}$…… are termed as a harmonic progression as a, a + d, a + 2d are in Arithmetic progression.

• First term of a HP is $\frac{1}{a}$
• There are many Application of Harmonic Progressions.

## Some Examples Using Above Formulas:

Question 1:  Varun Observed that from a  point 355 meters away from the foot of a tower, the top of the tower is at an angle of elevation of $30^{o}$, then the height (in meters) of the tower is?

= $\frac{1}{\sqrt{3}}$ = X/355

=X= 614.87 Question 2: A man is standing in his balcony, which is on the third floor of the building and is at the height of 10m. His angle of elevation at the top of the opposite building is $60^{o}$ ,and the angle of depression of the base is $30^{o}$ . Determine the height of the building opposite to him?

Answer: Let AB the building opposite to the man.
Let us assume he is standing at point C
Therefore, we can say that  ∠BCD = 600
∠ACD = 300
Let BD = h
In the given  ADC, tan 300 = 10/CD

• 1/√3= 10/CD
• CD = 10 √3
Also, in the given  ΔDBC, we can say that
Tan 600 = h/ CD
• √3= h/ 10 √3
• H= 30 mtrs
Therefore, the height of the opposite building is
AB = BD + DA
AB = 10 + 30
AB = 40 mtrs Question 3: A man stands at the corner of a farm, which is square. The angle of elevation of a scarecrow which presents diagonally opposite corner is$60^{o}$ . When he walks backward in the straight line fir 80ft. The angle of elevation of the scarecrow now becomes $30^{o}$. Calculate the field’s area.

Answer: Tan 600 = PQ/QR = √3
PQ= √3QR
Tan 300 = 1/√3 = PQ/SQ
|= PQ/ 80+ QR
80+ QR= √3PQ
80+ QR = 3QR
QR = 40ft It can also be seen from the figure that the man at point R and the scarecrow at point Q are diagonally opposite to one another.
Therefore, QR is the diagonal of the square farm.
Diagonal =side x  /√2
Therefore, 40 = side x  /√2
Side = 40 √2
Therefore, area = (side)2 = (40 √2)2
= 800 sq feet

Question 4Adam is standing in the Sun. He is 6ft tall and is casting a shadow of 4ft. A flag is also present near the man casting a shadow of 36ft. Calculate the height of the flag.
Answer: The angle of elevation for the Sun will be the same for both the cases.
Therefore, the ratio of object to shadow will be the same for both the cases as per the proportionality rule.
Object height/ Shadow height = 6/ 4 = H /36
Therefore, H = 54 ft = Height of the flag Question 5: When a boy looks from the foot and the top of a tower at the roof of a building, the angles of elevation and depression are $27^{o}$and $63^{o}$, The height of this building is 40m, then calculate the height of the tower given that tan 630 =2.

Answer: Let the tower be AB
Let the building of height 40m be CD
In the given triangle ACD, AC/DC = cot 270
= cot (90-63)
AC/40 = tan 630 = 2
AC = 80m
Now, DE = AC = 80m
Also, in triangle BED, tan 630 = BE/DE
2= BE/80
Therefore, BE = 160 m
Therefore, the height of the tower can be calculated as AE + EB
= 40 +160 = 200m ## Get over 200+ course One Subscription

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