In mathematics, trigonometry is one of the most important branches. Thus, the application of trigonometry includes, measuring the heights of towers, determining the distance of the object in the sea from the shore, finding the distance between celestial bodies, etc. Mainly, we use trigonometry to measure heights and distances. Therefore, we use trigonometric ratios to measure heights and distances of different objects.

**Heights and Distances**

Before we study height and distances we need to understand some terms.

**1. Angle**

An angle is a measure of rotation of a ray about its initial point. The original ray is called the initial side and the final position of the ray after rotation is called the terminal side of the angle. Vertex is the point of rotation.

If the ray rotates in an anticlockwise direction, the angle is said to be a positive angle and if the ray rotates in the clockwise direction, the angle is a negative angle.

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**Degree Measure**

If a rotation from the initial side to terminal side is 360^{th} of a revolution, the angle is said to have a measure of one degree, written as 1°.

**Radian Measure**

The angle subtended at the centre by an arc of length 1 unit in a unit circle whose radius is 1 unit is said to have a measure of 1 radian. It is another way of measuring any angle.

In a circle of radius **r**, subtends an angle θ radian at the centre and an arc of length **l**,

we have **l=** \(\frac{θ}{r}\)

**Relation Between Degrees and Radians : **2π radian = 360° or π radians = 180°** **

Degree | 30^{o} |
45^{o} |
60^{o} |
90^{o} |
180^{o} |
270^{o} |
360^{o} |

Radian | \(\frac{π}{6}\) | \(\frac{π}{4}\) | \(\frac{π}{3}\) | \(\frac{π}{2}\) | π | \(\frac{3π}{2}\) | 2π |

**Trigonometric Ratios**

In a right-angled \(\Delta\)BAO, where BOA = \(\Theta\),

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i. sin \(\Theta\) = \(\frac{Perpendicular}{Hypotenuse}\) = \(\frac{AB}{OB}\)

ii. cos \(\Theta\) = \(\frac{Base}{Hypotenuse}\) = \(\frac{OA}{OB}\)

iii. tan \(\Theta\) = \(\frac{Perpendicular}{Base}\) = \(\frac{AB}{OA}\)

iv. cosec \(\Theta\) = \(\frac{1}{sinθ}\) = \(\frac{OB}{AB}\)

v. sec \(\Theta\) = \(\frac{1}{cosθ}\) = \(\frac{OB}{OA}\)

vi. cot \(\Theta\) = \(\frac{1}{tanθ}\) = \(\frac{OA}{AB}\)

**Trigonometrical Identities**

- sin
^{2}\(\Theta\) + cos^{2}\(\Theta\) = 1 - 1+ tan
^{2}\(\Theta\) = sec^{2}\(\Theta\) - 1+ cot
^{2}\(\Theta\) = cosec^{2}\(\Theta\)

**Values of Trigonometric-Ratios**

\(\Theta\) | 0° | \(\frac{π}{6}\) 30° | \(\frac{π}{4}\) 45° | \(\frac{π}{3}\)
60° |
\(\frac{π}{2}\)
90° |

sin\(\Theta\) | 0 | \(\frac{1}{2}\) | \(\frac{1}{2}\) | \(\frac{3}{2}\) | 1 |

cos\(\Theta\) | 1 | \(\frac{3}{2}\) | \(\frac{1}{2}\) | \(\frac{1}{2}\) | 0 |

tan\(\Theta\) | 0 | \(\frac{1}{3}\) | 1 | 3 | not defined |

**Line of Sight**

The line which is drawn from the eyes of the observer to the point being viewed on the object is called the line of sight.

**The angle of Elevation**

It is the angle which is formed by the line of sight with the horizontal level and the point on the object (above horizontal level) viewed by the observer.

The angle of elevation of point P from O = ∠AOP.

**The angle of Depression**

It is the angle which is formed by the line of sight with the horizontal level and the point on the object (below horizontal level) viewed by the observer.

The angle of Depression = ∠AOP

**Points to remember :**

- The angle of elevation is numerically equal to the angle of depression. They are always acute angles.
- Both the angles are measured with the horizontal level.
- To determine the unknown side of the right angled triangle where one side is known and the acute angle is known. So we combine both the sides with trigonometric ratios of the triangle. i.e.

\(\frac{known side}{unknown side}\) = \(\frac{given side}{required side}\) = suitable t – ratio of Δ

**Solved on Heights and Distances**

**Q 1. **A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is 60^{0}. When he moves 40 m away from the bank, the angle of elevation measured is 30^{0}. Find the width of the river and the height of the tree.

**Solution:-** Let the height of the tree be y and width of the river be x and CD = 40 m

In ΔABD, tan 30^{0} = \(\frac{AB}{BD}\)

\(\frac{1}{\(\sqrt{3}\)}\) = \(\frac{y}{x + 40}\) ⇒ x+40 = y\(\sqrt{3}\)………..(i)

In ΔABC,

tan 60^{0} = \(\frac{AB}{BC}\)

\(\sqrt{3}\) = \(\frac{y}{x}\) ⇒ y = x\(\sqrt{3}\)……………(ii)

Putting the value of y from (ii) to (i)

x+40= [x\(\sqrt{3}\)] \(\sqrt{3}\) = 3x

40 = 2x

⇒ x= 20

y = x\(\sqrt{3}\) = 20\(\sqrt{3}\)

Height of the tree = 20\(\sqrt{3}\) m

Width of the river = 20 m.

5 7 31 283 ?

4533

3967

5×1+2=7

7×4+3=31

31×9+4=283

283×16+5=4533

4,8,24,28,84,88,_

264

4+4=8×3=24+4=28×3=84+4=88×3=264

1 5 20 ???

60

16,4,68,12,?,4,30,1,9 plz reply i need it