Today our topic of discussion is Horizontal Line, Vertical Line and Vertical Plane.
Horizontal Line, Vertical Line and Vertical Plane
Distance and Elevation
From very ancient times trigonometric ratios are applied to find the distance and height of any distant object. At present trigonometric ratios are of boundless importance because of its increasing usage.
The heights of the hills, mountains towers, trees and the widths of those rivers which cannot be measured in ordinary method are measured with the help of trigonometry. It is necessary to know the trigonometrical ratios values of acute angles in this regard.
At the end of this chapter, the students will be able to-
> explain the geoline, vertical plane and angles of elevation and declination.
> solve mathematical problem related to distance and height with the help of trigonometry.
> measure practically different types of distances and heights with the help of trigono etry.
Horizontal Line, Vertical Line and Vertical Plane
A horizontal line is a straight line on the horizontal plane. A straight line parallel to horizon is also called a horizontal line. A vertical line is a line perpendicular to the horizontal plane. It is also called a normal line. A horizontal line and a vertical line intersected at right angles on the plane define a plane. It is known as vertical plane.
In the figure, a tree with height of AB is standing vertically at a distance of CB from a point C on the horizontal plane. Here, CB is the horizontal line. BA is the vertical line and the plane ABC is perpendicular to the horizontal plane which is a vertical plane.
Angle of Elevation and Angle of Depression
Observe the figure, AB is a straight line parallel to the horizon. The points A, O, B, P and Q lie on the same vertical plane. The point P on the straight line AB makes angle ∠POB with the line AB. Here at A O, the angle of elevation of P is POB.
angle of elevation
So, the angle at any point above the plane with the straight line parallel to horizon is called the angle of elevation.
The points Q, lie at lower side of the straight line AB parallel to horizon. Here, the angle of depression at O of Q is angle QOB So, the angle at any point below the straight line parallel to the plane is called the angle of depression.
Note: For solving the problems in this chapter approximately right figure is needed. While drawing the figure, the following techniques are to be applied.
- While drawing 30 ° angle it is needed base > perpendicular.
- While drawing 45 °angle, it is needed base = perpendicular.
- While drawing 60° angle, it is needed base < perpendicular.
Example 1. The angle of elevation at the top of a tower at a point on the ground is 30° at a distance of 75 metre from the foot. Find the height of the tower.
Solution:
Let, the height of the tower is AB = h metre. The angle of elevation at C from the foot of the tower BC = 75 metre of A on the ground is ∠ ACB = 30 °
From ΔABC we get,tan ∠ACB = (AB)/(BC)
or, tan 30 °= h/75
or, 1/(√3)) = h/75 multiplying denominator and or, √3* h = 75 or, numerator by √3
or, h = 75/(√3)
or, h = (75s√3)/3
or, h = 25√3
h = 43.301 (approx.).
∴ Height of the tower is 43.30 metre (approx.)
Example 2. The height of a tree is 105 metre. If the angle of elevation of the tree at a point from its foot on the ground is 60°, find the distance of the point on the ground from the foot of the tree.
Solution:
Let, the distance of the point on the ground from the foot of tree is BC = x metre. Height of the tree AB = 105 metre and at C the angle of elevation of the vertex of tree is ∠ACB = 60 °From right-∠ABC we. get,
tan ∠ACB = (AB)/(BC)
or, tan 60 °= 105/x
or, √3 =105/x [:tan 60 = √3]
or, √3 * x = 105
x = 105/(√3))
or, x = (105√3)/3
or, x = 35√3
x = 60.622 (approx.)
∴The required distance of the point on the ground from the foot of the tree is 60.62 metre (approx.).
Example 3. A ladder of 18 metre long touches the roof of a wall and makes an angle of 45° with the horizon. Find the height of the wall.
Solution:
Let, the height of the wall AB = h metre, length of ladder AC =18m and makes angle ∠ACB = 45 °with the ground.
From ΔABC we get, sin ∠ACB = (AB)/(AC)
or, sin 45 °= h/18
or, or, 1/(√(2)) = h 18 [*; √(2) * h = 18 sin or, 45 °= 1/(√2)) ] h = 18/(√2))
or, h = (18√2))/2 [multiplying the numerator and denominator y * √2]
or, h = 12.728 (approx.)
Therefore, The required height of the wall is 12.73 (approx.)
Example 4. A tree leaned due to storm. The stick with height of 7 metre from its foot was leaned against the tree to make it straight. If the angle of depression at the point of contacting with the stick on the ground is 30°, find the length of the stick.
Solution:
Let, the height of the stick from the foot leaned against the tree of AB = 7 metre and angle of depression ∠ DBC = 30 °
: ∠ACB= ∠DBC = 30 °[alternate angle]
From right-angled ΔABC we get,
sin ∠ACB = (AB)/(BC) *
or ,sin 30° = 7 BC
or, 1 2 = 7 BC [*; sin 30 °= 1/2 ]
∴ BC = 14
∴The required height of the stick is 14 metre.
work:
In the figure, if depression angle ∠CAE = 60° elevation angle ∠ADB = 30° . AC = 36 metre, AB ⊥ DC and D, C, B lie on the same straight line, find the lengths of the sides AD, AB and CD.
Example 5. The angle of elevation at a point of the roof of a building is 60° in any point on the ground. Moving back 42 metre from the angle of elevation of the point of the place of the building becomes 45° . Find the height of the building.
Solution:
Let, the height of the building is AB = h metres. The angle of elevation at the top ∠ACB = 60° . The angle of elevation becomes ∠ADB = 45° moving back from C by CD = 42 metre.
Let, BC = x metre.
BD = BC + CD = (x + 42) metre.
From ΔABC we get,
tan ∠ACB = (AB)/(BC) *
or , tan 60 °= h/x
or, √3 = h/x [ tan 60 °= √(3) * J x = h/(√(3)) .(1)
Again, from ΔABD we get, tan ∠ADB = tan 45 °= (AB)/(BD)
or, sin * 45 °= h/(x + 42)
or, 1= h x+42 [*** tan 45° = 1 ]
or, h = x + 42
or, h = h/(√(3)) + 42 [from (1)]
or, √(3) * h = h + 42√(3)
or, √(3) * h – h = 42√(3)
or, √(3) – 1) * h = 42√(3) *
or h = (42√(3))/(√3) – 1)
h = 99.373 (approx.)
∴Height of the building is 99.37 metre (approx.).
Example 6. A pole is broken such that the undetached broken part makes an angle of 30° with the other and touches the ground at a distance of 10 metre form its foot. Find the length of the pole.
Solution:
Let, the total height of the pole is AB = h metre. It breaks at the height of BC = x metre without separation and makes an angle with the other, ∠BCD = 30° and touches the ground at a distance BD = 10 metre from the foot.
Here, CD = AC = AB – BC = (h – x) metre From ΔBCD we get,
∠BCD = (BD)/(BC) 1 10 √3 I
or, tan 30 °= 10/x
∴ z=10√3 or.
Again, sin ∠BCD = (BD)/(CD)
or, sin 30° = (BD)/(CD)
or, 1/2 = 10/(h – x)
or, h – x = 20
or , h = 20 + x
or, h = 20 + 10√(3) [putting the value of r]
h = 37.321 (approx.)
∴ Height of the pole is 37.32 metre (approx.).
See more:
