angle of elevation 5examples
Answers
Answer:
Which one is the angle of elevation?
The angle of elevation is the angle between a horizontal line from the observer and the line of sight to an object that is above the horizontal line. In the diagram below, AB is the horizontal line.
Answer:
What Is Angle Of Elevation?
The angle of elevation is the angle between a horizontal line from the observer and the line of sight to an object that is above the horizontal line.
Step-by-step explanation:
Example 1
student sitting in a classroom sees a picture on the black board at a height of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30°. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45°. Find the distance moved by the student.
Solution :
Distance moved by the student = BC
In triangle ABD :
∠ABD = 30°
tan θ = opposite side/Adjacent side
tan 30° = AD/BD
1/√3 = 1.5/BD
BD = 1.5 x √3 ==> 1.5 √3
In triangle ACD :
∠ACD = 45°
tan θ = opposite side/Adjacent side
tan 45° = AD/CD
1 = 1.5/CD
CD = 1.5
BC = BD - CD
BC = 1.5 √3 - 1.5
= 1.5 (√3 - 1) = 1.5(1.732 - 1)
= 1.5 (0.732) ==> 1.098 m
Hence the distance moved by the student is 1.098 m.
example 2
A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. 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 :
From the given information, we can draw a rough diagram
AD = 28.5 m
In triangle ABD
∠ABD = 30°
tan θ = opposite side/Adjacent side
tan 30° = AD/BD
1/√3 = 28.5/BD ==> BD = 28.5 √3 --(1)
In triangle ACD
∠ABD = 60°
tan θ = opposite side/Adjacent side
tan 60° = AD/CD
√3 = 28.5/CD ==> CD = 28.5/√3
Multiplying by √3 on both numerator and denominator, we get
CD = 28.5/√3 ==> 28.5√3/3 ==> 9.5√3 -->(2)
the distance he walked towards the building = BD - CD
= 28.5 √3 - 9.5√3 ==> 19√3
Example 3
From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a Yacht and a Barge along the same line of sight .The angles of depression for the Yacht and the Barge are 45° and 30° respectively. For safety purposes the two sea vessels should be at least 300 feet apart. If they are less than 300 feet , the keeper has to sound the alarm. Does the keeper have to sound the alarm ?
Solution :
From the above picture, we have to find the value of CD.
In triangle ABC
∠ACB = 45°
sin θ = opposite side/Hypotenuse side
sin 45° = AB/BC
1/√2 = 200/BC
BC = 200 √2
In triangle ABD
∠ADB = 30°
sin 30° = AB /BD
1/2 = 200/BD
BD = 200 x 2 ==> 400
CD = BD - BC ==> 400 - 200 √2 ==> 200(2 - √2)
= 200 (2 - 1.414)
= 200(0.586) ==> 117.2 m
From this, we come to know that the distance between Yacht and a Barge is less than 300 m. Hence the keeper has to sound the alarm.
Example 4
From a boat on the lake, the angle of elevation to the top of the cliff is 24°22'. If the base of the cliff is 747 feet from the boat, how high is the cliff (to the nearest foot)? 2) From a boat on the river below a dam, the angle of elevation to the top of the dam is 24°8'.
Example 5
A man who is 2 m tall stands on horizontal ground 30 m from a tree. The angle of elevation of the top of the tree from his eyes is 28˚. Estimate the height of the tree.
Solution:
Let the height of the tree be h. Sketch a diagram to represent the situation.
tan 28˚ =
h – 2 = 30 tan 28˚
h = (30 ´ 0.5317) + 2 ← tan 28˚ = 0.5317
h = 17.95
The height of the tree is approximately 18.0 m.