Example 1 :
A 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 m
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.
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