Question
Question: From a point on the ground, the angle of elevation of the top of a tower is observed to be 60 degree...
From a point on the ground, the angle of elevation of the top of a tower is observed to be 60 degrees. From a point, 40m vertically above the first point of observation, the angle of elevation of the top of the tower is 30 degrees. Find the horizontal distance from the point of observation
A.−203m
B.203m
C.−203m
D.203m
Solution
Hint : In this question we are supposed to find the horizontal distance from the point of observation which is AP. We will consider the triangle AQP and using angle QAP, we will find the values of required sides of the triangle.
Complete step-by-step answer :
Let point A is 40m above from point B.
Given: angle QAP = 60 degree
Angle QBR = 30 degree
AB = 40m So, PR will also be 40m
First, we will take the triangle QAP.
In △QAP,
Take the angle QAP as tan60
tan60∘=APPQ
Value of tan 60 is 3. So, replacing it in the above equation.
3=APPR+QR
Value of PR is 40. So, replacing it in the above equation.
3=AP40+QR
3AP=40+QR------(1)
Now we will take triangle QBR.
In △QBR
Take angle QBR as tan 30.
tan30∘=BRQR
Value of tan30 is 31. And BR is equal to AP. So, replacing it in the above equation.
31=APQR
AP=3QR-----(2)
Putting this value of AP in equation (1)
3(3QR)=40+QR
3QR=40+QR
2QR=40
QR=20m
Now, AP=3QR
Value of QR is 20. So, putting it in the above equation.
AP=3×20
AP=203m
Value of horizontal distance from the point of observation is 203m.
So, option (D) is the correct answer.
So, the correct answer is “Option D”.
Note : The angle of elevation is the angle formed by the horizontal line of sight and the horizontal plane for an object above the horizontal.
Values of trigonometric angles are used to solve these kinds of problems.