SEE OPT Maths: Height And Distance

1. The shadow of a tower standing on a plan is found to be 60 feet longer when the sun's altitude is 30° than when it is 45°. What will be the height of the tower? Find it.
Solution: Let AB be the height of the towers and BC and BD are the length of shadows. Let ∡ACB = 45° and
∡ADB = 30° be the angle of elevation or heights of sun at two points. 


From the right angled ∆ABC, 
     tan θ = pb
⇒   tan45° = ABBC
⇒   1 = ABBC
⇒   AB = BC.

Again, from the right angled triangle ABD.
      tan θ = pb
⇒   tan30° = ABBD
⇒    1√3 = ABBC+60
⇒    1√3 = ABAB+60                     [ Since, AB = BC.]
⇒    AB + 60 = AB √3
⇒    AB(1 - √3) = -60
⇒    AB = 603- 1
∴   AB = 81.96m
Thus, the height of the tower is 81.96m.

2. The angles of elevations of the root of a house and window are 45° and 30° respectively. If the window is 18 feet below the roof of the house, find the height of the house.
Solution: Here, let, AC= xm be the height of roof of a house.
Let AB = 18 m B is the point
from the figure, BC = (x -18)m.

From right angled ∆BDC
      tan 30° = BCDC= x - 18DC
⇒    1√3 = x – 18DC
∴    DC = √(3 )(x-18).

Again, from right angled ∆ADC,
       tan 45° = ACDC = x3x-18
⇒    1 = x3x-18
⇒    √3x- 18√3 = x
⇒    x(√3 -1) = 18√3      
⇒    x  = (18√3)/(√3-1) 
          = 42.59 feet .
Thus, the height of house is 42.59 feet. 
 
3. The angles of elevations of the top of a tower observed from 27 m and 75 m away from its foot on the same side are found to be complementary. Find the height of the tower. 
Solution: Let AB be the height of a tower. Let ∡ACB and ∡ADB are two complementary angles. 

So, ∡ADB = 0 and ∡ACB = 90° - θ.

From the right angled triangle ABC
     tan (90° - θ) = p/b = AB/BC
⇒   cot θ = AB/27 ..........(i)
From the right angled ∆ABD
  cot θ = b/p = BD/AB
          = 75/AB ........... (ii)
Equating (i) and (ii) then,
AB/27= 75/AB
⇒   AB² =27 ⨉ 75
∴ AB =45m.
​​
Thus the height of the tower is 45 m.

4. A vertical pole AB is divided by a point C such that AC: CB = 2:1. The angle of elevation of C from a point on the ground is 30° . What is the angle of elevation of A from the same point? Find it where B is the foot of the pole.
Solution: Here, let AB be a vertical pole which is divided by C in the ratio of 2:1.
So, AC = 2x and BC = x. Let ∡CDB= 30° be an angle of elevation. 

We know that, tan θ = pb
   tan 30° = xBD
⇒    1√3 = xBD
∴   BD = √3x.

Again, from right angled ∆ADB,
     tan θ = pb = ABBD
⇒    tan θ = 3xBD
⇒    tan θ = 3x3x
⇒    tan θ=√3
⇒    tan θ=tan⁡60 °
∴ θ = 60°.
Thus, the value of θ( angle of elevation of A ) is 60°.

5. A tower and flagstaff on its top subtend angles of 30° and 15° respectively at a point 100 meter away from the foot of the tower, find the height of flagstaff.
Solution: Let AB be the height of flagstaff and BC be a tower.

Let ∡BDC and ∡ADB be the angles whose values are 30° and 15° respectively.

From right angled ∆BCD,
     tan θ = pb = BCCD
⇒    tan 30° = BC100
⇒    1√3 = BC100
∴   BC = 100√3 = 57.73m.

Again,
∡ADC = 30° + 15° = 45°
From the right angled ∆ACD,
      tan θ = pb = AC100
⇒    tan 45° = AC100
⇒   1 = AC100
∴    AC = 100m.

Finally, AB = AC-BC
                    = 100m - 57.73m
                    = 42.26m.
Thus, the measure ofheight of flagstaff  is 42.26 m.

6. From the roof and basement of a house 20 m high, the angles of elevation of the top of a tower are 45° and 60° respectively. Find the height of the tower.
Solution: Here,  
Let CD = 20 m be the height of a house and AB = x m be the height of the tower.

Let ∡ADE = 45° and ∡ACB = 60 ° are the angles of elevation of the top of tower from top and bottom of the house respectively.

From right angled ∆AED,
     tan 45° = AEDE
⇒   1 = AEDE
⇒    DE= AE
⇒    BC = (x - 20).

Again, from right angled ∆ABC,
      tan 60° = ABBC
⇒   √3 = xx-20
⇒    x√3-20√3 = x
⇒   x(√3-1)=20√3
∴ x = 20√3√3-1
       = 47.32m.
Thus, the height of the tower is 47.32 m.

7. The angle of elevation of the top of a tower from a point was observed to be 45° and on walking 60 metre away from that point it was found to be 30°. Find the height of the tower.
Solution: Here,
Let AB = x m be the height of a tower. 

Let ∡ADC = 30° and ∡ACB = 45° are the angles of elevation.
Let CD = 60 m be the distance between two points.
From the right angled ∆ABC,
      tan θ = pb
⇒   tan 45° = xBC
⇒   1 = xBC
∴   BC = x m.

Again, from right angled ∆ABD,
     tan 30° = pb = ABBD
⇒    1√3 = xCD+BC
⇒    √3 x = CD+BC
⇒   √3 x = 60+x
⇒   x(√3-1) = 60
∴    x = 60√3-1 = 81.96.
Thus, the height of the tower is 81.96 m.

8. From the top of mountain 21 meter high, the angles of depression of the top and the bottom of a tower are observed to be 45° and 60° respectively. Find the height of the tower.
Solution: Here, let AB= 21 m be the height of a mountain.

Let ∡FAD = 45° and ∡FAC = 60° be the angle of depressions.
Let CD = x m be the height of the tower.
∴ ∡FAD = ∡ADE = 45° and ∡FAC = ∡ACB = 60°.

From the right angled ∆ABC
      tan θ = pb
⇒    tan 60° = 21BC
⇒    √3 = 21BC
∴   BC = 213 = 12.12 m.
Again, from the right angled ∆AED,
     tan θ = pb
⇒    tan 45° =  AEED
⇒    1 =  21-xBC
⇒   1 = 21-x12.12
⇒   12.12 = 21-x
⇒    x = 21-12.12
 ∴   x = 8.88 m .
Thus, the height of the tower is 8.88 m.
 
9. The angles of elevation of the top o a tower as observed from the points at the distances of 144 meter and 121 meter from the foot of the tower are found to be complementary. Find the height of the tower.
Solution: Here,  
Let AB be a tower whose height is required to find. 

Let, ∡ACB = (90° - θ) and ∡ADB are the angle of elevations.
BC = 121 m and BD = 144 m
From the right angled ∆ABC,
      tan (90° - θ) = ABBC
⇒    cot θ = AB121........(i)
From right angled ∆BAD,
     tan θ = ABBD
              = AB144. 
∴ cot θ = 144AB........ (ii)              [ Since,  Cot = 1 / Tan.]
From (i) and (ii) then,
 AB121= 144AB
⇒    AB² = 144m×121m
∴   AB = 12×11m=132m.

Thus, the height of the tower is 132 m.

10. The angles of depression and elevation of the pinnacle of a temple 20 meter high are found to be 60° and 30° from the top and foot of a tower respectively. Find the height of the tower.
Solution: ​​​​​​Here,
Let AB the height of a tower CD =20 m be the height of temple.

Let, ∡EAD = 60° and ∡DBC =30° be the angle of depression and angle of elevation of the pinnacle of the temple from the top of the tower and bottom of the tower.

From the figure, ∡EAD = ∡ADF = 60°

From right angled triangle BCD,
tan θ = pb = CDBC
⇒   tan 30° =  20BC
⇒    1√3  = 20BC
∴ BC = 20√3 m.

From the right angled triangle AFD,
tan θ = pb = AFFD
⇒    tan 60° = AFBC
⇒    √3 = AF203
∴   AF = 60.

Now,
   AB = AF + BF = 60m + 20m
∴ AB = 80 m.
Thus, the height of tower is 80 m.