In this section, we try to solve problems when Angles of depression are given.
Angle of Depression and Angle of Elevation are equal become they are alternative angles.

**Problems involving Angle of Depression**

In this section, we try to solve problems when Angles of
depression are given.

Angle of Depression and Angle of Elevation are equal become they
are alternative angles.

A player sitting on the top of a tower of height 20 m observes the
angle of epression of a ball lying on the ground as 60° . Find the distance between the foot of the
tower and the ball. (√3 = 1.732)

Let* **BC*** **be the height of the tower and

*BC *= 20* *m* *and* *∠*XCA *= 60° =* *∠*CAB*

Let *AB* = *x* metres.

In right triangle *ABC*,

tan 60° = *BC/AB*

Hence, the distance between the foot of the tower and the ball is
11.54 m.

**Example 6.27**

The horizontal distance between two buildings is 140** **m. The angle of** **depression of the top of
the first building when seen from the top of the second building is 30° . If the height of the first building is 60 m,
find the height of the second building. (√3 = 1.732)

*Solution*

The height of the first building *AB *=* *60* *m.
Now,* AB *=* MD *=* *60* *m Let the height of the second building
*CD *=* h*. Distance* BD *=* *140* *m

Now, *AM* = *BD* = 140 m

From the diagram,

∠*XCA *=* *30° =* *∠*CAM*

*CM *= 80.78

Now, *h* = *CD* = *CM* +*MD* = 80.78+60 =
140.78

Therefore the height of the second building is 140.78 m

**Example 6.28**

From the top of a tower** **50** **m** **high, the angles of depression of the top and** **bottom of a tree are
observed to be 30° and 45° respectively. Find the height of the tree. (√3
= 1.732)

*Solution*

The height of the tower* **AB*** **=

Let the height of the tree *CD* = *y* and *BD* = *x*

From the diagram, ∠*XAC* = 30° = ∠*ACM* and ∠*XAD* = 45° = ∠*ADB*

Therefore, height of the tree = *CD* = *MB* = *AB*
−*AM* = 50 −28. 85 = 21. 15 m

**Example 6.29**

As observed from the top of a** **60** **m** **high light house from the sea level,** **the angles of depression
of two ships are 28° and 45° . If one ship is exactly behind the other on
the same side of the lighthouse, find the distance between the two ships. (tan 28° =0.5317)

*Solution*

Let the observer on the** **lighthouse

Height of the lighthouse *CD* = 60 m

From the diagram,

∠*XDA *=* *28° =* *∠*DAC *and

∠*XDB *=* *45° =* *∠*DBC*

Distance between the two ships *AB* = *AC* −*BC* =
52.85 m

**Example 6.30**

A man is watching a boat speeding away from the top of a tower.
The boat**
**makes an angle of
depression of 60° with the man’s eye when at a distance of 200 m from the
tower. After 10 seconds, the angle of depression becomes 45°. What is the
approximate speed of the boat (in km / hr), assuming that it is sailing in
still water? (√3 = 1.732)

*Solution*

Let* **AB*** **be the tower.

Let *C* and *D* be the positions of the boat.

From the diagram,

∠*XAC *=* *60° =* *∠*ACB *and

∠*XAD *=* *45° =* *∠*ADB *,* BC *=* *200* *m

In right triangle *ABC*, tan 60° = *AB/BC*

gives √3 = AB /200

we get *AB* = 200√3 ...(1)

In right triangle *ABD*, tan 45° = *AB/BD*

Gives 1 = 200√3 / *BD* [by (1) ]

we get, BD = 200√3

Now, CD = BD − BC

CD = 200√3 − 200 = − 200(√3
– 1) = 146 4.

It is given that the distance *CD* is covered in 10 seconds.

That is, the distance of 146.4 m is covered in 10 seconds.

Therefore, speed of the boat = distance / time

= 146.4 / 10 = 14. 64 m/s
gives 14. 64 × (3600/1000) km/hr = 52. 704 km/hr

Tags : Solved Example Problems | Trigonometry | Mathematics Solved Example Problems | Trigonometry | Mathematics

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