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

**Problems involving Angle of Elevation and
Depression**

Let us consider the following situation.

A man standing at a top of lighthouse located in a beach watch on
aeroplane flying above the sea. At the same instant he watch a ship sailing in
the sea. The angle with which he watch the plane correspond to angle of
elevation and the angle of watching the ship corresponding to angle of
depression. This is one example were one oberseves both angle of elevation and
angle of depression.

In the Fig.6.26, *x*° is the angle of elevation and *y*°
is the angle of depression.

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

From the top of a** **12** **m** **high building, the angle of elevation of the top
of** **a cable tower is 60° and
the angle of depression of its foot is 30°. Determine the height of the tower.

As shown in Fig.6.27,* **OA*** **is the building,

*PP*’ is the cable tower
with *P* as the top and *P* ' as the bottom.

Then the angle of elevation of *P*, ∠*MOP* = 60°

And the angle of depression of *P*’ , ∠*MOP*^{′} = 30°.

Suppose, height of the cable tower *PP* ' = *h* metres.

Through *O*, draw *OM* ┴ *PP* '

*MP *=* PP*′* *−* MP*′* *=* h *−* OA *=*
h *−12

Hence, the required height of the cable tower is 48 m.

A pole 5** **m** **high is fixed on the top of a tower. The angle of elevation of** **the top of the pole
observed from a point ‘*A*’ on the ground is 60° and the angle of
depression to the point ‘*A*’ from the top of the tower is 45°. Find the
height of the tower. (√3=1.732)

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

Let ‘*A*’ be the point of observation.

Let *BC* = *x* and *AB* = *y*.

From the diagram,

∠*BAD *=* *60°* *and ∠*XCA *=* *45° =* *∠*BAC*

Hence, height of the tower is 6.83 m.

**Example 6.33**

From a window (*h*** **metres high above the ground) of a house in a
street,**
**the angles of elevation
and depression of the top and the foot of another house on the opposite
side of the street are *θ*_{1} and *θ*_{2}
respectively. Show that the height of the opposite house is .

*Solution*

Let* **W*** **be the point on the window where the angles of elevation and
depression

Then *WA* is the width of the street.

Height of the window = *h* metres

=
AQ (WR = AQ)

Let *PA* = *x* metres.

Therefore,
height of the opposite house = *PA*+*AQ* = *x*
+* h* =

Hence Proved.

Tags : Solved Example Problems | Trigonometry | Mathematics , 10th Mathematics : UNIT 6 : Trigonometry

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10th Mathematics : UNIT 6 : Trigonometry : Problems involving Angle of Elevation and Depression | Solved Example Problems | Trigonometry | Mathematics

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