SOME APPLICATIONS OF TRIGONOMETRY-Notes
Maths - Notes
Height and Distance Concepts
- Angle of Elevation:
The angle of elevation is the angle formed between the horizontal line of sight and the line of sight drawn upward from an observer to an object located above eye level.
Angle of Elevation - The observer looks upward at an object.
- The line from the observer’s eyes to the object slopes upward.
- The angle made between this upward line of sight and the horizontal is called the angle of elevation.
- Angle of Depression:
The angle of depression is the angle formed between the horizontal line of sight and the line of sight drawn downward from an observer to an object located below eye level.
Angle of Depression - The observer looks downward at an object.
- A straight line drawn from the observer’s eyes to the object slopes downward.
- The angle between this downward-sloping line and the horizontal line at the observer’s eye level is called the angle of depression.
Important points:
- The angle of depression is always measured from the horizontal, never from the vertical.
- It commonly appears in problems involving balconies, towers, hills, airplanes, or elevated positions.
- Line of Sight:
Line of sight is the straight, imaginary line drawn from the observer’s eye to the object being viewed.
It represents the direct visual connection between the observer and the object.
Line of Sight - If the object is above the horizontal level, the line of sight rises and forms an angle of elevation.
- If the object is below the horizontal level, the line of sight lowers and forms an angle of depression.
Single Right-Triangle Applications
- Height of poles, towers, trees, or buildings
- Distance between two points on level ground
- Length of shadows at given times of day
Example-1
A tower stands vertically on the ground. From a point on the ground, which is 15 m away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower.
Solution:
Height of the tower AB \[\begin{aligned}\tan\ 60^\circ &= \dfrac{AB}{BC}\\\\ AB&=BC\times \tan\ 60^\circ\end{aligned}\] \[\color{cyan}\boxed{\tan\ 60^\circ=\sqrt{3}}\] \[\begin{aligned}&=15\times\sqrt{3}\\ &=15\times 1.732\\ &\approx 26\ m \end{aligned}\]Example-2
An electrician has to repair an electric fault on a pole of height 5 m. She needs to reach a point 1.3m below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of 60° to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take \(\sqrt3\) = 1.73)
Solution:
Height of the Pole=5 mHeight of pole where electrician need to reach = 5-1.3=3.7 m Distance of Foot of ladder from pole BC \[ \begin{aligned} \tan 60^\circ&=\dfrac{AB}{BC}\\\\ BC&=\dfrac{AB}{\tan 60^\circ}\\\\ BC&=\dfrac{3.7}{\sqrt3}\\\\ BC&=\dfrac{3.7}{1.73}\\\\ &\approx 2.14\ m \end{aligned} \] Distance of foot of the ladder from pole is 2.14 m away
Examle-3
An observer 1.5 m tall is 28.5 m away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45°. What is the height of the chimney?
Solution:
Height of the Chimney would be AB + Height of Observer = AB+1.5 \[ \begin{aligned} \tan\ 45^\circ&=\dfrac{AB}{BC}\\\\ AB&= BC\times \tan\ 45^\circ\\\\ AB&=28.5\ m\\ \end{aligned} \] Height of the Chimney =\[28.5+1.5=30\ m\]
From a point P on the ground the angle of elevation of the top of a 10 m tall building is 30°. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from P is 45°. Find the length of the flagstaff and the distance of the building from the point P. (You may take 3 = 1.732)
Solution:
Given: AB=10 m, \(\angle DPA=45^\circ) and \(\angle BPA=30^\circ)
$$\begin{aligned}\tan 30^{\circ }&=\dfrac{AB}{AP}\\\\ AP&=\dfrac{AB}{\tan 30^{\circ }}\\\\ &\color{cyan}\boxed{\tan\ 30^\circ=\frac{1}{\sqrt3}}\\\\ AP&=10\times \sqrt{3}\\\\ AP&=10\times 1.732\\ AP&=17\cdot 32\end{aligned}$$Let \(DB=x\)
$$\begin{aligned} AD&=10+x\\\\ \tan 45^{\circ }&=\dfrac{AD}{AP}\\\\ &\color{cyan}\boxed{\tan\ 45^\circ = 1}\\\\ 1&=\dfrac{AD}{AP}\\\\ AP&=AD\\ 17\cdot 32&=10+x\\ x&=17.32-10\\ x&=7.32\end{aligned}$$
Length of the flagstaff =7.32 m
Distance of the building from the point P=17.32 m
Examle-5
The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it is 60°. Find the height of the tower.
Solution:
Let height of the tower is \(h\)
and BC =\(x\)
DC = 40M
Substituting Value of \(x\) in Equation-(1)
$$\begin{align}h&=x\sqrt{3}\\ h&=20\sqrt{3}\\ &=20\times 1\cdot 732\\ &=34.64m\end{align}$$Height of the tower is 34.64 m.
Example-6
The angles of depression of the top and the bottom of an 8 m tall building from the top of a multi-storeyed building are 30° and 45°, respectively. Find the height of the multi storeyed building and the distance between the two buildings.
Solution:
Let multistoryed building is a \(x\ m\) higher than the other building
Hence, height of the multistoreyed building will be \(x+8\ m\)
\(QP\parallel CD\parallel AB\)
therefore,
Equating value of AB from Equation-(1) and Equation-(2)
$$\begin{align} x+8&=x\sqrt{3}\\\\ x-\sqrt{3}x&=-8\\\\ \sqrt{3}x-x&=8\\\\ x\left( \sqrt{3}-1\right) &=8\\\\ x&=\dfrac{8}{\sqrt{3}-1}\\\\ &=\dfrac{8}{\sqrt{3}-1}\times \dfrac{\sqrt{3}+1}{\sqrt{3}+1}\\\\ &=\dfrac{8\left( \sqrt{3}+1\right) }{3-1}\\\\ &=\dfrac{8\left( \sqrt{3}+1\right) }{2}\\\\ x&=4\left( \sqrt{3}+1\right) \end{align}$$Height of multi-storeyed builiding is \(PB=BC+CP\Rightarrow 8+x\)
$$\begin{align}PB&=x+8\\\\ &=4\sqrt{3}+4+8\\\\ &=12+4\sqrt{3}\\\\ &=4\left( 3+\sqrt{3}\right) \end{align}$$Height of multi-storeyed building is \(4\left( 3+\sqrt{3}\right)\ m\)
