Angles of Elevation and Depression: When a person stands on the ground and looks up at an object, it creates an elevation between the line of sight to the object and the horizontal line of sight. The angle formed between the horizontal line of sight and the line of sight to the object is known as the Angle of Elevation.

Similarly, when an observer stands at a height and looks down at an object, it creates depression between the line of sight to the object and the horizontal line of sight. Thus, the Angle of Depression is formed between the horizontal line of sight and the line of sight to the object. Check out the article to know more about Angles of Elevation and Depression.

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Define Angles of Elevation and Depression

Angle of Elevation: The angle of elevation is an angle formed between the horizontal line and the line of sight. If the line of sight is skywards from the horizontal line, then the angle formed is an elevation angle.

In the above figure, an observer looks at the object, standing on the ground, forming an angle of elevation with the line of sight and horizontal line. Here, if we join an imaginary line between the object and the horizontal line, a right-angled triangle is formed.

Thus we can use here trigonometry concept to find the distance of the observer from the tower or building. The tower’s height or structure or the length at which the object is kept will be considered perpendicular. The horizontal line will regard as an adjacent side of the triangle formed.

Angle of Depression: The angle which the line of sight makes with the horizontal line passing through the observer’s eye when the object is below the observer is called the angle of depression.

An observer looks at the object standing at the top of a tower or tree or something in the above figure, forming a depression angle with the line of sight and horizontal line. Here, if we join an imaginary line between the object and the horizontal line, a right-angled triangle is formed. Thus we can use here trigonometry concept to find the distance of the observer from the tower or building.

The angle of depression is just the opposite scenario of the angle of elevation. In this case, the observer is standing at some height, and the object is kept below the observer’s horizontal line of sight. Therefore, we can define it as if an object/ thing is kept below the observer’s eye level; the angle formed between the horizontal line and the observer’s line of sight is called the angle of depression.

Angles of Elevation and Depression Examples

We have several examples of both angles of elevation and depression. Some of them are given below:
The two poles are placed apart at some distance. The length of the shorter pole and the angle of depression from the long pole is given. We can sketch a diagram to represent the situation as below.

1. Draw two vertical lines to represent the shorter pole and the long pole.

2. Draw a line from the top of the long pole to the top of the shorter pole. (This is the line of sight)

3. Draw a horizontal line to the top of the long pole and mark the angle of depression.

Now, using the trigonometric concepts and ratios, we can find the unknown parameters in this situation.

The concept can be used to find the approximate heights and distances of objects like the height of a mountain, the height of a tall building, the height at which an aeroplane is flying, and a boat’s position from a lighthouse.

Trigonometric Ratios of Some Specific Angles of \(\sin ,\cos \) & \(\tan \)

	\({0^{\rm{o}}}\)	\({30^{\rm{o}}}\)	\({45^{\rm{o}}}\)	\({60^{\rm{o}}}\)	\({90^{\rm{o}}}\)
\(\sin \)	\(0\)	\(\frac{1}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{{\sqrt 3 }}{2}\)	\(1\)
\(\cos \)	\(1\)	\(\frac{{\sqrt 3 }}{2}\)	\(\frac{1}{{\sqrt 2 }}\)	\(\frac{1}{2}\)	\(0\)
\(\tan \)	\(0\)	\(\frac{1}{{\sqrt 3 }}\)	\(1\)	\[\sqrt 3 \]	\(\infty \)

Angles of Elevation and Depression Facts

We can use the angles of elevation and depression in measuring heights and distances in trigonometric applications using right triangles. When we look up or down to view objects, we can find these angles. Devices are available to measure angles of elevation and depression. We can use these measured angles in measuring heights and distance, which are either tedious or impractical to measure, by modelling the situation into right triangles.

An angle of elevation of one location relative to the other is always congruent (equal in measure) to the angle of depression of the first location close to the second.

Solved Examples on Angles of Elevation and Depression

Let us understand the concept through Angles of elevation and depression practices problem.

Example.1. An angle of elevation of a tower top from a point on the ground, \(30\,{\rm{m}}\) away from the foot of the tower, is \({30^{\rm{o}}}.\) Find the height of the tower.
Ans: Let \(PQ\) be the height of the tower and \(R\) be a point on the ground such that the angle of elevation of the top \(P\) of tower \(PQ\) is \({30^{\rm{o}}}.\) In \(\Delta PQR,\,\angle R = {30^{\rm{o}}}\) and base \(QR = 30\,{\rm{m}}\) and we have to find the measure of the perpendicular \(PQ.\) So, we use trigonometrical ratios, which contain base and perpendicular.

In \(\Delta PQR,\,\tan \,R = \frac{{PQ}}{{QR}}\)
\( \Rightarrow \tan \,{30^{\rm{o}}} = \frac{{PQ}}{{30}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{PQ}}{{30}}\)
\( \Rightarrow PQ = \frac{{30}}{{\sqrt 3 }}\)
\( \Rightarrow PQ = 10\sqrt 3 \,{\rm{m}}\)
Therefore, the height of the tower is \(10\sqrt 3 \,{\rm{m}}{\rm{.}}\)

Example.2. A kite is flying at the height of \(60\,{\rm{m}}\) above the ground. The string attached to the kite is temporarily fixed to a point on the ground. The inclination of the string with the ground is \({60^{\rm{o}}}.\) Find the length of the string, assuming there is no slack in the string.
Ans: Let \(A\) be the kite and \(CA\) be the string attached to the kite such that its one end is tied to a point \(C\) on the ground. The inclination of the string \(CA\) with the ground is \({60^{\rm{o}}}.\) In \(\Delta ABC,\,\angle C = {60^{\rm{o}}}\) and perpendicular \(AB = 60\,{\rm{m}},\) and we have to find hypotenuse \(AC.\) So, we use the trigonometric ratio involving perpendicular and hypotenuse.

In \(\Delta ABC,\,\sin \,C = \frac{{AB}}{{AC}}\)
\( \Rightarrow \sin \,{60^{\rm{o}}} = \frac{{AB}}{{AC}}\)
\( \Rightarrow \frac{{\sqrt 3 }}{2} = \frac{{60}}{{AC}}\)
\( \Rightarrow AC = \frac{{120}}{{\sqrt 3 }}\)
\( \Rightarrow AC = 40\sqrt 3 \,{\rm{m}}\)
Therefore, the length of the string is \(40\sqrt 3 \,{\rm{m}}{\rm{.}}\)

Example.3. A circus artist is climbing a \(20\,{\rm{m}}\) long rope, tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the angle made by the rope with the ground level is \({30^{\rm{o}}}.\)
Ans: Let \(PQ\) be the vertical pole and \(PR\) be the \(20\,{\rm{m}}\) long rope such that one end is tied from the top of the vertical pole \(PQ\) and the other end is tied to a point \(R\) on the ground.

In \(\Delta PQR,\,\sin \,{30^{\rm{o}}} = \frac{{PQ}}{{PR}}\)
\( \Rightarrow \frac{1}{2} = \frac{{PQ}}{{20}}\)
\( \Rightarrow PQ = 10\,{\rm{m}}\)
Therefore, the height of the pole is \(10\,{\rm{m}}{\rm{.}}\)

Example.4. From the top of a hill, the angles of depression of two consecutive kilometer stones due east are found to be \({30^{\rm{o}}}\) and \({45^{\rm{o}}}.\) Find the height of the hill. \(\left( {\sqrt 3 = 1.732} \right)\)
Ans: Let \(AB\) be the height of the hill. Let \(C\) and \(D\) be the two stones due east of the hill at a distance of \(1\,{\rm{km}}\) from each other such that angles of depression of \(C\) and \(D\) be \({45^{\rm{o}}}\) and \({30^{\rm{o}}},\) respectively. Let \(AC = x\,{\rm{km}}.\)

In \(\Delta CAB,\) we have
\(\tan \,{45^{\rm{o}}} = \frac{{AB}}{{AC}}\)
\( \Rightarrow 1 = \frac{{AB}}{x}\)
\( \Rightarrow AB = x\)…..(i)
In \(\Delta DAB,\) we have
\(\tan \,{30^{\rm{o}}} = \frac{{AB}}{{AD}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{AB}}{{x + 1}}\)
\( \Rightarrow \sqrt 3 \,AB = x + 1\)…(ii)

Substituting the value of \(x\) from equation (i) in equation (ii), we get
\(\sqrt 3 \,AB = AB + 1\)
\(\sqrt 3 \,AB = AB + 1\)
\(AB = \frac{1}{{\sqrt 3 – 1}}\)

Rationalising the denominator, we get
\(AB = \frac{1}{{\sqrt 3 – 1}} \times \frac{{\sqrt 3 + 1}}{{\sqrt 3 + 1}}\)
\( \Rightarrow AB = \frac{{\sqrt 3 + 1}}{{\left( {\sqrt 3 – 1} \right)\left( {\sqrt 3 + 1} \right)}}\)
\( \Rightarrow AB = \frac{{\sqrt 3 + 1}}{2}\)
\( \Rightarrow AB = \frac{{2.73}}{2}\)
\( \Rightarrow AB = 1.365\,{\rm{km}}\)
Therefore, the height of the hill is \(1.365\,{\rm{km}}{\rm{.}}\)

Example.5. An aeroplane at an altitude of \(1200\,{\rm{m}}\) finds that two ships are sailing towards it in the same direction. The angles of depression of the ships as observed from the aeroplane are \({60^{\rm{o}}}\) and \({30^{\rm{o}}}\) respectively. Find the distance between the two ships. \(\left( {\sqrt 3 = 1.732} \right)\)
Ans: Let the aeroplane be at \(B\) and let the two ships be at \(C\) and \(D,\) such that their angles of depression from \(B\) are \({30^{\rm{o}}}\) and \({60^{\rm{o}}},\) respectively.

We have, \(AB = 1200\,{\rm{m}}.\) Let \(AC = x\) and \(CD = y\)
In \(\Delta CAB,\) we have \(\tan \,{60^{\rm{o}}} = \frac{{AB}}{{CA}}\)
\( \Rightarrow \sqrt 3 = \frac{{1200}}{x}\)
\( \Rightarrow x = \frac{{1200}}{{\sqrt 3 }} = 400\sqrt 3 \)
In \(\Delta BAD,\) we have
\(\tan \,{30^{\rm{o}}} = \frac{{AB}}{{AD}}\)
\( \Rightarrow \frac{1}{{\sqrt 3 }} = \frac{{1200}}{{x + y}}\)
\( \Rightarrow x + y = 1200\sqrt 3 \)
\( \Rightarrow y = 1200\sqrt 3 – x\)
\( \Rightarrow y = 1200\sqrt 3 – 400\sqrt 3 \)
\( \Rightarrow y = 800\sqrt 3 \)
\( \Rightarrow y = 800 \times 1.732\)
\( \Rightarrow y = 1385.6\,{\rm{m}}\)
Hence the distance between the two ships is \(1385.6\,{\rm{m}}{\rm{.}}\)

Summary

In the above article, we learned the definition of angle of elevation and angle of depression and their applications in real life. We solved some examples on the angle of elevation and depression to understand the concept pragmatically.

FAQs on Angles of Elevation and Depression

Check some commonly asked questions about Angles of Elevation and Depression below:

Q.1. How do you find the angle of elevation and depression?
Ans: The elevation and depression angles are used in finding distances, heights of buildings or towers, etc. We can find the angles with the help of trigonometric ratios, such as sine, cosine, and tangent.

Q.2. Explain angles of elevation and depression with an example?
Ans: The angle which the line of sight makes with the horizontal line passing through the observer’s eye when the object is above the observer is called the angle of elevation.
Example: A kid watching the kite flying from the ground forms an angle of elevation.
The term angle of depression represents the angle from the horizontal downward to an object. Thus, an observer’s line of view would be below the horizontal.
Example: A man from the top of a tower watching an object on the ground forms the angle of depression.

Q.3. What is the formula for the angle of depression?
Ans: There is no particular formula to calculate the angle of depression. It is entirely based on the measurements which are given in a specific question. Based on these measurements, we will find the angle of depression using trigonometric ratios, such as sine, cosine, and tangent.

Q.4. Why is the angle of elevation and depression the same?
Ans: The angle of depression is the same as the angle of elevation. Since the horizontal lines are parallel, alternate interior angles are created by a transversal (the line of sight) cutting across two parallel lines (the horizontals).

Q.5. Is the angle of elevation equal to the angle of depression?
Ans: An angle of elevation of one location relative to the other is always congruent (equal in measure) to the angle of depression of the first location close to the second.

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