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Trigonometry Word Problems

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 Sub Topics Trigonometry has been used since ancient times to calculate distances and heights. Ancient astronomers have used it to calculate the distances of planets and stars from earth. Trigonometry can be used to calculate height of very tall buildings and mountains without actually measuring it. Do you know, that in 1952 ,the height of the tallest mountain- Mount Everest was calculated from a distance of 160Km applying trigonometry? Trigonometry word problems reflect the real life situations are extensively seen in various applications like  engineering, architecture and even in geology. Trigonometry word problems mainly deal with the height and distance calculations.  Trigonometry word problems may look intimidating at first look  .Let us see how to understand and solve trigonometric word problems.  The best way to solve trigonometric word problems is using diagrammatic representation of the problem. Once you have the diagram which is mostly a right angled triangle ,it becomes very easy to set up the trigonometric ratios and solve the problem.  Let us understand some of the terms widely used in trigonometry word problems.  1) Angle of Elevation:  The angle of elevation is always measured in upward direction. It is the angle between line of sight and the horizontal line.  In the accompanying diagram, $AC$ is the line of sight (movement of your eyes as you look up at the top of the tower). The $angle AB$ formed by the line of sight with the horizontal is the angle of elevation. The angle of elevation of the is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object. 2) Angle of Depression: The angle of depression is measured in downward direction. Suppose a boy sitting in the balcony of 5th floor is looking down at a car on the road . He has to look down to look at the car on the road.  The angle formed by the line of sight with the horizontal is the angle of depression. The horizontal line is parallel to the ground level. Thus, the angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level.

Word Problems

Example 1:

Miranda is standing $20$ ft away from the tree. If the angle of elevation is $60^{\circ}$ , what is the height of the tree?

Solution:

Let us first draw the diagram.  Its okay if the diagram is not perfect to scale.

We can use a simple line diagram for the same illustration:

Set up the ratio.

Here,in triangle $ABC,\ AC$ is  the horizontal distance between Miranda and the object.

$BC$ represents the tree. $AB$ is the line of sight .Thus $\angle ACB$  is the angle of elevation.

We need to solve for the height of the tree i.e. $BC$ = $h$.

Use the trigonometric ratio involving the sides $AC, \ BC$ and the  $\angle A$.

$tan\ A$ = $\frac{BC}{AC}$

$tan\ 60^{\circ}$ = $\frac{h}{20}$

$h$ = $20 \times tan\ 60^0$

$h$ = $20 \times \sqrt 3$

$h$ = $34.64$

The height of the tree is $34.64$ ft (nearest to hundredth place).
Example 2 :

An observer $1.5$ m tall is $26$ m away from a chimney. The angle of elevation of the top of the chimney from her eyes is $40^{\circ}$. What is the height of the chimney?

Solution:

Draw the diagram.

Let  $CD$ be the Chimney and $AE$ be the observer.

$AC$  be the distance between the observer and the chimney.

In $\Delta ABC,\ BC$ = $h - 1.5m$

Use $tan\ A$ = $\frac{BC}{AB}$

$tan 40^{\circ}$ = $\frac{((h-1.5))}{26}$

So,  $(h - 1.5)$ = $26\ tan\ 40^o$

$(h - 1.5)$ =  $26 \times 0.84$

$h-1.5$ = $21.8$

$h$ = $21.8 + 1.5$ = $23.3$ m

The observer is $23$.5 m away from the Chimney.
Example 3:

A lighthouse operator at point $P\ ,21$ m above the sea level sights a sailboat at point $R$. The angle of depression of sighting is $15^{\circ}$ . How far is the sailboat from the base of the lighthouse? Give your answer to nearest meter.

Solution:

Let  $P$ be the position of top of the lighthouse operator and $R$ be the position of the sailboat.

Here, the $\angle RPA$ and $\angle PRQ$ are alternate angles. So $\angle PRA$ = $15^{\circ}$

In $\Delta PQR,\ tan\ R$ = $\frac{PQ}{QR}$

$tan15^{\circ}$ = $\frac{21}{x}$

$x$ = $\frac{21}{tan 15^o}$

$x$ = $78.37$ m = $78$ m rounded to nearest meter.
Example 4:

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$, respectively. If the bridge is at a height of $10$ ft  from the banks, find the width of the river .

Solution:

Check the diagram. Let $A$ and $B$ be the points on the opposite sides of banks of the river.

Let $C$ be the point on the bridge at height of $10$ ft. We have to determine the width of the river, $AB$

$AB$ = $BD + DA$

Consider two triangle $\Delta BDC$ and  $\Delta ADC$  to solve for $BD$ and $AD$.

$tan\ 30^{\circ}$ = $\frac{CD}{BD}$

$BD$  = $\frac{CD}{ tan 30^{\circ}}$

$BD$ = $\frac{10}{(\frac{1}{\sqrt 3})}$

$BD$ = $10 \times \sqrt 3$  =  $17.32$ ft

Now ,consider  $\Delta ADC$,

$tan\ 45^{\circ}$ = $\frac{CD}{AD}$

$AD$ = $\frac{CD}{tan 45^o}$

$AD$ = $\frac{10}{1}$ = $10$ ft.

So, the width of the river = $AB$ = $BD + AD$

$AB$ = $17.32 + 10$ = $27.32$ ft .

Width of the river is $27.32$ ft.