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# Trigonometry Problems

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 1 Introduction 2 Trig Values Chart 3 Trigonometry Practice Problems 4 Trigonometry Word problems 5 Trigonometry worksheets 6 Summary

24nd September 2020

## INTRODUCTION

Trigonometry is the branch of mathematics that deals with certain functions and ratios called trigonometric ratios or functions of angles and their application in various real-life areas. There are 6 trig functions of angles that are used majorly. They have been named and accordingly abbreviated as sine function (sin x), cosine function (cos x), tangent function (tan x), cotangent function (cot x), secant function (sec x), and cosecant function (csc or cosec x).

Trigonometry has a unique attribute of having multiple real-life applications across various fields, considering it involves the measurement of different situational lengths and angles. Surveyors may use it during the surveying of a property, cartography, and so on, and it has long been used in astronomy and navigation to name a few applications.

So, learning how to solve problems right from the basic level to more real-life complex applications is important. And that will be covered entirely with this blog.

## Trig Values Chart:

The trigonometric formula given below will help you to find the values of trigonometric ratios for standard angles 0°, 30°, 45°, 60°, 90°, 180°, 270°, and 360. If you memorize these ratios the calculations will be very easy for you to solve the trigonometry questions in exams. That is why it becomes very important for you to remember these values.

 $$\theta$$ $$0^{\circ}$$ $$30^{\circ}$$ $$45^{\circ}$$ $$60^{\circ}$$ $$90^{\circ}$$ $$180^{\circ}$$ $$270^{\circ}$$ $$360^{\circ}$$ $$\text{sin } \theta$$ $$0$$ $$\frac{1}{2}$$ $$\frac{1}{\sqrt2}$$ $$\frac{\sqrt3}{2}$$ $$1$$ $$0$$ $$-1$$ $$0$$ $$\text{cos } \theta$$ $$1$$ $$\frac{\sqrt3}{2}$$ $$\frac{1}{\sqrt2}$$ $$\frac{1}{2}$$ $$0$$ $$-1$$ $$0$$ $$1$$ $$\text{tan } \theta$$ $$0$$ $$\frac{1}{\sqrt3}$$ $$1$$ $$\sqrt{3}$$ Not defined $$0$$ Not defined $$0$$ $$\text{cosec } \theta$$ Not defined $$2$$ $$\sqrt{2}$$ $$\frac{2}{\sqrt3}$$ $$1$$ Not defined $$-1$$ Not defined $$\text{sec } \theta$$ $$1$$ $$\frac{1}{2}$$ $$\sqrt{2}$$ $$2$$ Not defined $$-1$$ Not defined $$1$$ $$\text{cot } \theta$$ Not defined $$\sqrt{3}$$ $$1$$ $$\frac{1}{\sqrt3}$$ $$0$$ Not defined $$0$$ Not defined

The trig table is made up of the following of trigonometric ratios that are interrelated to each other – sin, cos, tan, cos, sec, cot.

• sin (reciprocal of cosecant)  = opposite /hypotenuse
• cos (reciprocal of secant)  = adjacent  / hypotenuse
• tan (reciprocal of cotangent)  = opposite  /adjacent
• cot (reciprocal of tangent)  = adjacent  / opposite
• cosec (reciprocal of sine)  = hypotenuse  / opposite
• sec (reciprocal of cosine)  = hypotenuse  / adjacent

The calculations can easily be figured out by memorizing a table of functions most commonly known as the Trigonometric Table. This finds use in several areas. Some of them include navigation video games, aviation, science, geography, engineering, geometry, etc. The trigonometric table helped in many developments and in the field of Mechanical Engineering for first innovation.

These values of standard trigonometric angles hold increased precedence as compared to others as the most important problems employ these ratios. It is therefore very important to know and remember the ratios of these standard angles. Remembering the trigonometry table will be useful as it finds many applications, and there are many methods to remember the table. Knowing the Trigonometry formulas, ratios and identities automatically will lead to figuring out the table and the values. The Trigonometric ratio table is dependent upon the trigonometry formulas in the same way all the functions of trigonometry are interlinked with each other.

Before attempting to begin, it is better to try and remember these values, and know the following trigonometric  ratios of complementary angles and their Reciprocal relation:

Trigonometric ratios of complementary angles.

Reciprocal relations of Trigonometric Ratios
sin x = cos (90∘−x) 1 / sin x = cosec x
cos x = sin (90∘−x) 1 / cos x= sec x
tan x = cot (90∘−x) 1 / sec x= cos x
cot x = tan (90∘−x) 1 / tan x= cot x
sec x = cot (90∘−x) 1 / cot x= tan x
cot x = sec (90∘−x) 1/ cosec x = sin x

## Trigonometry Practice Problems:

These trigonometry problems provided will be very helpful in clearing all concepts and learning step by step procedure of solving all types of trigonometry problems.

1. How to solve trigonometry problems or questions?

Step 1: Check for a diagram, if there is one missing try to draw it yourself.

Step 2: Mark the right angles in the diagram.

Step 3: Label and write all the measurements of the other angles and the lengths of any lines that are given or can be inferred from the trigonometry problem.

Step 4: Mark out the elements that are to be found which could be angles or sides you have to calculate.

Step 5: Check if you have a right triangle to work with, else make a decision of whether you need to somehow create the right triangles by extra development.

For example, dividing an isosceles triangle into two congruent right triangles.

Step 6: Decide which concept is first applicable which could be the Pythagorean theorem, sine function, cosine function, or tangent function.

For example, it is known that hypotenuse always has to be the longest, so if that condition is not met you need to check the calculation or method again.

2. How can Cosine function be used to calculate the side of a right triangle?

 Example

Calculate the value of cos theta in the following triangle.

Solution:

Use the Pythagorean theorem to evaluate the length of PR.

3. How to use Tangent to calculate the side of a triangle?

 Example

Calculate the length of the side x, given that tan θ = 0.4

Solution:

4. How to use Sine to calculate the side of a triangle?

 Example

Calculate the length of the side x, given that sin θ = 0.6

Solution:

Using the Pythagorean theorem:

## Trigonometry Word Problems

A practical application of the trigonometric functions is to find the measure of lengths that you cannot measure. One of the most common types of trigonometry word problems is on the angle of elevation and depression.

Angle of Elevation

The angle of depression is the angle formed by a horizontal line and the line of sight down to an object when the image of an object is located beneath the horizontal line.

Angle of Depression

The angle of elevation is the angle formed by a horizontal line and the line of sight up to an object when the image of an object is located above the horizontal line.

### → the angle of elevation = the angle of depression ←

Solving trigonometry word problems on the Angle of depression and angle of elevation.

### How to find angle of elevation? How to find the angle of depression?

While one has to solve a trigonometry problem with an angle of depression which generally has be used for questions of “how to find the height of a triangle” type. you need to find the measure of an angle INSIDE the triangle. There are two options:

Option 1: find the angle inside the triangle that is adjacent (next door) to the angle of depression. This adjacent angle will always be the complement of the angle of depression since the horizontal line and the vertical line are perpendicular (90º). In the diagram at the left, the adjacent angle is 52º.

Option 2:You can use the equation:

angle of depression = angle of elevation and thus label ∠BAC as 38º inside the triangle.

Notice that both options, the answer is the same.

Let's see how to put these skills to work in trig word problems.

How to find the height of a triangle?

 Example 1

A nursery plants a new tree and attaches a guy wire to help support the tree while its roots take hold. An 8-foot wire is attached to the tree and to a stake in the ground. From the stake in the ground, the angle of elevation of the connection with the tree is 42º. Find to the nearest tenth of a foot, the height of the connection point on the tree.

Solution:

• The "guy" wire is used to support a newly planted tree to hold in place and prevent it from bending or falling down during high winds.
• The "angle of elevation" as we have learned earlier, will be taken the ground up.
• Assume that the tree standing vertical is perpendicular to the ground.
• Hence the problem deals with "opposite" and "hypotenuse" making it a sine problem.

 Example 2

From the top of a fire tower, a forest ranger sees his partner on the ground at an angle of depression of 40º. If the height of the tower is 45 feet, how far is his partner from the tower base to the nearest tenth of a foot?

Solution:

•  Always the "angle of depression" is from a horizontal line of sight downward.
• Assume that the vertical standing tower is perpendicular to the ground.
• This solution will use alternate interior angles from the parallel horizontal lines, so place 40º inside the triangle by the partner (bottom right).
•  This solution deals with "opposite" and "adjacent" making it a tangent problem.

 Example 3

Find the shadow cast by a lamppost of height 10 feet when the angle of elevation of the sun is 58º.

Solution:

• Remember our discussion previously on how the "angle of elevation" is taken upward from the horizontal ground line.
• It is assumed that the lamp post is vertical, making it perpendicular to the ground.
• Shadows are on the ground!  If you place the "shadow" on the hypotenuse you have created an apparition ( a "ghost"), not a shadow!
• This solution deals with "opposite" and "adjacent" making it a tangent problem.

There is a possibility that trigonometry word problems won't use the terms “angle of elevation” and “angle of depression” directly. You may need to pick up the language or the diagram to infer it from the trigonometry word problem. Let us look at such an example.

 Example 4

A ladder leans against a brick wall. The foot of the ladder is at a distance of 6 feet from the wall. The ladder reaches a height of 15 feet on the wall. Find to the nearest degree, the angle the ladder makes with the wall.

 Example 5

A radio station tower was built in two sections. From a point, 87 feet from the base of the tower, the angle of elevation of the top of the first section is 25º, and the angle of elevation of the top of the second section is 40º. To the nearest foot, what is the height of the top section of the tower?

Solution:

•Let us work out this trigonometry word problem using two different triangles,

(1) the larger triangle with the 40º angle and a vertical side that represents the ENTIRE height, b, of the tower, and

(2) the smaller triangle with the 25º angle and a vertical side, a, that represents the height of the first (bottom) section of the tower.

• Solve and find values of a and b that are the vertical heights in the two different triangles.

• x is our required height which is of the second (top) section of the tower and will come out to the difference between b and a that is the ENTIRE height and the height of the first (bottom) section respectively. Subtracting will give us the answer.

• Observing the calculations done in both the triangles, this trigonometry problem deals with the tangent function since we have to work with the "opposite" and "adjacent".

• Larger triangle with height b:

• Smaller triangle with height a:

• Subtracting a from b will give us,

\begin{align}(b - a) = 73.00166791 - 40.56876626 = 32.43290165 \approx 32\,\,\rm{feet\,approx}\end{align}

 Example 6

The Washington Monument

An inquisitive math student is standing 25 feet from the base of the Washington Monument. The angle of elevation from her horizontal line of sight is 87.4. If her “eye height” is 5ft, how tall is the monument?

Solution:

Upon observing what is provided in the question we can come to the decision that using a tangent function will give us the required answer for the height of the monument. Remember to add the eye height of the student.

\begin{align}&\tan {87.4^ \circ } = h \div 25\\&h = 25.\tan {87.4^ \circ } = 550.54\end{align}

Adding 5 ft of the eye level, the total net height of the Washington Monument comes out to be 555.54 ft.

 Example 7

Solving for an Angle Measurement

A 25-foot tall flagpole casts a 42-foot shadow. What is the angle that the sun hits the flagpole?

The angle that the sun hits the flagpole is x. We need to use the inverse tangent ratio.

\begin{align}&\tan x = 42 \div 25\\&{\tan ^{ - 1}}42 \div 25 \approx 59.2 = x\end{align}

 Example 8

Solving for the Angle of Depression

Elise is standing on top of a 50-foot building and sees her friend, Molly. If Molly is 30 feet away from the base of the building, what is the angle of depression from Elise to Molly? Elise’s eye height is 4.5 feet.
Because of parallel lines, the angle of depression is equal to the angle at Molly, or x. We can use the inverse tangent ratio.

\begin{align}{\tan ^{ - 1}}(54.5 \div 30) = {61.2^ \circ } = x\end{align}

 Example 9

Ramp Problem Revisited

To find the horizontal length and the actual length of the ramp, we need to use the tangent and sine.

\begin{align}&\tan {5^ \circ } = 2 \div x\\&x = 2 \div \tan {5^ \circ } = 22.86\end{align}
\begin{align}&\sin {5^ \circ } = 2 \div y\\&y = 2 \div \sin {5^ \circ } = 22.95\end{align}

## Trigonometry worksheets

There are various Trigonometry worksheets provided by cuemath which can be availed on the following link. https://www.cuemath.com/trigonometry/

## Summary

Trigonometry is very useful in determining the unknown side of a specific triangle. The concept of the right triangle in trigonometry is very well used in determining the height of a building. Trigonometry doesn’t end in passing the Trigonometry subject and we will use it always in our engineering careers

We realized that trigonometry can be used in many areas such as astronomy and architecture they can aid in calculating many things and they can also be used in car desks and benches. Without really climbing a tree, you can find the height of a tree easily with trigonometry. They can be so widely used in real-life applications and are very useful for most architects and astronomers.

We can conclude that without trigonometry, life would be much more difficult. Without going through the troubles, you can easily find something so we think that it was a good invention by Archimedes, and thanks to this many architects need not go through the trouble to calculate things, so it really helps in real-life applications and not only in our tests and exams.

Written by Asha M.

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