Right Triangle Trigonometry

Learning Objectives

After completing this section, you should be able to:

1. Apply the Pythagorean Theorem to find the missing sides of a right triangle.
2. Apply the \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) and \({45}^{∘}\text{-}{45}^{∘}\text{-}{90}^{∘}\) right triangle relationships to find the missing sides of a triangle.
3. Apply trigonometric ratios to find missing parts of a right triangle.
4. Solve application problems involving trigonometric ratios.

Pythagorean Theorem

The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle).

For example, given that side \(a=6,\) and side \(b=8,\) we can find the measure of side \(c\) using the Pythagorean Theorem. Thus, \[\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {(6)}^{2}+{(8)}^{2} & = & {c}^{2} \\ 36+64 & = & {c}^{2} \\ 100 & = & {c}^{2} \\ \sqrt{100} & = & \sqrt{{c}^{2}} \\ 10 & = & c\end{array}\]

Using the Pythagorean Theorem

Try it.

Find the length of the missing side of the triangle ().

Solution

Using the Pythagorean Theorem, we have \[\begin{array}{lll}{(6)}^{2}+{b}^{2} & = & {(14)}^{2} \\ 36+{b}^{2} & = & 196 \\ {b}^{2} & = & 196-36 \\ {b}^{2} & = & 160 \\ b & = & \pm \sqrt{160} \\ & = & 4\sqrt{10}=12.65\end{array}\]

When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.

Distance

The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this:

In , the problem is to find the distance between the points \((-3,-1)\) and \((3,2).\) We call the length from point \((-3,-1)\) to point \((3,-1)\) side \(a\), and the length from point \((3,-1)\) to point \((3,2)\) side \(b\). To find side \(c\), we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is \(d=\sqrt{{({x}_{2}-{x}_{1})}^{2}+{({y}_{2}-{y}_{1})}^{2}},\) such that \(({x}_{2}-{x}_{1})\) is a substitute for \(a\) in the Pythagorean Theorem and is equal to \(3-(-3)=6;\) and \(({y}_{2}-{y}_{1})\) is a substitute for \(b\) in the Pythagorean Theorem and is equal to \(2-(-1)=3.\) When we plug in these numbers to the distance formula, we have \[\begin{array}{lll}d & = & \sqrt{{(3-(-3))}^{2}+{(2-(-1))}^{2}} \\ & = & \sqrt{{(6)}^{2}+{(3)}^{2}}=\sqrt{36+9} \\ & = & \sqrt{45}=3\sqrt{5}=6.7\end{array}\]

Thus, \(d=c\), the hypotenuse, in the Pythagorean Theorem.

Calculating Distance Using the Distance Formula

Try it.

You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive (). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.

Solution

You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then \(9(200)+7(200)=1,800\ \text{ft}+1,400\ \text{ft}=3,200\ \text{ft}\).

As the crow flies, use the distance formula. We have \[\begin{array}{lll}d & = & \sqrt{{(1,800-0)}^{2}+{(1,400-0)}^{2}} \\ & = & \sqrt{3,240,000+1,960,000} \\ & = & \sqrt{5,200,000} \\ & = & 2280.4\ \text{ft}\end{array}\]

Condensed — the full section is in OpenStax Contemporary Mathematics.

Right Triangle Trigonometry

In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like \({(2x+5y)}^{2},\) we do not have to expand it the long way: \[\begin{array}{lll}{(2x+5y)}^{2} & = & (2x+5y)(2x+5y) \\ & = & {(2x)}^{2}+10xy+10xy+{(5y)}^{2} \\ & = & 4{x}^{2}+20xy+25{y}^{2}\end{array}\]

If we know the perfect square formula, given as \[{(a+b)}^{2}={a}^{2}+2ab+{b}^{2},\] we can skip the middle step and just start writing down the answer. This may seem trivial with problems like \({(a+b)}^{2}.\) However, what if you have a problem like \({(2\sqrt{3}+3\sqrt[3]{31.8c})}^{2}?\) That is a different story. Nevertheless, we use the same perfect square formula. The same idea applies in geometry. There are special formulas and procedures to apply in certain types of problems. What is needed is to remember the formula and remember the kind of problems that fit. Sometimes we believe that because a formula is labeled special, we will rarely have use for it. That assumption is incorrect. So, let us identify the \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle and find out why it is special. See .

We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle given as \(1:\sqrt{3}:2,\) or \(x:x\sqrt{3}:2x.\) Thus, you only need to know the length of one side to find the other two sides in a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle.

Finding Missing Lengths in a

Try it.

Find the measures of the missing lengths of the triangle ().

Solution

We can see that this is a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle because we have a right angle and a \({30}^{∘}\) angle. The remaining angle, therefore, must equal \({60}^{∘}.\) Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side \(a\) is the shortest side, as it is opposite the smallest angle \({30}^{∘},\) and we can substitute \(a=x.\) The ratios are \(x:x\sqrt{3}:2x.\) We have the hypotenuse equaling 10, which corresponds to side \(c\), and side \(c\) is equal to 2\(x\). Now, we must solve for \(x\): \[\begin{array}{lll}2x & = & 10 \\ x & = & 5\end{array}\]

Side \(b\) is equal to \(x\sqrt{3}\) or \(5\sqrt{3}.\) The lengths are \(5,5\sqrt{3},10.\)

Condensed — the full section is in OpenStax Contemporary Mathematics.

Right Triangle Trigonometry

The \({45}^{∘}\text{-}{45}^{∘}\text{-}{90}^{∘}\) triangle is another special triangle such that with the measure of one side we can find the measures of all the sides. The two angles adjacent to the \({90}^{∘}\) angle are equal, and each measures \({45}^{∘}.\) If two angles are equal, so are their opposite sides. The ratio among sides is \(1:1:\sqrt{2},\) or \(x:x:x\sqrt{2},\) as shown in .

Finding Missing Lengths of a

Try it.

Find the measures of the unknown sides in the triangle ().

Solution

Because we have a \({45}^{∘}\text{-}{45}^{∘}\text{-}{90}^{∘}\) triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and \(\sqrt{2}.\) One leg measures 3, so the other leg, \(a\), measures 3. Remember the ratio of \(x:x:x\sqrt{2}.\) Then, the hypotenuse, \(c\), equals \(3\sqrt{2}.\)

Trigonometry Functions

Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications.

Trigonometry is based on three functions. We title these functions using the following abbreviations:

• \(\sin =\text{sine}\)
• \(\cos =\text{cosine}\)
• \(\tan =\text{tangent}\)

Letting \(r=\sqrt{{x}^{2}+{y}^{2}},\) which is the hypotenuse of a right triangle, we have . The functions are given in terms of \(x\), \(y\), and \(r\), and in terms of sides relative to the angle, like opposite, adjacent, and the hypotenuse.

\(\sin \theta =\frac{y}{r}=\frac{opp}{hyp}\)

\(\cos \theta =\frac{x}{r}=\frac{adj}{hyp}\)

\(\tan \theta =\frac{y}{x}=\frac{opp}{adj}\)

We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In , we have an illustration mainly to identify \(r\) and the sides labeled \(x\) and \(y\).

An angle \(\theta\) sweeps out in a counterclockwise direction from the positive \(x\)-axis and stops when the angle reaches the desired measurement. That ray extending from the origin that marks \({\theta }^{∘}\) is called the terminal side because that is where the angle terminates. Regardless of the information given in the triangle, we can find all missing sides and angles using the trigonometric functions. For example, in , we will solve for the missing sides.

Let’s use the trigonometric functions to find the sides \(x\) and \(y\). As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for \(y\).

We have \(\sin \theta =\frac{y}{r},\) and \(\theta ={60}^{∘}.\) Then, \[\begin{array}{lll}\sin {60}^{∘} & = & \frac{y}{2} \\ 2\sin {60}^{∘} & = & y \\ 1.732 & = & y \\ \sqrt{3} & = & y\end{array}\]

\(\sin {0}^{∘}=0\)	\(\cos {0}^{∘}=1\)
\(\sin {30}^{∘}=\frac{1}{2}\)	\(\cos {30}^{∘}=\frac{\sqrt{3}}{2}\)
\(\sin {45}^{∘}=\frac{\sqrt{2}}{2}\)	\(\cos {45}^{∘}=\frac{\sqrt{2}}{2}\)
\(\sin {60}^{∘}=\frac{\sqrt{3}}{2}\)	\(\cos {60}^{∘}=\frac{1}{2}\)
\(\sin {90}^{∘}=1\)	\(\cos {90}^{∘}=0\)

Condensed — the full section is in OpenStax Contemporary Mathematics.

Angle of Elevation and Angle of Depression

Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression. These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level.

Finding the Angle of Elevation

Try it.

A guy wire of length 110 meters runs from the top of an antenna to the ground (). If the angle of elevation of an observer to the top of the antenna is \({43}^{∘},\) how high is the antenna?

Solution

We are looking for the height of the tower. This corresponds to the \(y\)-value, so we will use the sine function: \[\begin{array}{lll}\sin {43}^{∘} & = & \frac{y}{110} \\ 110\sin {43}^{∘} & = & y \\ 75 & = & y\end{array}\]

The tower is 75 m high.

Finding Angle of Elevation

Try it.

You are sitting on the grass flying a kite on a 50-foot string (). The angle of elevation is \({60}^{∘}.\) How high above the ground is the kite?

Solution

We can solve this using the sine function, \(\sin \theta =\frac{opp}{hyp}.\) \[\begin{array}{lll}\sin {60}^{∘} & = & \frac{x}{50} \\ 50\sin {60}^{∘} & = & x \\ & = & 43.3\ \text{ft}\end{array}\]

Condensed — the full section is in OpenStax Contemporary Mathematics.

Key Concepts

• The Pythagorean Theorem is applied to right triangles and is used to find the measure of the legs and the hypotenuse according the formula \({a}^{2}+{b}^{2}={c}^{2},\) where c is the hypotenuse.
• To find the measure of the sides of a special angle, such as a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle, use the ratio \(x:x\sqrt{3}:2x,\) where each of the three sides is associated with the opposite angle and 2\(x\) is associated with the hypotenuse, opposite the \({90}^{∘}\) angle.
• To find the measure of the sides of the second special triangle, the \({45}^{∘}\text{-}{45}^{∘}\text{-}{90}^{∘}\) triangle, use the ratio \(x:x:x\sqrt{2},\) where each of the three sides is associated with the opposite angle and \(x\sqrt{2}\) is associated with the hypotenuse, opposite the \({90}^{∘}\) angle.
• The primary trigonometric functions are \(\sin \theta =\frac{opp}{hyp},\) \(\cos \theta =\frac{adj}{hyp},\) and \(\tan \theta =\frac{opp}{adj}.\)
• Trigonometric functions can be used to find either the length of a side or the measure of an angle in a right triangle, and in applications such as the angle of elevation or the angle of depression formed using right triangles.

Formula

The Pythagorean Theorem states \[{a}^{2}+{b}^{2}={c}^{2}\] where \(a\) and \(b\) are two sides (legs) of a right triangle and \(c\) is the hypotenuse.

Projects

1. One of the reasons so many formulas in geometry were discovered was because of the importance in finding measurements of lengths, areas, perimeter, and angles. Find at least five examples of how geometry can be used in practical applications today.
2. Who were the Pythagoreans? Why did this society exist? Explore what they did and discuss some of their beliefs.

Practice (12)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

1. Find the length of the missing side of the triangle ().

   ഉത്തരം വെളിപ്പെടുത്തുക

   Using the Pythagorean Theorem, we have \[\begin{array}{lll}{(6)}^{2}+{b}^{2} & = & {(14)}^{2} \\ 36+{b}^{2} & = & 196 \\ {b}^{2} & = & 196-36 \\ {b}^{2} & = & 160 \\ b & = & \pm \sqrt{160} \\ & = & 4\sqrt{10}=12.65\end{array}\]

   When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.

2. You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive (). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.

   ഉത്തരം വെളിപ്പെടുത്തുക

   You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then \(9(200)+7(200)=1,800\ \text{ft}+1,400\ \text{ft}=3,200\ \text{ft}\).

   As the crow flies, use the distance formula. We have \[\begin{array}{lll}d & = & \sqrt{{(1,800-0)}^{2}+{(1,400-0)}^{2}} \\ & = & \sqrt{3,240,000+1,960,000} \\ & = & \sqrt{5,200,000} \\ & = & 2280.4\ \text{ft}\end{array}\]

3. The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches (). What should the vertical height be?

   ഉത്തരം വെളിപ്പെടുത്തുക

   The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a, is 130 in. The length of c, or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b.

   Then, by the Pythagorean Theorem, we have: \[\begin{array}{lll}{a}^{2}+{b}^{2} & = & {c}^{2} \\ {(130)}^{2}+{b}^{2} & = & {(132.4)}^{2} \\ 16,900+{b}^{2} & = & 17,529.76 \\ {b}^{2} & = & 17,529.76-16,900 \\ {b}^{2} & = & 629.8 \\ b & = & \sqrt{629.8}=25\end{array}\]

   If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change?

   The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be \(12(25)=300\ \text{in}.\) So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length: \[\begin{array}{lll}{(300)}^{2}+{(25)}^{2} & = & {c}^{2} \\ 90,625 & = & {c}^{2} \\ \sqrt{90,625} & = & c=301\ \text{in}\end{array}\]

   The new ramp will look like .

4. Find the measures of the missing lengths of the triangle ().

   ഉത്തരം വെളിപ്പെടുത്തുക

   We can see that this is a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle because we have a right angle and a \({30}^{∘}\) angle. The remaining angle, therefore, must equal \({60}^{∘}.\) Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side \(a\) is the shortest side, as it is opposite the smallest angle \({30}^{∘},\) and we can substitute \(a=x.\) The ratios are \(x:x\sqrt{3}:2x.\) We have the hypotenuse equaling 10, which corresponds to side \(c\), and side \(c\) is equal to 2\(x\). Now, we must solve for \(x\): \[\begin{array}{lll}2x & = & 10 \\ x & = & 5\end{array}\]

   Side \(b\) is equal to \(x\sqrt{3}\) or \(5\sqrt{3}.\) The lengths are \(5,5\sqrt{3},10.\)

5. A city worker leans a 40-foot ladder up against a building at a \({30}^{∘}\) angle to the ground (). How far up the building does the ladder reach?

   ഉത്തരം വെളിപ്പെടുത്തുക

   We have a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle, and the hypotenuse is 40 ft. This length is equal to 2\(x\), where \(x\) is the shortest side. If \(2x=40\), then \(x=20\). The ladder is leaning on the wall 20 ft up from the ground.

6. Find the measures of the unknown sides in the triangle ().

   ഉത്തരം വെളിപ്പെടുത്തുക

   Because we have a \({45}^{∘}\text{-}{45}^{∘}\text{-}{90}^{∘}\) triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and \(\sqrt{2}.\) One leg measures 3, so the other leg, \(a\), measures 3. Remember the ratio of \(x:x:x\sqrt{2}.\) Then, the hypotenuse, \(c\), equals \(3\sqrt{2}.\)

7. Find the lengths of the missing sides for the triangle ().

   ഉത്തരം വെളിപ്പെടുത്തുക

   We have a \({55}^{∘}\) angle, and the length of the triangle on the \(x\)-axis is 6 units.

   Step 1: To find the length of \(r\), we can use the cosine function, as \(\cos \theta =\frac{x}{r}.\) We manipulate this equation a bit to solve for \(r\): \[\begin{array}{lll}\cos ({55}^{∘}) & = & \frac{6}{r} \\ r\cos ({55}^{∘}) & = & 6 \\ r & = & \frac{6}{\cos ({55}^{∘})} \\ r & = & \frac{6}{0.5736}=10.46\end{array}\]
   

   Step 2: We can use the Pythagorean Theorem to find the length of \(y\). Prove that your answers are correct by using other trigonometric ratios: \[\begin{array}{lll}{6}^{2}+{y}^{2} & = & {10.46}^{2} \\ {y}^{2} & = & 109.4-36 \\ y & = & 8.57\end{array}\]
   

   Step 3: Now that we have \(y\), we can use the sine function to prove that \(r\) is correct. We have \(\sin \theta =\frac{y}{r}.\) \[\begin{array}{lll}\sin ({55}^{∘}) & = & \frac{8.57}{r} \\ r\sin ({55}^{∘}) & = & 8.57 \\ r & = & \frac{8.57}{\sin ({55}^{∘})} \\ & = & \frac{8.57}{0.819}=10.46\end{array}\]

8. Solve for the lengths of a right triangle in which \(\theta ={30}^{∘}\) and \(r=6\) ().

   ഉത്തരം വെളിപ്പെടുത്തുക

   Step 1: To find side \(a\), we use the sine function: \[\begin{array}{lll}\sin {30}^{∘} & = & \frac{a}{6} \\ 6\sin {30}^{∘} & = & a=3\end{array}\]
   

   Step 2: To find \(b\), we use the cosine function: \[\begin{array}{lll}\cos {30}^{∘} & = & \frac{b}{6} \\ 6\cos {30}^{∘} & = & b=5.196\end{array}\]
   

   Step 3: Since this is a \({30}^{∘}\text{-}{60}^{∘}\text{-}{90}^{∘}\) triangle and side \(b\) should equal \(x\sqrt{3},\) if we input 3 for \(x\), we have \(b=3\sqrt{3}.\) Put this in your calculator and you will get \(3\sqrt{3}=5.196.\)

9. A small plane takes off from an airport at an angle of \({31.3}^{∘}\) to the ground. About two-thirds of a mile (3,520 ft) from the airport is an 1,100-ft peak in the flight path of the plane (). If the plane continues that angle of ascent, find its altitude when it is above the peak, and how far it will be above the peak.

   ഉത്തരം വെളിപ്പെടുത്തുക

   To solve this problem, we use the tangent function: \[\begin{array}{lll}\tan {31.3}^{∘} & = & \frac{x}{3,520} \\ 3,520\tan {31.3}^{∘} & = & 2,140\end{array}\]

   The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak.

10. Suppose you have two known sides, but do not know the measure of any angles except for the right angle (). Find the measure of the unknown angles and the third side.

    ഉത്തരം വെളിപ്പെടുത്തുക

    Step 1: We can find the third side using the Pythagorean Theorem: \[\begin{array}{lll}{6}^{2}+{4}^{2} & = & {c}^{2} \\ 52 & = & {c}^{2} \\ 2\sqrt{13} & = & c\end{array}\]

    Now, we have all three sides.

    Step 2: To find \(\theta ,\) we will first find \(\sin \theta .\) \[\begin{array}{lll}\sin \theta & = & \frac{opp}{hyp} \\ & = & \frac{4}{2\sqrt{13}} \\ & = & \frac{2}{\sqrt{13}}.\end{array}\]

    The angle \(\theta\) is the angle whose sine is \(\frac{2}{\sqrt{13}}.\)

    Step 3: To find \(\theta\), we use the inverse sine function: \[\begin{array}{lll}\theta & = & {\sin }^{-1}(\frac{2}{\sqrt{13}}) \\ & = & {33.7}^{∘}\end{array}\]
    

    Step 4: To find the last angle, we just subtract: \({180}^{∘}-{90}^{∘}-{33.7}^{∘}={56.3}^{∘}\).

11. A guy wire of length 110 meters runs from the top of an antenna to the ground (). If the angle of elevation of an observer to the top of the antenna is \({43}^{∘},\) how high is the antenna?

    ഉത്തരം വെളിപ്പെടുത്തുക

    We are looking for the height of the tower. This corresponds to the \(y\)-value, so we will use the sine function: \[\begin{array}{lll}\sin {43}^{∘} & = & \frac{y}{110} \\ 110\sin {43}^{∘} & = & y \\ 75 & = & y\end{array}\]

    The tower is 75 m high.

12. You are sitting on the grass flying a kite on a 50-foot string (). The angle of elevation is \({60}^{∘}.\) How high above the ground is the kite?

    ഉത്തരം വെളിപ്പെടുത്തുക

    We can solve this using the sine function, \(\sin \theta =\frac{opp}{hyp}.\) \[\begin{array}{lll}\sin {60}^{∘} & = & \frac{x}{50} \\ 50\sin {60}^{∘} & = & x \\ & = & 43.3\ \text{ft}\end{array}\]

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\theta

theta
The usual name for an angle.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

^\circ

degrees
1/360 of a full turn. 180° = π radians.

\angle ABC,\ \triangle ABC

angle, triangle
The angle at B between BA and BC; the triangle with those vertices.

\parallel,\ \perp,\ \cong,\ \sim

parallel, perpendicular, congruent, similar
Never meet; meet at 90°; identical shape and size; same shape.

How to: Right Triangle Trigonometry

1. Apply the Pythagorean Theorem to find the missing sides of a right triangle.
2. Apply the
3. Apply trigonometric ratios to find missing parts of a right triangle.
4. Solve application problems involving trigonometric ratios.
5. right triangle
6. sine
7. cosine
8. tangent

Questions people ask

Why does every triangle have angles adding to 180°?

Draw a line through one vertex parallel to the opposite side: the two alternate angles equal the other two angles of the triangle, and the three angles at the vertex lie on a straight line. That is Euclid's proof, and it is why the fact holds only on a flat plane.

When do I use the law of sines versus the law of cosines?

Cosines when you know three sides, or two sides and the angle between them. Sines when you know an angle and the side opposite it, plus one more piece.

What is the difference between area and perimeter?

Perimeter is the length of the boundary (one dimension, measured in metres); area is the amount of surface inside (two dimensions, square metres). Doubling every side doubles the perimeter but quadruples the area.

നീ സ്വയം ശ്രമിക്ക്.

⌨ Type ✎ Draw

നിങ്ങളുടെ കണക്കുകൂട്ടലുകൾ

പരിഹരിക്കുക

x²√÷×π( )=∫d/dxlim[ ]sinlogeˣ

Parts of this page are adapted from OpenStax Contemporary Mathematics (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

കൂടുതല്‍ Geometry

TrianglesPythagorean theoremCirclesPolygonsSolidsCoordinate geometryEuclid's axioms and the structure of proofCongruent and similar trianglesCircle theoremsTransformations: translation, rotation, reflection, dilationSolid geometry: prisms, pyramids, spheres