RIGHT TRIANGLES

Right Triangles

This section is intended as review for some of you and an introduction to angles for those of you who never had trigonometry.

The window to the right displays a right triangle. Some triangles contain a right angle while others do not. This section will focus on those that contain a right angle. The other types of triangles will be covered later in this course. With right triangles we will introduce several useful definitions.

Hypotenuse of a right triangle is the longest side and it is opposite the right angle in the triangle.
The Pythagorean formula is applicable: Hypotenuse² = Side1² + Side2².
Sine(angle) , abbreviated sin(angle), defined as the side opposite the angle divided by the hypotenuse of the right triangle
Cosine(angle), abbreviated cos(angle), defined as the side adjacent to the angle divided by the hypotenuse of the right triangle.
Tangent(angle), abbreviated tan(angle), defined as the side opposite the angle divided by the side adjacent to the angle.
Cosecant(angle), abbreviated as csc(angle), reciprocal of the sine so it is hyoptenuse divided by side opposite the angle.
Secant(angle), abbreviated as sec(angle), reciprocal of the cosine so it is hypotenuse divided by side adjacent to the angle.
Cotangent(angle), abbreviated as cot(angle), reciprocal of the tangent so it is adjacent side divided by side opposite the angle.

When you talk about "solving" a right triangle you mean finding the values of all of the angles and the lengths of all of the sides. Using trigonometry you can solve a right triangle if you are given:

Any two sides
Any side and either of the acute angles.

Note you cannot solve a right triangle if you are only given angles. Angles tell you about the shape, but not the size of a right triangle. You need additional information to determine the size of the triangle. The length of any side furnishes that information.

Let's look at the three triangles shown to the right and determine the values of Sin(A), Cos(A) and Tan(A).

In the first case we know the lengths of two sides. Using the Pythagorean Formula we can find the length of the third side.

H² = 4² + 3², H² = 16 + 9 = 25, H = Sqrt(25) = 5
Sin(A) = 3/5, Cos(A) = 4/5 and Tan(A) = 3/4

In the next case we know the hypotenuse and one side but need the value of the other side.

We will use the Pythagorean Formula again but with a slight variation. This time we are looking for the value of a side not the hypotenuse. So we will write it as:
Side² = 13² - 5²
Side² = 169 - 25 = 144
Side = Sqrt(144) = 12
Sin(A) = 12/13, Cos(A) = 5/13 and Tan(A) = 13/5

For the last case we again know the one angle and one side. In the first two cases we used the Pythagorean Theorem to determine the lengths of the sides of the triangle and the values of the trigonometric funcitons. In this case we will use the definitions of the trigonometric funcitons, along with the calculator values of the trigonometric functions to determine the lengths of the hypotenuse and the one unknow side of the triangle.

sin(25^o) = 6.23/H, H = 6.23/sin(25^o) = 14.74
cos(25^o) = S1/14.74, S1 = 14.74 * cos(25^o) = 13.36

Now it is your turn to try it. Here is a sample problem - Problem 1. The answer is at the end of this page.

Now let's look at a slightly different problem. What if you have the trigonometric functions but want to find the angles? To solve this problem we will need to introduce the idea of the INVERSE FUNCTION. On your calculator there are probably three buttons labled with trigonometric buttons. They probably look something like this:

^sin-1

^cos-1 ^tan-1

When you input some data and then press the Sin key the calculator computes the value of the angle that you input. For example if you calculate the sin(30^o) the calculator returns 0.5. This is useful for finding sides when we know angles. But what about the case we are discussing where we know sides but not angles? To the left of the keys printed on the calculator you will see sin^-1, cos^-1 and tan^-1. These are the inverse functions and they are secondary functions of the sin, cos and tan keys. To activate them you first hit the 2^nd key and then the sin, cos or tan key. Here is the difference between the button key and the secondary key:

The sin key gives you the sin of an angle, so sin(30^o) = 0.5. The input is the angle and the output is the sin of the angle.
The sin^-1 key function gives you the angle so sin^-1(0.5) = 30^o. The input is the value of the sin of the function and the output is the angle! Input and output have been reversed.
Each key has an inverse that performs the above function.

Let's try several to make certain that you understand the concept.

Find the sine of 83.7^o. sin(83.7^o) = 0.99396 Rounded to four places 0.9940
Find the cosine of 274.9^o. cos(274.9^o) = 0.0854
Find the tangent of 315.2^o. tan(315.2^o = -0.9930 (Results can be negative.)

Now let's go the other way, with the value of the trigonometric function as input and the angle as the answer.

If sin(A) = 0.3864 find the value of A. sin^-1(0.3864) = 22.73^o.
If cos(A) = 0.9971 find the value of A. cos^-1(0.9971) = 4.36^o.
If tan(A) = 18.567 find the value of A. tan^-1(18.567) = 86.92^o.

What do we do if the problem use the secant, cosecant or cotangent? There aren't any keys for these trigonometric functions so we have to convert them to functions that do have keys. Remember they have a reciprocal relationship with cos, sin and tan (respectively).

Given sec(A), sec(A) = 1/cos(A) so cos(A) = 1/sec(A). Sec(A) = 2.5063, cos(A) = 1/2.5063 = 0.3990
Given csc(A), csc(A) = 1/sin(A) so sin(A) = 1/csc(A). Csc(A) = 3.0955, sin(A) = 1/3.0955 = 0.3230
Given cot(A), cot(A) = 1/tan(A) so tan(A) = 1/cot(A). Cot(A) = 14.509, tan(A) = 1/14.509 = 0.0689.

The picture on the right presents a problem that involves a right triangle. A Scud missile is approaching a friendly surface-to-air missile site. The Scud is flying at an altitude of 1,500 meters. The angle of elevation from the friendly radar to the Scud is 10^o. Your job is to determine the ground range and slant range from the radar site to the missile.

Sin(10^o) = 1,500/Slant Range, Slant Range = 1,500/Sin(10^o) = 8,638 m or 8.638 km.
Tan(10^o) = 1,500/Ground Range, Ground Range = 1,500/Tan(10^o) = 8,507 m or 8.507 km.

Problem 2 - Using the same setting but new data here is a problem for you to try. An airplane is flying across the country at an altitude of 40,000 feet. A traffic control radar picks the aircraft up when the radars elevation angle (from the local horizontal) is 8^o. determine the slant range and the ground range from the radar to the aircraft. The answer is at the end of this section.

Answer to Problems

S2 = Sqrt[10² - 6²] = Sqrt(100 - 36) = Sqrt(64) = 8
sin(A) = 8/10 = 4/5, cos(A) = 6/10 = 3/5, tan(A) = 8/6 = 4/3

sin(8^o) = 40,000/Slant Range, Slant Range = 40,0000/sin(8^o) = 287,412 ft. (about 54.4 miles)
cos(8^o) = Ground Range/Slant Range, Ground Range = Slant Range * cos(8^o) = 284,615 ft (about 53.9 miles)
An alternative calculation method for Ground Range would be tan(8^o) = 40,000/Ground Range
Ground Range = 40,000/tan(8^o) = 284,615 ft. (about 53.9 miles).