Right Triangle Trigonometry can be used to help solve for the missing side of a right triangle in a physics problem. For this to work, the triangle must have one right angle, and you must have at least the measure of one other angle and one side.

To start a Right Triangle Trig. problem you must first understand vectors,vector addition, and force diagrams and how they work. A vector is something that has both magnitude, size, and direction. To add vectors, (vector addition), you simply place the "tip" of one vector to the "tail" of another, like so. The last connecting line to complete the triangle is the sum of the two vectors you previously placed "tip to tail". A force diagram is a diagram of all of the forces acting on your chosen object (see Force Diagrams).

Once you understand vectors, vector addition, and how to apply them to force diagrams, you can begin to apply the three different Right Triangle Trigonometry equations to your vectors. The three different equations used for solving right tirangle trig problems are

Sin Theta= Opposite / Hypotenuse
Cos Theta= Adjacent / Hypotenuse
Tan Theta= Opposite / Adjacent

These equations can be easily remembered by using SOH- CAH- TOA
Sin, Cos, and Tan are all buttons on a scientific calculator. Theta is the second angle in your triangle (not right angle). The opposite side in a triangle is the side directly above the angle you're using. You can also remember it by it being on the "opposite" side of the angle. The adjacent side is the side next to the angle. Finally, the hypotenuse is the longest side of the triangle and is the only side that does not help for a right angle.

Trig functions can be used to find sides or angles.
If the opposite side and hypotenuse are given, we can find the angle x by saying x=sin^-1 opposite / hypotenuse
If the opposite side and adjacent side are given, we can find the angle x by saying x= tan^-1 opposite / adjacent
If the adjacent side and hypotenuse are given, we can find the angle x by saying x= cos^-1 adjacent / hypotenuse

Sample Problem

Because right triangle trig is used to find the value of a force in a situation, (for example force of friction), you must start your problem by drawing a force diagram of your object. In this sample problem you are solving for the force of gravity in the y direction acting on a box that has a constant velocity sliding down a ramp at a 30 degree angle, (solving for adjacent side of right triangle). The box weighs 10kg so you can determine that Fg=98N (9.8m/s2 x 10kg= 98N). Theta= 30 degrees.

Once you have your force diagram drawn, you then have to draw an axis to fit the diagram as shown by the downwards facing dotted line above. To complete the right triangle you must add another vector "tip to tail" to your fitted axis and force diagram.

Next you have to apply one of the three Right Triangle trig equations to solve for Fgy. Because we have the value of the hypotenuse (98N), we need to use Cos Theta= Adjacent/ hypotenuses

Cos 30= adj./ 98N
Adj= 84.87 N

In the conclusion, the force of gravity acting on the box in the y direction is 84.87 N.

Right Triangle Trigonometry can be used to help solve for the missing side of a right triangle in a physics problem. For this to work, the triangle must have one right angle, and you must have at least the measure of one other angle and one side.

To start a Right Triangle Trig. problem you must first understand

vectors,vector addition,andforce diagramsand how they work. A vector is something that has both magnitude, size, and direction. To add vectors, (vector addition), you simply place the "tip" of one vector to the "tail" of another, like so. The last connecting line to complete the triangle is the sum of the two vectors you previously placed "tip to tail". A force diagram is a diagram of all of the forces acting on your chosen object (see Force Diagrams).Right Triangle Trig EquationsOnce you understand vectors, vector addition, and how to apply them to force diagrams, you can begin to apply the three different Right Triangle Trigonometry equations to your vectors. The three different equations used for solving right tirangle trig problems are

Sin Theta= Opposite / Hypotenuse

Cos Theta= Adjacent / Hypotenuse

Tan Theta= Opposite / Adjacent

These equations can be easily remembered by using SOH- CAH- TOA

Sin, Cos, and Tan are all buttons on a scientific calculator. Theta is the second angle in your triangle (not right angle). The opposite side in a triangle is the side directly above the angle you're using. You can also remember it by it being on the "opposite" side of the angle. The adjacent side is the side next to the angle. Finally, the hypotenuse is the longest side of the triangle and is the only side that does not help for a right angle.

Trig functions can be used to find sides or angles.

If the opposite side and hypotenuse are given, we can find the angle x by saying x=sin^-1 opposite / hypotenuse

If the opposite side and adjacent side are given, we can find the angle x by saying x= tan^-1 opposite / adjacent

If the adjacent side and hypotenuse are given, we can find the angle x by saying x= cos^-1 adjacent / hypotenuse

Sample ProblemBecause right triangle trig is used to find the value of a force in a situation, (for example force of friction), you must start your problem by drawing a force diagram of your object. In this sample problem you are solving for the force of gravity in the y direction acting on a box that has a constant velocity sliding down a ramp at a 30 degree angle, (solving for adjacent side of right triangle). The box weighs 10kg so you can determine that Fg=98N (9.8m/s2 x 10kg= 98N). Theta= 30 degrees.

Once you have your force diagram drawn, you then have to draw an axis to fit the diagram as shown by the downwards facing dotted line above. To complete the right triangle you must add another vector "tip to tail" to your fitted axis and force diagram.

Next you have to apply one of the three Right Triangle trig equations to solve for Fgy. Because we have the value of the hypotenuse (98N), we need to use Cos Theta= Adjacent/ hypotenuses

Cos 30= adj./ 98N

Adj= 84.87 N

In the conclusion, the force of gravity acting on the box in the y direction is 84.87 N.