Trigonometric Ratio 

Mathematics is the language of physics. It serves as instrument to explain and test many physics theories and principles. It is also used to solve many physics problems.

Many of physics problems deal with angles. To solve them, we use the trigonometry concepts of math. Trigonometry is the study of triangles and the relationship among their sides and angles. This is also important in dealing with vectors.

Let us review first the different parts of a right triangle. The longest side of the right triangle is called the hypotenuse. The other two sides are the sides opposite and adjacent to the indicated angle. The opposite and adjacent sides may change depending on which angle is referred to, whereas, the hypotenuse is always the side which is the longest. The Greek letter \(\theta\) is commonly used to denote angle in physics problems.

Consider the right triangle below.

Based on the triangle above, we obtain the following relationship:

\({BC \over AB}=\frac 24;\quad {DE \over AD}=\frac 48\;\text{and}\; {FG \over AF}=\frac {6}{12}\)


\({BC \over AB}={DE \over AD}={FG \over AF}=\frac 12\).

These relationships show that the ratio of the length of the shorter leg or the opposite side to the angle to the length of the hypotenuse has a value the same value regardless of the size of the triangle. This pattern is referred to as a trigonometric ratio. This is the ratio of the two sides in a right triangle. In this case, the ratio we obtain is the sine ratio. Specifically, the sine ratio of 30º since the given angle is 30º. Thus, the sine ratio can be expressed as

\(\sin \theta={\text{length of the opposite side to the angle}\;\theta \over \text{length of the hypotenuse}}.\)

The other important trigonometric ratios are the cosine ratio and tangent ratio.

Cosine ratio is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

\(\cos \theta={\text{length of the adjacent side to the angle}\;\theta \over \text{length of the hypotenuse}}.\)

Tangent ratio is the ratio of the length of the side opposite to the angle to the length of the side adjacent to the angle.

\(\tan \theta={\text{length of the opposite side to the angle}\;\theta \over \text{length of the adjacent side to the angle}}.\)

Mnemonics SOH CAH TOA are used to remember trigonometric ratios easier. SOH stands for sine of the angle as the ratio of the length of the opposite side to the angle to the length of the hypotenuse. CAH for cosine of the angle as the ratio of the length of the side adjacent to the angle to the length of the hypotenuse. TOA stands for tangent is equal to the length of the opposite side to the angle over the length of the side adjacent to the angle.

Example 1.

Refer to the figure below to find:

  1. sin 30º
  2. sin 60º
  3. cos 30º
  4. cos 60º
  5. tan 30º
  6. tan 60º


  1. \(\sin 30^\circ={1 \over 2}=0.5\)
  2. \(\sin 60^\circ={\sqrt 3 \over 2}\approx0.87\)
  3. \(\cos 30^\circ={\sqrt 3 \over 2}\approx0.87\)
  4. \(\cos 60^\circ={1 \over 2}=0.5\)
  5. \(\tan 30^\circ={1\over \sqrt 3}={\sqrt 3 \over 3}\approx0.58\)
  6. \(\tan 60^\circ={\sqrt 3 \over 1}=\sqrt 3\approx1.73\)

Example 2.

Solve the area of the given triangle below.



hypotenuse=20 m


1. We will use the sine ratio to solve for the height of the triangle.

\(\sin 30^\circ={\text{opposite side}\over\text{hypotenuse}}={\text{BC}\over20\;m}\)\(\implies\)\(\text{BC}=20\;m(\sin 30^\circ)\)

then, \(\text{BC}=20\;m({1 \over 2})=10\;m\).

2. Using the formula for the area of the triangle, \(A={1 \over 2}bh\) and referring to the figure, side BC=AC=10 m, thus the area of the triangle is:

\(A={1\over 2}(10\;m)(10\;m)=100\;m^2\).

Example 3.

Alphonso leaned a ladder against the wall at the side of their house to reach the rooftop and get his ball that flew over their when he was playing with it. How long was the ladder that he is using if he was able to reach the rooftop 15-ft from the ground by leaning the ladder 70º with the ground?


height of the house (groud to the rooftop) = 15-ft

angle \(\theta\) between the ladder and the ground = 70º


1. First, we can draw the given problem to visualize the situation.

2. In the drawing, a right triangle is formed, thus a trigonometric ratio can be used to solve the problem. Given only an angle and opposite side, we can only use the sine ratio.

\(\sin 70^\circ={\text{height of the house}\over \text{length of the ladder}}={15\;ft \over \text{length of the ladder}}\)\(\implies\)\(\text{length of the ladder}={15\;ft \over \sin 70^\circ}\)


\(\text{length of the ladder}={15\;ft \over \sin 70^\circ}={15\;ft \over 0.94}=15.96\;ft\approx16\;ft\).

The ladder he used was 16 ft. long.

Example 4. 

Joe often walked to school taking the road x meters East then 10 meters North. Sometimes, he took a short cut passing through a river which is d meters, 35º North of East.

a. Solve for the x meters East Joe walked.

b. Solve for the d meters he walked as short cut.


First, draw the problem.

a. To solve for the x meters, use the tangent ratio.

\(\tan 35^\circ={10\;m \over x}\)\(\implies \)\(x={10\;m \over \tan 35^\circ}\approx 14\;m\)

b. To solve for d, we can use either the cosine ratio or the sine ratio.

b.1. Using cosine ratio,

\(\cos 35^\circ={14\;m \over d}\)\(\implies\)\(d={14\;m \over \cos 35^\circ}\approx 17\;m\)

b.2. Using sine ratio,

\(\sin 35^\circ={10\;m \over d}\)\(\implies\)\(d={10\;m \over \sin 35^\circ}\approx 17\;m\)

Example 5.

A ball is thrown at 40º angle with the horizontal. What are the and y components of the velocity vector given that the initial velocity of the ball is 25 m/s?


a. To solve for \(\vec v_x\), use cosine ratio.

\(\cos 40^\circ={\vec v_x \over \vec v}\)\(\implies\)\(\vec v_x=\vec v\; \cos 40^\circ=(25\;m/s)(\cos 40^\circ)\approx 19\;m/s\)

b. To solve for \(\vec v_y\), use sine ratio.

\(\sin 40^\circ={\vec v_y \over \vec v}\)\(\implies\)\(\vec v_y=\vec v\; \sin 40^\circ=(25\;m/s)(\sin 40^\circ)\approx 16\;m/s\)