The Sine Law

Hi there. This page is on the sine law for finding side lengths and angles of non-right angle triangles.

Table Of Contents


Refer to the picture below.


Remember that SOH CAH TOA applies to right angle triangles where we have:

    \[\sin(\theta) = \dfrac{\text{Opposite Side Length To } \theta}{\text{Hypotenuse Side Length}}\]

    \[\cos(\theta) = \dfrac{\text{Adjacent Side Length To } \theta}{\text{Hypotenuse Side Length}}\]

    \[\tan(\theta) = \dfrac{\text{Opposite Side Length To } \theta}{\text{Adjacent Side Length To }\theta}\]

In the case where there is no right angle in the triangle, SOH CAH TOA does not work and another strategy/technique/formula is needed.

The Sine Law

Suppose we have a triangle ABC with side lengths a, b, and c. This triangle may not be a right angle triangle.


The side length a is opposite to angle A, the side length b is opposite to the angle B and c is opposite to the angle C.

The sine law formula is actually a ratio of side lengths over the sine of the angle opposite to the lengths (and vice versa). One can use the formula

    \[\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)} = \dfrac{c}{\sin(C)}\]


    \[\dfrac{\sin(A)}{a} = \dfrac{\sin(B)}{b} = \dfrac{\sin(C)}{c}\]

The first one of the two above is best suited for finding unknown side lengths while the second one is better suited for finding unknown angles. If you use the “wrong” formula, it just means that more algebra will be needed to achieve the (same) desired answer.

SOH CAH TOA As A Special Case Of The Sine Law


The sine law can be used for any triangle. What happens if the sine law is (accidentally) used for right angle triangles? This section will investigate what would happen.

Suppose all the angles in the triangle above are known and side lengths c and b are known as well. Side length a is unknown. If the sine law is used we have:

    \[\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)} = \dfrac{c}{\sin(C)}\]

To solve for the side length a we have two cases. We can solve for a using either

    \[\dfrac{a}{\sin(A)} = \dfrac{c}{\sin(C)}\]


    \[\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)}\]

Case One

In \dfrac{a}{\sin(A)} = \dfrac{c}{\sin(C)} we have the angle C as 90 degrees. The sine of 90 degrees is 1. This simplifies the equation to:

    \[\dfrac{a}{\sin(A)} = c\]

Solving for a gives:

    \[a = c \sin(A)\]

From the above, we can have \sin(A) = \dfrac{a}{c}. This formula is the SOH CAH TOA representation where a is the side length opposite to angle A and c is the hypotenuse side length.

Case Two

In the second case, we have \dfrac{a}{\sin(A)} = \dfrac{b}{\sin(B)}. It is known that angle C is 90 degrees from the 180 degrees in the triangle. This means that angle B has to 90 degrees minus angle A. We can re-write the equality formula as:

    \[\dfrac{a}{\sin(A)} = \dfrac{b}{\sin(90 - A)}\]

There is a trigonometric identity which states that \sin(90 - \theta) = \cos(\theta). So \sin(90 - A) = \cos(A).

    \[\dfrac{a}{\sin(A)} = \dfrac{b}{\cos(A)}\]

Rearranging for the side length a yields:

    \[a= \dfrac{b \sin(A) }{\cos(A)} = b \tan(A)\]

From the above formula, we can have \tan(A) = \dfrac{a}{b} which is another SOH CAH TOA representation. In this case a is opposite to angle A and b is the adjacent side length to angle A.

A similar procedure can be done for finding an unknown angle or side length assuming all else is known. This can be done as an exercise.

As you can see from the above, using the sine law on right angle triangles instead of using the usual SOH CAH TOA can be slightly more difficult as trigonometric identities will be needed to obtain the desired results.


1) Suppose we have a triangle ABC with angle A as 55^{\circ} and side lengths BC = 8 units, AC = 9 units and AB = 4 units. What is the angle of B?



Angle A of 55 degrees is opposite to the side length BC of 8 units. We can use the fact that we also know the side length of AC which is opposite to the unknown angle B. With this information in mind, we use the sine law and have the equation set up as:

    \[\dfrac{\sin(A)}{BC} = \dfrac{\sin(B)}{AC}\]

When we substitute the values, the above equation is now:

    \[\dfrac{\sin(55^{\circ})}{8} = \dfrac{\sin(B)}{9}\]

We solve for \sin(B) and then the angle B.

    \[\sin(B) = \dfrac{9 \sin(55^{\circ})}{8} = 0.9215\]

Since we are finding the angle B, we use the inverse cosine (\sin^{-1}(x)).

    \[\sin^{-1}(\sin(B)) = \sin^{-1}(0.9215)\]

    \[B = 67.15^{\circ}\]

2) Suppose we have a triangle ABC with angle A as 50^{\circ}, angle B as 70^{\circ} with a known side length of b = 10 What is the side length of c = AB?


We have the pair of a known angle B and the side length b= AC = 10. The angle C can be determined as 180^{\circ} - 50^{\circ} - 70^{\circ} = 60^{\circ}.

The sine law can be set up as:

    \[\dfrac{10}{\sin(70^{\circ})} = \dfrac{c}{\sin(60^{\circ})}\]

Isolating for the side c = AB gives us:

    \[c = \dfrac{10 \sin(60^{\circ})}{\sin(70^{\circ})} = \dfrac{10(0.8660)}{0.9397} = 9.22\]

Practice Problems

1) In the triangle ABC with side lengths a = 2, b = 5, c = 3, and the angle A as 30^{\circ}, what is the angle of C?

2) You are given the triangle EFG with side length e = 7.5, f = 8.5, g = 5, and the angle G as 30^{\circ}, what are the remaining angles?

3) Suppose there is a triangle ABC with side lengths a = 5.2, b = 4.5, and an unknown side length c. Angle A is 60^{\circ} and angle B is 50^{\circ}. What is the side length c?

Solutions To Practice Problems


    \[\dfrac{\sin(30^{\circ})}{2} = \dfrac{\sin(C)}{3}\]

    \[\sin(C) = \dfrac{3 \sin(30^{\circ})}{2} = 0.75\]

    \[C = \sin^{-1}(0.75) = 48.59^{\circ}\]

2) I am finding angle E first.

    \[\dfrac{\sin(30^{\circ})}{5} = \dfrac{\sin(E)}{7.5}\]

    \[\sin(E) = \dfrac{7.5 \sin(30^{\circ})}{5} = 0.75\]

    \[E = \sin^{-1}(0.75) = 48.59^{\circ}\]

The third remaining angle F with the two other angles add up to 180^{\circ}. Angle F would be 180^{\circ} - 48.59^{\circ} - 30^{\circ} = 101.41^{\circ}

3) Since the three angles in a triangle add up to 180^{\circ}, angle C would be 70^{\circ}.

The sine law set up for finding the side length c would be:

    \[\dfrac{c}{\sin(70^{\circ} )} = \dfrac{5.2}{\sin(60^{\circ} )}\]

Solving for c:

    \[c = \dfrac{5.2 \sin(70^{\circ})}{\sin(60^{\circ} )}\]

    \[c = \dfrac{5.2 (0.9397)}{0.8660} = 5.64\]