# Law of Sines

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

A triangle has the sides a = 30, c = 50, and angle α = 25°. Find the angle γ if it is the largest angle of the triangle.

### Example 2

A triangle has the side a = 5 cm and angles α = 50° and γ = 100°. Calculate the length of the side b.

### Example 3

Find the distance $$x$$ across the lake shown in Figure $$7,$$ if $$BC = a = 100\,\text{m},$$ $$\beta = 85^\circ,$$ and $$\gamma = 75^\circ.$$

### Example 4

The side lengths of a triangle are $$a = 30,$$ $$b = 40,$$ and $$c = 50.$$ Find the angles of the triangle if its area is $$S = 600.$$

### Example 5

Find the height $$h$$ of a tree if the angle of elevation of its top changes from $$\alpha = 25^\circ$$ to $$\beta = 40^\circ$$ degrees as the observer advances $$a = 10\,\text{meters}$$ towards the tree.

### Example 6

The angles of a triangle are in the proportion $$2:3:4.$$ Find the unknown sides of the triangle if its least side is $$10\,\text{cm}.$$

### Example 1.

A triangle has the sides $$a = 30,$$ $$c = 50,$$ and angle $$\alpha = 25^\circ.$$ Find the angle $$\gamma$$ if it is the largest angle of the triangle.

Solution.

To calculate the angle $$\gamma,$$ we use the Law of Sines:

$\frac{a}{{\sin \alpha }} = \frac{c}{{\sin \gamma }}, \Rightarrow \sin \gamma = \frac{{c\sin \alpha }}{a} = \frac{{50\sin 25^\circ }}{{30}} = \frac{{50 \times 0,4226}}{{30}} = 0,7043.$

If $$\sin \gamma = 0,7043,$$ the angle $$\gamma$$ can be equal either to $$\gamma = 44,77^\circ,$$ or to $$\gamma = 135,23^\circ.$$

In the first case, the third angle $$\beta$$ will be equal to

$\beta = 180^\circ - \alpha - \gamma = 180^\circ - 25^\circ - 44,77^\circ = 110,23^\circ,$

that is, the angle $$\beta$$ is the largest angle of the triangle, which contradicts to the condition of the problem.

Hence, the answer is $$\gamma = 135,23^\circ.$$

### Example 2.

A triangle has the side $$a = 5\,\text{cm}$$ and angles $$\alpha = 50^\circ$$ and $$\gamma = 100^\circ.$$ Calculate the length of the side $$b.$$

Solution.

By the Sine Rule:

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }}.$

The angle $$\beta$$ is written as

$\beta = 180^\circ - \alpha - \gamma .$

Then

$b = \frac{{a\sin \beta }}{{\sin \alpha }} = \frac{{a\sin \left( {180^\circ - \alpha - \gamma } \right)}}{{\sin \alpha }}.$

Use the reduction identity:

$\sin \left( {180^\circ - \alpha - \gamma } \right) = \sin \left( {\alpha + \gamma } \right).$

Hence,

$b = \frac{{a\sin \left( {180^\circ - \alpha - \gamma } \right)}}{{\sin \alpha }} = \frac{{a\sin \left( {\alpha + \gamma } \right)}}{{\sin \alpha }} = \frac{{5\sin \left( {50^\circ + 100^\circ } \right)}}{{\sin 50^\circ }} = \frac{{5\sin 150^\circ }}{{\sin 50^\circ }} = \frac{{5 \times 0,5}}{{0,7660}} \approx 3,26\,\text{cm}.$

### Example 3.

Find the distance $$x$$ across the lake shown in Figure $$7,$$ if $$BC = a = 100\,\text{m},$$ $$\beta = 85^\circ,$$ and $$\gamma = 75^\circ.$$

Solution.

The angle $$\alpha$$ is equal to

$\alpha = 180^\circ - \left( {\beta + \gamma } \right).$

Using the Sine Rule, we get

$\frac{x}{{\sin \gamma }} = \frac{a}{{\sin \alpha }}, \;\Rightarrow \frac{x}{{\sin \gamma }} = \frac{a}{{\sin \left( {180^\circ - \left( {\beta + \gamma } \right)} \right)}}.$

By the reduction identity,

$\sin \left( {180^\circ - \left( {\beta + \gamma } \right)} \right) = \sin \left( {\beta + \gamma } \right).$

Hence, the distance $$x$$ between the points $$A$$ and $$B$$ is given by

$x = \frac{{a\sin \gamma }}{{\sin \left( {\beta + \gamma } \right)}} = \frac{{100\sin 75^\circ }}{{\sin \left( {85^\circ + 75^\circ } \right)}} = \frac{{100\sin 75^\circ }}{{\sin 160^\circ }} = \frac{{100 \times 0,9659}}{{0,3420}} = 282,43\,\text{m} \approx 282\,\text{m}.$

### Example 4.

The side lengths of a triangle are $$a = 30,$$ $$b = 40,$$ and $$c = 50.$$ Find the angles of the triangle if its area is $$S = 600.$$

Solution.

The area of a triangle is expressed by the formula

$S = \frac{{abc}}{{4R}},$

where $$R$$ is the radius of the circumscribed circle.

We can determine the radius $$R$$ from this formula:

$R = \frac{{abc}}{{4S}} = \frac{{30 \times 40 \times 50}}{{4 \times 600}} = 25.$

Now, using the Extended Law of Sines we find the angles $$\alpha,$$ $$\beta,$$ and $$\gamma:$$

$\frac{a}{{\sin \alpha }} = 2R, \;\Rightarrow \sin \alpha = \frac{a}{{2R}} = \frac{{30}}{{2 \times 25}} = \frac{{30}}{{50}} = \frac{3}{5}, \;\Rightarrow \alpha = \arcsin \frac{3}{5} \approx 36,87^\circ ;$
$\frac{b}{{\sin \beta }} = 2R, \;\Rightarrow \sin \beta = \frac{b}{{2R}} = \frac{{40}}{{2 \times 25}} = \frac{{40}}{{50}} = \frac{4}{5}, \;\Rightarrow \beta = \arcsin \frac{4}{5} \approx 53,13^\circ ;$
$\frac{c}{{\sin \gamma }} = 2R, \;\Rightarrow \sin \gamma = \frac{c}{{2R}} = \frac{{50}}{{2 \times 25}} = \frac{{50}}{{50}} = \frac{5}{5} = 1, \;\Rightarrow \gamma = 90^\circ .$

### Example 5.

Find the height $$h$$ of a tree if the angle of elevation of its top changes from $$\alpha = 25^\circ$$ to $$\beta = 40^\circ$$ degrees as the observer advances $$a = 10\,\text{meters}$$ towards the tree.

Solution.

Consider the triangle $$\triangle BCA.$$ In this triangle, the angle $$\angle CBA$$ is equal to $$180^\circ - \beta.$$ Then the angle $$\angle BCA$$ is given by

$\angle BCA = 180^\circ - (180^\circ - \beta + \alpha ) = \cancel{{180^\circ }} - \cancel{{180^\circ }} + \beta - \alpha = \beta - \alpha .$

Apply the Law of Sines to the triangle $$\triangle BCA:$$

$\frac{{AB}}{{\sin \left( {\beta - \alpha } \right)}} = \frac{{CB}}{{\sin \alpha }}, \;\Rightarrow \frac{a}{{\sin \left( {\beta - \alpha } \right)}} = \frac{{CB}}{{\sin \alpha }}, \;\Rightarrow CB = \frac{{a\sin \alpha }}{{\sin \left( {\beta - \alpha } \right)}}.$

The height of the tree is determined from the right triangle $$\triangle BDC:$$

$h = CD = CB\sin \beta = \frac{{a\sin \alpha \sin \beta }}{{\sin \left( {\beta - \alpha } \right)}}.$

Substitute the given values and calculate the height of the tree:

$h = \frac{{10\sin 25^\circ \sin 40^\circ }}{{\sin \left( {40^\circ - 25^\circ } \right)}} = \frac{{10\sin 25^\circ \sin 40^\circ }}{{\sin 15^\circ }} = \frac{{10 \cdot 0,4226 \cdot 0,6428}}{{0,2588}} \approx 10,50\,\text{m}.$

### Example 6.

The angles of a triangle are in the proportion $$2:3:4.$$ Find the unknown sides of the triangle if its least side is $$10\,\text{cm}.$$

Solution.

Let's assume that the angles of the triangle are given by

$\alpha = 2x,\;\beta = 3x,\;\gamma = 4x.$

Since the sum of angles of a triangle is $$180^\circ,$$ we get:

$\alpha + \beta + \gamma = 180^\circ , \;\Rightarrow 2x + 3x + 4x = 180^\circ , \;\Rightarrow 9x = 180^\circ , \;\Rightarrow x = 20^\circ ,$

so the angles are measured in degrees as follows:

$\alpha = 2x = 40^\circ ,\;\beta = 3x = 60^\circ ,\;\gamma = 4x = 80^\circ .$

By the Law of Sines, we can write

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }} = \frac{c}{{\sin \gamma }},$

where $$a,b,c$$ are the lengths of the sides opposite to the angles $$\alpha, \beta, \gamma,$$ respectively.

The shortest side is opposite the shortest angle, that is $$a = 10\,\text{cm}.$$ Find the two other sides:

$\frac{a}{{\sin \alpha }} = \frac{b}{{\sin \beta }}, \Rightarrow b = \frac{{a\sin \beta }}{{\sin \alpha }} = \frac{{a\sin 60^\circ }}{{\sin 40^\circ }} = \frac{{10 \times 0,8660}}{{0,6428}} = 13,47\,\text{cm};$
$\frac{a}{{\sin \alpha }} = \frac{c}{{\sin \gamma }}, \;\Rightarrow c = \frac{{a\sin \gamma }}{{\sin \alpha }} = \frac{{a\sin 80^\circ }}{{\sin 40^\circ }} = \frac{{2a\cancel{{\sin 40^\circ }}\cos 40^\circ }}{{\cancel{{\sin 40^\circ }}}} = 2a\cos 40^\circ = 20 \times 0,7660 = 15,32\,\text{cm}.$