Precalculus

Trigonometry

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Law of Sines

Solved Problems

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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.

A triangle given by the side a and angles alpha and gamma.
Figure 6.

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.

A triangle given by the side a and angles alpha and gamma.
Figure 7.

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.

Finding the distance across a lake using the law of sines
Figure 8.

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.

A triangle with sides a,b,c and angles alpha, beta, gamma
Figure 9.

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.

Determining the height of a tree using the law of sines
Figure 10.

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:

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