The question asks for the value of the expression \(\frac{\text{AB}}{\sin \text{C}}\) in a triangle ABC with given side lengths AB = 16 m, BC = 10 m, and CA = 10 m.

Understanding the Sine Rule in Triangle ABC

This expression, \(\frac{\text{AB}}{\sin \text{C}}\), is directly related to the Sine Rule (also known as the Law of Sines). The Sine Rule for any triangle ABC states:

\(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\)

Where:

• \(a\) is the length of the side opposite vertex A (which is BC).
• \(b\) is the length of the side opposite vertex B (which is CA).
• \(c\) is the length of the side opposite vertex C (which is AB).

In our specific triangle ABC:

• Side \(a\) = BC = 10 m
• Side \(b\) = CA = 10 m
• Side \(c\) = AB = 16 m

The expression we need to find is \(\frac{\text{AB}}{\sin \text{C}}\), which is \(\frac{c}{\sin C}\). According to the Sine Rule, this ratio is equal to \(\frac{a}{\sin A}\) and \(\frac{b}{\sin B}\).

So, \(\frac{\text{AB}}{\sin \text{C}} = \frac{16}{\sin C} = \frac{10}{\sin A} = \frac{10}{\sin B}\).

Applying Sine Rule and Properties of Triangle ABC

From the Sine Rule ratios, we see that \(\frac{10}{\sin A} = \frac{10}{\sin B}\). This implies that \(\sin A = \sin B\). In a triangle, if the sines of two angles are equal, the angles themselves are equal (since angles in a triangle are between 0° and 180°, and \(\sin x = \sin y\) implies \(x=y\) or \(x+y=180°\); \(A+B=180°\) would mean C=0°, which is not a valid triangle). Therefore, angle A = angle B.

A triangle with two equal angles (A = B) is an isosceles triangle. This matches the given side lengths: CA = BC = 10 m. The sides opposite equal angles are equal.

We can use the Sine Rule expression \(\frac{\text{AB}}{\sin \text{C}} = \frac{\text{CA}}{\sin \text{B}}\) to find the required value. We know AB = 16, CA = 10. We need to find \(\sin B\).

Calculating sin B using the Cosine Rule

We can find the cosine of angle B using the Cosine Rule (Law of Cosines) in triangle ABC. The Cosine Rule relating to angle B is:

\(b^2 = a^2 + c^2 - 2ac \cos B\)

Substituting the values:

\(10^2 = 10^2 + 16^2 - 2(10)(16) \cos B\)

\(100 = 100 + 256 - 320 \cos B\)

\(100 - 100 - 256 = -320 \cos B\)

\(-256 = -320 \cos B\)

\(\cos B = \frac{-256}{-320} = \frac{256}{320}\)

Simplifying the fraction \(\frac{256}{320}\):

\(\cos B = \frac{256 \div 64}{320 \div 64} = \frac{4}{5}\)

Now that we have \(\cos B\), we can find \(\sin B\) using the identity \(\sin^2 B + \cos^2 B = 1\):

\(\sin^2 B + \left(\frac{4}{5}\right)^2 = 1\)

\(\sin^2 B + \frac{16}{25} = 1\)

\(\sin^2 B = 1 - \frac{16}{25} = \frac{25 - 16}{25} = \frac{9}{25}\)

\(\sin B = \sqrt{\frac{9}{25}} = \frac{3}{5}\) (Since B is an angle in a triangle, \(\sin B\) is positive).

Finding AB/sin C using the Sine Rule and sin B

Now we use the Sine Rule equality \(\frac{\text{AB}}{\sin \text{C}} = \frac{\text{CA}}{\sin \text{B}}\):

\(\frac{\text{AB}}{\sin \text{C}} = \frac{10}{\sin B}\)

Substitute the value of \(\sin B = \frac{3}{5}\):

\(\frac{\text{AB}}{\sin \text{C}} = \frac{10}{3/5}\)

\(\frac{\text{AB}}{\sin \text{C}} = 10 \times \frac{5}{3}\)

\(\frac{\text{AB}}{\sin \text{C}} = \frac{50}{3}\)

Alternative Calculation using sin C

We could also directly calculate \(\sin C\). First find \(\cos C\) using the Cosine Rule:

\(c^2 = a^2 + b^2 - 2ab \cos C\)

\(16^2 = 10^2 + 10^2 - 2(10)(10) \cos C\)

\(256 = 100 + 100 - 200 \cos C\)

\(256 = 200 - 200 \cos C\)

\(256 - 200 = -200 \cos C\)

\(56 = -200 \cos C\)

\(\cos C = -\frac{56}{200} = -\frac{7}{25}\)

Now find \(\sin C\): \(\sin^2 C + \cos^2 C = 1\)

\(\sin^2 C + \left(-\frac{7}{25}\right)^2 = 1\)

\(\sin^2 C + \frac{49}{625} = 1\)

\(\sin^2 C = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625}\)

\(\sin C = \sqrt{\frac{576}{625}} = \frac{24}{25}\) (Angle C in an isosceles triangle with sides 10, 10, 16 is obtuse, but its sine is still positive).

Finally, calculate \(\frac{\text{AB}}{\sin \text{C}}\):

\(\frac{16}{\sin C} = \frac{16}{24/25} = 16 \times \frac{25}{24} = \frac{16 \times 25}{24} = \frac{2 \times 25}{3} = \frac{50}{3}\)

Both methods yield the same result.

The information about the lamp post is extra information and not required to solve the specific question about \(\frac{\text{AB}}{\sin \text{C}}\).

Final Answer

The value of \(\frac{\text{AB}}{\sin \text{C}}\) is \(\frac{50}{3}\) m.

Revision Table: Key Concepts

Additional Information: Circumradius Connection

The ratio \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\) is also equal to \(2R\), where \(R\) is the circumradius of the triangle (the radius of the circle that passes through all three vertices of the triangle). In this problem, \(\frac{\text{AB}}{\sin \text{C}} = \frac{c}{\sin C}\) represents \(2R\). So, the circumradius of triangle ABC is \(R = \frac{1}{2} \times \frac{50}{3} = \frac{25}{3}\) m.

The lamp post information likely pertains to another part of a larger problem, possibly involving heights, distances, or angles of elevation/depression, but it is not needed for this specific calculation using the triangle's sides and angles.