Solving Right-Angled Triangle Problems using Trigonometry

This problem involves a right-angled triangle, which can be solved using basic trigonometric ratios. We are given some information about the sides and angles of the triangle and asked to find the length of a specific side.

Understanding the Given Information

We have a triangle named ABC, which is right-angled at vertex B. This means that the angle at B is \(90^\circ\). We are given the following lengths and angles:

• Side AB = 4 cm
• Angle \(\angle \text{C} = 60^\circ\)

We need to find the length of side AC. In a right-angled triangle ABC, with the right angle at B:

• AC is the side opposite the right angle (B), so it is the hypotenuse.
• AB is the side opposite angle C.
• BC is the side adjacent to angle C.

Choosing the Correct Trigonometric Ratio

We know the length of the side opposite angle C (AB) and we want to find the length of the hypotenuse (AC). The trigonometric ratio that relates the opposite side and the hypotenuse is the sine function. The definition of sine for an angle \(\theta\) in a right-angled triangle is:

\[\sin(\theta) = \frac{\text{Opposite Side}}{\text{Hypotenuse}}\]

In our triangle, for angle C:

\[\sin(\text{C}) = \frac{\text{Side Opposite to C}}{\text{Hypotenuse}} = \frac{\text{AB}}{\text{AC}}\]

Applying the Sine Ratio

We are given \(\angle \text{C} = 60^\circ\) and AB = 4 cm. Substituting these values into the sine equation:

\[\sin(60^\circ) = \frac{4 \text{ cm}}{\text{AC}}\]

Finding the Value of \(\sin(60^\circ)\)

The sine of \(60^\circ\) is a standard trigonometric value for special angles. It is known that:

\[\sin(60^\circ) = \frac{\sqrt{3}}{2}\]

Here is a quick table of trigonometric values for common special angles:

Solving for AC

Now substitute the value of \(\sin(60^\circ)\) into our equation:

\[\frac{\sqrt{3}}{2} = \frac{4}{\text{AC}}\]

To solve for AC, we can cross-multiply:

\[\text{AC} \times \sqrt{3} = 4 \times 2\]

\[\text{AC} \times \sqrt{3} = 8\]

Now, divide both sides by \(\sqrt{3}\) to isolate AC:

\[\text{AC} = \frac{8}{\sqrt{3}}\]

Rationalizing the Denominator

It is standard practice to rationalize the denominator if it contains a square root. To rationalize \(\frac{8}{\sqrt{3}}\), we multiply both the numerator and the denominator by \(\sqrt{3}\):

\[\text{AC} = \frac{8}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}\]

\[\text{AC} = \frac{8 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}\]

\[\text{AC} = \frac{8\sqrt{3}}{3}\]

So, the length of AC is \(\frac{8\sqrt{3}}{3}\) cm.

Conclusion

Based on the trigonometric calculations for the right-angled triangle with AB = 4 cm and \(\angle \text{C} = 60^\circ\), the length of the hypotenuse AC is \(\frac{8\sqrt{3}}{3}\) cm.

Revision Table: Right-Angled Triangle Trigonometry

Additional Information: Special Angle Trigonometry

Understanding the trigonometric values for special angles like \(30^\circ\), \(45^\circ\), and \(60^\circ\) is crucial for solving many trigonometry problems quickly without a calculator. These values come from studying special right-angled triangles: the \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle and the \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle.

30-60-90 Triangle

Consider an equilateral triangle with side length 2. If you draw an altitude from one vertex to the opposite side, it bisects that side and the angle at the vertex. This creates a right-angled triangle with angles \(30^\circ\), \(60^\circ\), and \(90^\circ\). The sides will be in the ratio \(1 : \sqrt{3} : 2\) (opposite \(30^\circ\) : opposite \(60^\circ\) : opposite \(90^\circ\)).

• Side opposite \(30^\circ\) = 1
• Side opposite \(60^\circ\) = \(\sqrt{3}\)
• Hypotenuse (opposite \(90^\circ\)) = 2

Using these side lengths:

• \(\sin(30^\circ) = \frac{1}{2}\), \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), \(\tan(30^\circ) = \frac{1}{\sqrt{3}}\)
• \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), \(\cos(60^\circ) = \frac{1}{2}\), \(\tan(60^\circ) = \sqrt{3}\)

45-45-90 Triangle

Consider an isosceles right-angled triangle. The two acute angles are \(45^\circ\). If the equal sides have length 1, the hypotenuse can be found using the Pythagorean theorem: \(1^2 + 1^2 = c^2 \Rightarrow c^2 = 2 \Rightarrow c = \sqrt{2}\). The sides are in the ratio \(1 : 1 : \sqrt{2}\) (opposite \(45^\circ\) : opposite \(45^\circ\) : opposite \(90^\circ\)).

• Side opposite \(45^\circ\) = 1
• Side opposite \(45^\circ\) = 1
• Hypotenuse (opposite \(90^\circ\)) = \(\sqrt{2}\)

Using these side lengths:

• \(\sin(45^\circ) = \frac{1}{\sqrt{2}}\), \(\cos(45^\circ) = \frac{1}{\sqrt{2}}\), \(\tan(45^\circ) = 1\)

These special triangle ratios are fundamental for solving problems involving these specific angles quickly and accurately.