Solving for the Length of CD in a Right Triangle

The question asks us to find the length of the segment CD in a right-angled triangle ABC, where D is a point on the hypotenuse AB, and we are given the lengths of the two perpendicular sides and the measure of angle BCD.

Understanding the Triangle ABC

We have a right-angled triangle ABC. Let's assume the right angle is at C. We are given:

• BC = 5 cm
• AC = 12 cm

By the Pythagorean theorem, the length of the hypotenuse AB is:

\(AB^2 = BC^2 + AC^2\)

\(AB^2 = 5^2 + 12^2 = 25 + 144 = 169\)

\(AB = \sqrt{169} = 13\) cm.

So, the hypotenuse AB has a length of 13 cm.

Analyzing Point D and Angles

Point D lies on the hypotenuse AB. We are given that \(\angle \text{BCD} = 30^\circ\). Since \(\triangle \text{ABC}\) is right-angled at C, \(\angle \text{ACB} = 90^\circ\).

The angle \(\angle \text{ACD}\) can be found by subtracting \(\angle \text{BCD}\) from \(\angle \text{ACB}\):

\(\angle \text{ACD} = \angle \text{ACB} - \angle \text{BCD}\)

\(\angle \text{ACD} = 90^\circ - 30^\circ = 60^\circ\).

Now we have partitioned the triangle ABC into two smaller triangles, \(\triangle \text{ACD}\) and \(\triangle \text{BCD}\), sharing the common side CD.

Using the Area Method to Find CD

The area of the larger triangle ABC is equal to the sum of the areas of the two smaller triangles ACD and BCD.

Area(\(\triangle \text{ABC}\)) = Area(\(\triangle \text{ACD}\)) + Area(\(\triangle \text{BCD}\))

Calculate Area(\(\triangle \text{ABC}\))

Since \(\triangle \text{ABC}\) is a right-angled triangle, its area is half the product of its perpendicular sides:

Area(\(\triangle \text{ABC}\)) \( = \frac{1}{2} \times \text{BC} \times \text{AC}\)

Area(\(\triangle \text{ABC}\)) \( = \frac{1}{2} \times 5 \times 12 = \frac{60}{2} = 30\) square cm.

Express Area(\(\triangle \text{ACD}\)) in terms of CD

The area of \(\triangle \text{ACD}\) can be calculated using the formula: Area \( = \frac{1}{2} ab \sin C\), where a and b are the lengths of two sides and C is the angle between them.

Area(\(\triangle \text{ACD}\)) \( = \frac{1}{2} \times \text{AC} \times \text{CD} \times \sin(\angle \text{ACD})\)

Area(\(\triangle \text{ACD}\)) \( = \frac{1}{2} \times 12 \times \text{CD} \times \sin(60^\circ)\)

Area(\(\triangle \text{ACD}\)) \( = 6 \times \text{CD} \times \frac{\sqrt{3}}{2}\)

Area(\(\triangle \text{ACD}\)) \( = 3\sqrt{3} \times \text{CD}\).

Express Area(\(\triangle \text{BCD}\)) in terms of CD

Similarly, for \(\triangle \text{BCD}\):

Area(\(\triangle \text{BCD}\)) \( = \frac{1}{2} \times \text{BC} \times \text{CD} \times \sin(\angle \text{BCD})\)

Area(\(\triangle \text{BCD}\)) \( = \frac{1}{2} \times 5 \times \text{CD} \times \sin(30^\circ)\)

Area(\(\triangle \text{BCD}\)) \( = \frac{5}{2} \times \text{CD} \times \frac{1}{2}\)

Area(\(\triangle \text{BCD}\)) \( = \frac{5}{4} \times \text{CD}\).

Set up the Equation and Solve for CD

Using the area relationship:

Area(\(\triangle \text{ABC}\)) = Area(\(\triangle \text{ACD}\)) + Area(\(\triangle \text{BCD}\))

\(30 = 3\sqrt{3} \times \text{CD} + \frac{5}{4} \times \text{CD}\)

Factor out CD:

\(30 = \text{CD} \left( 3\sqrt{3} + \frac{5}{4} \right)\)

Combine the terms inside the parenthesis by finding a common denominator:

\(3\sqrt{3} + \frac{5}{4} = \frac{3\sqrt{3} \times 4}{4} + \frac{5}{4} = \frac{12\sqrt{3} + 5}{4}\)

Substitute this back into the equation:

\(30 = \text{CD} \left( \frac{12\sqrt{3} + 5}{4} \right)\)

Now, solve for CD:

\(\text{CD} = \frac{30}{\left( \frac{12\sqrt{3} + 5}{4} \right)}\)

\(\text{CD} = 30 \times \frac{4}{12\sqrt{3} + 5}\)

\(\text{CD} = \frac{120}{12\sqrt{3} + 5}\)

Rearranging the denominator for better readability:

\(\text{CD} = \frac{120}{5 + 12\sqrt{3}}\) cm.

Conclusion

The length of the segment CD is \(\frac{120}{5 + 12\sqrt{3}}\) cm. This method, using the sum of areas of the two smaller triangles being equal to the area of the larger triangle, is effective when a segment divides a triangle into two, and the angles formed by the segment with one side are known.

Revision Table: Right Triangle Geometry

Additional Information: Other Approaches

While the area method is straightforward here, other trigonometric methods could also be used, although they might be more complex. For example, one could use the Sine Rule in \(\triangle \text{BCD}\) and \(\triangle \text{ACD}\).

In \(\triangle \text{ABC}\), we can find \(\sin A\), \(\cos A\), \(\sin B\), \(\cos B\). \(\tan A = 5/12\), \(\tan B = 12/5\).

Using Sine Rule in \(\triangle \text{BCD}\): \(\frac{\text{CD}}{\sin B} = \frac{\text{BC}}{\sin(\angle \text{BDC})}\). So, \(\text{CD} = \frac{5 \sin B}{\sin(\angle \text{BDC})}\).

Using Sine Rule in \(\triangle \text{ACD}\): \(\frac{\text{CD}}{\sin A} = \frac{\text{AC}}{\sin(\angle \text{ADC})}\). So, \(\text{CD} = \frac{12 \sin A}{\sin(\angle \text{ADC})}\).

Note that \(\angle \text{BDC}\) and \(\angle \text{ADC}\) are supplementary, so \(\sin(\angle \text{BDC}) = \sin(\angle \text{ADC})\). Also, \(\angle \text{ADC} = 180^\circ - (\angle A + \angle \text{ACD}) = 180^\circ - (A + 60^\circ)\). \(\angle \text{BDC} = 180^\circ - (\angle B + \angle \text{BCD}) = 180^\circ - (B + 30^\circ)\).

This approach involves finding \(\sin A\) and \(\sin B\) from the original triangle and dealing with the unknown angles at D, which might be more involved than the area method.

The area method simplifies the problem significantly by leveraging the known angles at C and the known side lengths AC and BC.