Question

$$ \begin{aligned} &\text { The lengths of the sides } \mathrm{BC}, \mathrm{CA}, \mathrm{AB} \text { of a triangle } \mathrm{ABC}, \text { are } 5,12 \text { and } 15 \text { respectively. If } \mathrm{x}, \mathrm{y}, \mathrm{z} \text { are lengths of the internal }\\\ &\text { bisectors of angles } A, B, C \text { of the triangle, obtain the value of } \frac{1}{x} \cos \frac{A}{2}+\frac{1}{y} \cos \frac{B}{2}+\frac{1}{z} \cos \frac{C}{2} \end{aligned} $$

Short answer

Answer: The value of the given expression is \(\frac{1}{2}\).

Step-by-step solution

Create a relationship between the angle bisectors and sides of the triangle

Using the angle bisector theorem, we can find the relationship between angle bisectors and sides of the triangle. The theorem states that the ratio of sides of the triangle is equal to the ratio of angle bisectors. Therefore, we can write the relationship as: $$\frac{x}{y}=\frac{BC}{CA}\text{ or }\frac{x}{y}=\frac{5}{12}$$ Similarly, we can find the relationships for the other angle bisectors: $$\frac{y}{z}=\frac{CA}{AB}\text{ or }\frac{y}{z}=\frac{12}{15} \quad \text{and} \quad \frac{z}{x}=\frac{AB}{BC}\text{ or }\frac{z}{x}=\frac{15}{5}$$

Find the value of ratio of sin function angles

Using the sine rule, we can find the ratio of sines function angles to the respective angle bisectors. The sine rule for angle bisectors is: $$\frac{\sin \frac{A}{2}}{x} = \frac{\sin \frac{B}{2}}{y} = \frac{\sin \frac{C}{2}}{z}$$

Use the sine-cosine relationship

We know that \(\cos \frac{A}{2} = \sqrt{\frac{1+\cos A}{2}}\). The same relationship can be used for the other two angles. We can rewrite the given expression as follows: $$\frac{1}{x} \cos \frac{A}{2}+\frac{1}{y} \cos \frac{B}{2}+\frac{1}{z} \cos \frac{C}{2}=\frac{\sin \frac{A}{2}\cos\frac{A}{2}}{\sin \frac{A}{2}x}+\frac{\sin \frac{B}{2}\cos\frac{B}{2}}{\sin \frac{B}{2}y}+\frac{\sin \frac{C}{2}\cos\frac{C}{2}}{\sin \frac{C}{2}z}$$ Now use the sine-cosine relationship for the angle bisector for all the angles: $$\sin \frac{A}{2}=\frac{2\sin\frac{B}{2}\sin\frac{C}{2}}{sin(B+C)}$$ $$\sin \frac{B}{2}=\frac{2\sin\frac{A}{2}\sin\frac{C}{2}}{sin(A+C)}$$ $$\sin \frac{C}{2}=\frac{2\sin\frac{A}{2}\sin\frac{B}{2}}{sin(A+B)}$$

Calculate the value of the given expression

Now let's substitute the sine expressions into the equation in Step 3: $$\frac{\sin \frac{A}{2}\cos \frac{A}{2}}{2\sin \frac{B}{2}\sin \frac{C}{2}}+\frac{\sin \frac{B}{2}\cos \frac{B}{2}}{2\sin \frac{A}{2}\sin \frac{C}{2}}+\frac{\sin \frac{C}{2}\cos \frac{C}{2}}{2\sin \frac{A}{2}\sin \frac{B}{2}}=\frac{1}{2}(\frac{\sin 2\frac{A}{2}}{\sin{B+C}}+\frac{\sin 2\frac{B}{2}}{\sin{C+A}}+\frac{\sin 2\frac{C}{2}}{\sin{A+B}})$$ We know from trigonometry that, $$\sum_{cyclic} \frac{\sin {2\theta}}{\sin {(\pi-2\theta)}} = \frac{\sin {2\theta}}{\sin {\pi}\sin {2(\pi-\theta)}} = \frac{1}{2}$$ Therefore, the value of the given expression is, $$\frac{1}{x} \cos \frac{A}{2}+\frac{1}{y} \cos \frac{B}{2}+\frac{1}{z} \cos \frac{C}{2}=\frac{1}{2}$$

Key concepts

Angle Bisector Theorem

The Angle Bisector Theorem is a fundamental principle in triangle geometry. It describes a relationship between the angles and sides of a triangle. The theorem states that in any triangle, the bisector of an angle divides the opposite side into two segments. These segments are proportional to the other two sides of the triangle. Here's how it works:

• If you have a triangle ABC, and you draw a bisector from angle A, it will meet the side BC at some point D.
• According to the theorem, \( \frac{BD}{DC} = \frac{AB}{AC} \).

This theorem helps in solving various geometric problems, especially when the side lengths and angle bisectors come into play. In our specific problem, we use it to find relationships between the internal bisectors x, y, z, and the sides of the triangle.
These relations were shown as \( \frac{x}{y} = \frac{5}{12} \), and similarly for other bisectors.

Sine Rule

The Sine Rule, also known as the Law of Sines, is another key concept in solving triangle problems. It expresses a relationship between the lengths of the sides of a triangle and the sines of its angles.

• The rule states that for any triangle ABC, \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \).
• This means that the ratio of the length of a side to the sine of its opposite angle is the same for all three sides and angles in the triangle.

In the context of our problem, we adapted the sine rule to apply to the internal bisectors.
This gives us \( \frac{\sin \frac{A}{2}}{x} = \frac{\sin \frac{B}{2}}{y} = \frac{\sin \frac{C}{2}}{z} \), helping further to connect trigonometric properties with the geometry of the triangle.

Trigonometric Identities

Trigonometric Identities are formulas involving trigonometric functions that are true for any value of the occurring variables. They are essential in simplifying expressions involving angles. Some basics include the Pythagorean identities, angle sum and difference identities, and double angle identities.

For our problem, the double angle identity for cosine helps greatly:

• \( \cos \frac{A}{2} = \sqrt{\frac{1 + \cos A}{2}} \).
• It allows the expression \( \frac{1}{x} \cos \frac{A}{2} + \frac{1}{y} \cos \frac{B}{2} + \frac{1}{z} \cos \frac{C}{2} \) to be simplified.

Because the sine and cosine functions are interconnected through identities, the sine-cosine relationship \( \sin^2 \theta + \cos^2 \theta = 1 \) gets involved in calculations. This deepens understanding of how angle measures relate to side measures, providing smooth pathways to simplify and solve complex trigonometric equations.