Solution of Triangles: Distance of orthocentre from the vertices
Solution of Triangles: Distance of Orthocentre from the Vertices
In the study of triangles, particularly in trigonometry, the concept of the orthocentre is a fundamental one. The orthocentre of a triangle is the point where the three altitudes of the triangle intersect. An altitude of a triangle is a perpendicular line segment from a vertex to the line containing the opposite side. The orthocentre can lie inside the triangle, on the triangle, or outside the triangle, depending on whether the triangle is acute, right, or obtuse, respectively.
Properties of the Orthocentre
Before we delve into the distances from the orthocentre to the vertices, let's review some important properties of the orthocentre:
- The orthocentre is the common point of intersection of the altitudes of the triangle.
- In an acute triangle, the orthocentre lies inside the triangle.
- In a right triangle, the orthocentre is the vertex at the right angle.
- In an obtuse triangle, the orthocentre lies outside the triangle.
Distance of Orthocentre from the Vertices
The distance of the orthocentre from the vertices of a triangle can be calculated using various methods, including trigonometric formulas and coordinate geometry. Here, we will focus on the trigonometric approach.
Trigonometric Formulas
To find the distance of the orthocentre (H) from the vertices (A, B, and C) of a triangle ABC, we can use the following formulas:
Distance from vertex A to the orthocentre (AH): $$ AH = 2R \cos A $$
Distance from vertex B to the orthocentre (BH): $$ BH = 2R \cos B $$
Distance from vertex C to the orthocentre (CH): $$ CH = 2R \cos C $$
Where:
- ( A, B, C ) are the angles at vertices A, B, and C, respectively.
- ( R ) is the circumradius of the triangle (the radius of the circumscribed circle).
Table of Differences and Important Points
| Property | Acute Triangle | Right Triangle | Obtuse Triangle |
|---|---|---|---|
| Location of Orthocentre | Inside the triangle | At the right-angled vertex | Outside the triangle |
| AH (Distance from A to H) | ( 2R \cos A ) | 0 (since H is at A) | ( 2R \cos A ) (A is obtuse) |
| BH (Distance from B to H) | ( 2R \cos B ) | Hypotenuse (since H is at A) | ( 2R \cos B ) |
| CH (Distance from C to H) | ( 2R \cos C ) | Hypotenuse (since H is at A) | ( 2R \cos C ) |
Examples
Let's go through some examples to understand how to apply these formulas.
Example 1: Acute Triangle
Consider an acute triangle ABC with angles ( A = 60^\circ ), ( B = 50^\circ ), and ( C = 70^\circ ). Let the circumradius ( R ) be 10 units.
Distance from A to H: $$ AH = 2 \times 10 \times \cos 60^\circ = 20 \times \frac{1}{2} = 10 \text{ units} $$
Distance from B to H: $$ BH = 2 \times 10 \times \cos 50^\circ \approx 20 \times 0.6428 \approx 12.856 \text{ units} $$
Distance from C to H: $$ CH = 2 \times 10 \times \cos 70^\circ \approx 20 \times 0.3420 \approx 6.840 \text{ units} $$
Example 2: Right Triangle
Consider a right triangle ABC with angle ( A = 90^\circ ), ( B = 60^\circ ), and ( C = 30^\circ ). Let the circumradius ( R ) be 5 units.
Distance from A to H: $$ AH = 2 \times 5 \times \cos 90^\circ = 0 \text{ units (since H is at A)} $$
Distance from B to H: $$ BH = 2 \times 5 \times \cos 60^\circ = 10 \times \frac{1}{2} = 5 \text{ units (hypotenuse)} $$
Distance from C to H: $$ CH = 2 \times 5 \times \cos 30^\circ = 10 \times \frac{\sqrt{3}}{2} \approx 8.660 \text{ units (hypotenuse)} $$
Example 3: Obtuse Triangle
Consider an obtuse triangle ABC with angle ( A = 120^\circ ), ( B = 30^\circ ), and ( C = 30^\circ ). Let the circumradius ( R ) be 7 units.
Distance from A to H: $$ AH = 2 \times 7 \times \cos 120^\circ = 14 \times (-\frac{1}{2}) = -7 \text{ units (outside the triangle)} $$
Distance from B to H: $$ BH = 2 \times 7 \times \cos 30^\circ = 14 \times \frac{\sqrt{3}}{2} \approx 12.124 \text{ units} $$
Distance from C to H: $$ CH = 2 \times 7 \times \cos 30^\circ = 14 \times \frac{\sqrt{3}}{2} \approx 12.124 \text{ units} $$
In conclusion, the distances from the orthocentre to the vertices of a triangle can be calculated using trigonometric formulas involving the angles at the vertices and the circumradius of the triangle. These distances provide insight into the geometric properties and relationships within the triangle.