A circle is inscribed in a triangle ABC. It touches the sides AB and AC at M and N respectively. If O is the centre of the circle and ∠A = 70°, then what is ∠MON equal to?
110°
Finding the Angle at the Center of an Inscribed Circle
This problem involves a circle inscribed within a triangle, a common topic in geometry. We are given a triangle ABC with a circle inscribed inside it. The circle touches two sides, AB and AC, at points M and N respectively. O is the center of this inscribed circle. We know the angle at vertex A is 70°, and we need to find the measure of the angle ∠MON.
Understanding the Geometry
When a circle is inscribed in a triangle, the sides of the triangle are tangent to the circle. The points where the circle touches the sides are called points of tangency. In this case, M is the point of tangency on AB, and N is the point of tangency on AC.
A key property of circles and tangents is that the radius drawn to the point of tangency is perpendicular to the tangent line. Since O is the center of the circle and OM and ON are radii drawn to the points of tangency M and N, we know that:
- OM is perpendicular to AB. This means ∠AMO = 90°.
- ON is perpendicular to AC. This means ∠ANO = 90°.
Analyzing Quadrilateral AMON
Consider the quadrilateral formed by the points A, M, O, and N. The sum of the interior angles in any quadrilateral is 360°. The four angles in quadrilateral AMON are:
- ∠A (given as 70°)
- ∠AMO (which is 90°)
- ∠MON (which we need to find)
- ∠ANO (which is 90°)
So, the sum of these angles is:
\[ \angle A + \angle AMO + \angle MON + \angle ANO = 360^\circ \]
Calculating ∠MON
Now we can substitute the known values into the equation:
\[ 70^\circ + 90^\circ + \angle MON + 90^\circ = 360^\circ \]
Combine the known angles:
\[ 70^\circ + 180^\circ + \angle MON = 360^\circ \]
\[ 250^\circ + \angle MON = 360^\circ \]
To find ∠MON, subtract 250° from 360°:
\[ \angle MON = 360^\circ - 250^\circ \]
\[ \angle MON = 110^\circ \]
Therefore, the measure of angle MON is 110°.
Step-by-Step Solution
- Identify the quadrilateral AMON formed by vertex A, points of tangency M and N, and the center O of the inscribed circle.
- Recall that radii OM and ON are perpendicular to the tangents AB and AC at the points of tangency M and N, respectively. Thus, ∠AMO = 90° and ∠ANO = 90°.
- Recall that the sum of interior angles in a quadrilateral is 360°.
- Write the equation for the sum of angles in quadrilateral AMON: ∠A + ∠AMO + ∠MON + ∠ANO = 360°.
- Substitute the known values: 70° + 90° + ∠MON + 90° = 360°.
- Solve for ∠MON: ∠MON = 360° - (70° + 90° + 90°) = 360° - 250° = 110°.
Summary Table of Angles
| Angle | Value | Reason |
|---|---|---|
| ∠A | 70° | Given |
| ∠AMO | 90° | Radius is perpendicular to tangent at point of tangency |
| ∠ANO | 90° | Radius is perpendicular to tangent at point of tangency |
| ∠MON | 110° | Calculated |
| Sum of angles in AMON | \(70^\circ + 90^\circ + 110^\circ + 90^\circ = 360^\circ\) | Property of quadrilateral |
The calculated value of ∠MON is 110°.
Revision Table: Circle Inscribed in a Triangle
| Concept | Description | Relevance to Problem |
|---|---|---|
| Inscribed Circle | A circle tangent to all three sides of a triangle. Its center is the incenter (intersection of angle bisectors). | The problem involves an inscribed circle. |
| Point of Tangency | The single point where a tangent line touches a circle. | M and N are points of tangency. |
| Radius to Tangent | The radius drawn from the circle's center to the point of tangency is perpendicular to the tangent line. | Used to determine ∠AMO and ∠ANO are 90°. |
| Quadrilateral Angle Sum | The sum of interior angles in any quadrilateral is 360°. | Used to find ∠MON in quadrilateral AMON. |
Additional Information: Incenter and Angle Bisectors
The center O of the inscribed circle is also known as the incenter of the triangle. The incenter is the point where the angle bisectors of the triangle intersect.
In this problem, AO is the angle bisector of ∠A. However, this property was not necessary to solve for ∠MON using the quadrilateral method.
The relationship between ∠A and ∠MON is also a specific property related to the incenter. For an inscribed circle touching sides AB and AC at M and N, the angle subtended by the points of tangency at the center (∠MON) is supplementary to the angle at the opposite vertex (∠A), provided M and N are points on the tangents from A. More generally, for tangents from a point A to a circle at M and N, the angle ∠MAN and ∠MON (where O is the center) are supplementary. Since AMON is a cyclic quadrilateral in this specific scenario (two right angles opposite each other), this is a useful property. However, the quadrilateral angle sum method is a more general approach that applies here.
Let's verify the supplementary property: ∠A + ∠MON = 70° + 110° = 180°. This confirms the supplementary relationship holds, as expected.
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