In ∆ABC, ∠B = 68° and ∠C = 32°. Sides AB and AC are produced to points D and E respectively. The bisectors of ∠DBC and ∠BCE meet at F. what is the measure of ∠BFC?
50°
Understanding the Geometry Problem: Finding Angle BFC
This problem asks us to find the measure of an angle formed by the bisectors of two exterior angles of a triangle. We are given a triangle $\triangle ABC$ with the measures of two interior angles, $\angle B$ and $\angle C$. Sides AB and AC are extended to form exterior angles, and the bisectors of these exterior angles meet at point F. We need to determine the measure of $\angle BFC$.
Step-by-Step Solution for Triangle Angles
First, let's find the measure of the third interior angle, $\angle A$, in $\triangle ABC$. The sum of the interior angles in any triangle is always $180^\circ$.
We are given:
- $\angle B = 68^\circ$
- $\angle C = 32^\circ$
Using the angle sum property of a triangle:
\(\angle A + \angle B + \angle C = 180^\circ\)
\(\angle A + 68^\circ + 32^\circ = 180^\circ\)
\(\angle A + 100^\circ = 180^\circ\)
\(\angle A = 180^\circ - 100^\circ\)
\(\angle A = 80^\circ\)
Calculating the Exterior Angles of Triangle ABC
When a side of a triangle is produced, it forms an exterior angle. An exterior angle and its adjacent interior angle are supplementary (they add up to $180^\circ$).
The side AB is produced to point D, forming the exterior angle $\angle DBC$. This angle is supplementary to the interior angle $\angle B$.
\(\angle DBC = 180^\circ - \angle B\)
\(\angle DBC = 180^\circ - 68^\circ\)
\(\angle DBC = 112^\circ\)
The side AC is produced to point E, forming the exterior angle $\angle BCE$. This angle is supplementary to the interior angle $\angle C$.
\(\angle BCE = 180^\circ - \angle C\)
\(\angle BCE = 180^\circ - 32^\circ\)
\(\angle BCE = 148^\circ\)
Finding Angles using Exterior Angle Bisectors
We are told that BF is the bisector of $\angle DBC$ and CF is the bisector of $\angle BCE$. A bisector divides an angle into two equal halves.
Since BF bisects $\angle DBC$, the angle $\angle FBC$ is half of $\angle DBC$:
\(\angle FBC = \frac{1}{2} \angle DBC\)
\(\angle FBC = \frac{1}{2} \times 112^\circ\)
\(\angle FBC = 56^\circ\)
Since CF bisects $\angle BCE$, the angle $\angle FCB$ is half of $\angle BCE$:
\(\angle FCB = \frac{1}{2} \angle BCE\)
\(\angle FCB = \frac{1}{2} \times 148^\circ\)
\(\angle FCB = 74^\circ\)
Determining the Measure of Angle BFC
Now, consider the triangle $\triangle BFC$. The sum of the interior angles in $\triangle BFC$ is $180^\circ$. The angles in $\triangle BFC$ are $\angle BFC$, $\angle FBC$, and $\angle FCB$.
\(\angle BFC + \angle FBC + \angle FCB = 180^\circ\)
We have calculated $\angle FBC = 56^\circ$ and $\angle FCB = 74^\circ$. Substitute these values into the equation:
\(\angle BFC + 56^\circ + 74^\circ = 180^\circ\)
\(\angle BFC + 130^\circ = 180^\circ\)
Subtract $130^\circ$ from both sides to find $\angle BFC$:
\(\angle BFC = 180^\circ - 130^\circ\)
\(\angle BFC = 50^\circ\)
Summary of Calculations
| Angle | Calculation | Measure |
|---|---|---|
| $\angle A$ | $180^\circ - (\angle B + \angle C)$ | $180^\circ - (68^\circ + 32^\circ) = 80^\circ$ |
| $\angle DBC$ (Exterior) | $180^\circ - \angle B$ | $180^\circ - 68^\circ = 112^\circ$ |
| $\angle BCE$ (Exterior) | $180^\circ - \angle C$ | $180^\circ - 32^\circ = 148^\circ$ |
| $\angle FBC$ (Bisector of $\angle DBC$) | $\frac{1}{2} \angle DBC$ | $\frac{1}{2} \times 112^\circ = 56^\circ$ |
| $\angle FCB$ (Bisector of $\angle BCE$) | $\frac{1}{2} \angle BCE$ | $\frac{1}{2} \times 148^\circ = 74^\circ$ |
| $\angle BFC$ | $180^\circ - (\angle FBC + \angle FCB)$ | $180^\circ - (56^\circ + 74^\circ) = 50^\circ$ |
The measure of $\angle BFC$ is $50^\circ$.
Revision Table: Key Concepts
| Concept | Description | Formula/Property |
|---|---|---|
| Sum of Interior Angles of a Triangle | The sum of the measures of the three interior angles of any triangle is $180^\circ$. | $\angle A + \angle B + \angle C = 180^\circ$ |
| Exterior Angle of a Triangle | An exterior angle is formed by extending one side of a triangle. It is supplementary to the adjacent interior angle. | Exterior Angle = $180^\circ$ - Adjacent Interior Angle |
| Angle Bisector | A line segment or ray that divides an angle into two equal angles. | If BF bisects $\angle DBC$, then $\angle FBC = \angle FBD = \frac{1}{2} \angle DBC$. |
| Angle Sum Property in $\triangle BFC$ | The sum of angles in the triangle formed by the bisectors. | $\angle BFC + \angle FBC + \angle FCB = 180^\circ$ |
Additional Information: Exterior Angle Bisector Theorem
There is a theorem that directly relates the angle formed by the bisectors of two exterior angles of a triangle (like $\angle BFC$) to the third interior angle ($\angle A$).
The theorem states that the angle formed by the bisectors of two exterior angles of a triangle is equal to $90^\circ$ minus half of the third interior angle.
In our case, this means:
\(\angle BFC = 90^\circ - \frac{1}{2} \angle A\)
We calculated $\angle A = 80^\circ$. Let's use this formula to verify our result:
\(\angle BFC = 90^\circ - \frac{1}{2} \times 80^\circ\)
\(\angle BFC = 90^\circ - 40^\circ\)
\(\angle BFC = 50^\circ\)
This formula confirms our step-by-step calculation is correct. This theorem is useful for solving such problems quickly if you remember the formula.
Important Questions from Lines and Angles
If angles of a triangle are in the ration of 2 : 3 : 4, then the measure of the smallest angle is:
In the triangle, if AB = AC and ∠ABC = 72°, then ∠BAC is:
The angles of a triangle are (8x - 15)°,(6x - 11)° and ( 4x – 10)°. What is the value of x ?
In a ΔABC, the bisectors of ∠B and ∠C meet at point O, inside the triangle. If ∠BOC = 122°, then the measure of ∠A is:
In ΔABC, D is a point on side BC such that ∠ADC = 2∠BAD. If ∠A = 80° and ∠C = 38°, then what is the measure of ∠ADB?
