Calculating the Angle at the Incentre of a Triangle

The question asks us to find the measure of the angle ∠BIC, where I is the incentre of triangle ABC and the measure of ∠A is given as 88°.

The incentre of a triangle is the point where the three angle bisectors of the triangle meet. An angle bisector divides an angle into two equal parts. The incenter is also the center of the inscribed circle (incircle) of the triangle.

There is a standard formula that relates the angle formed at the incenter by two angle bisectors (like ∠BIC) to the angle of the triangle at the third vertex (in this case, ∠A). The formula for ∠BIC is:

$$ \text{∠BIC} = 90^\circ + \frac{1}{2} \text{∠A} $$

Let's use this formula to calculate ∠BIC given that ∠A = 88°.

Step-by-Step Incentre Angle Calculation

We are given:

• Triangle ABC
• I is the incentre
• ∠A = 88°

We need to find ∠BIC.

Using the formula: $$ \text{∠BIC} = 90^\circ + \frac{1}{2} \text{∠A} $$

Substitute the value of ∠A into the formula: $$ \text{∠BIC} = 90^\circ + \frac{1}{2} \times 88^\circ $$

First, calculate half of ∠A: $$ \frac{1}{2} \times 88^\circ = \frac{88}{2}^\circ = 44^\circ $$

Now, add this value to 90°: $$ \text{∠BIC} = 90^\circ + 44^\circ $$ $$ \text{∠BIC} = 134^\circ $$

Therefore, the measure of ∠BIC is 134°.

Revision Table: Key Incenter Formulas

Additional Information on Incenter Properties

• Definition: The incenter (I) is the intersection point of the three angle bisectors of a triangle.
• Equidistance: The incenter is equidistant from the sides of the triangle. The distance from the incenter to each side is the radius of the incircle.
• Location: The incenter always lies inside the triangle, regardless of whether the triangle is acute, right, or obtuse.
• Angle Bisectors: Lines segment BI and CI are angle bisectors of ∠B and ∠C respectively. Thus, ∠IBC = ∠ABI = $\frac{1}{2}\text{∠B}$ and ∠ICB = ∠ACI = $\frac{1}{2}\text{∠C}$.
• Derivation of Formula: The formula ∠BIC = $90^\circ + \frac{1}{2}\text{∠A}$ can be derived using the angle sum property of triangle BIC and the fact that BI and CI are angle bisectors. In ▵BIC, ∠BIC + ∠IBC + ∠ICB = 180°. Substituting ∠IBC = $\frac{1}{2}\text{∠B}$ and ∠ICB = $\frac{1}{2}\text{∠C}$, we get ∠BIC + $\frac{1}{2}\text{∠B}$ + $\frac{1}{2}\text{∠C}$ = 180°. From triangle ABC, ∠A + ∠B + ∠C = 180°, so $\frac{1}{2}(\text{∠B} + \text{∠C}) = \frac{1}{2}(180^\circ - \text{∠A}) = 90^\circ - \frac{1}{2}\text{∠A}$. Substituting this back, ∠BIC + $90^\circ - \frac{1}{2}\text{∠A}$ = 180°. Rearranging gives ∠BIC = 180° - 90° + $\frac{1}{2}\text{∠A}$ = $90^\circ + \frac{1}{2}\text{∠A}$.