Question

If the bisectors of the acute angles of a right triangle meet at O, then the angle at O, between the two bisectors is :

• A. \(45^\circ\)
• B. \(95^\circ\)
• C. \(135^\circ\)
• D. \(90^\circ\)

Hint:

In any triangle, let the angles be A, B, C. If bisectors of \(\angle A\) and \(\angle C\) meet at O, then \(\angle AOC = 180^\circ - (\frac{\angle A}{2} + \frac{\angle C}{2})\). In a right triangle, one angle is \(90^\circ\). Let this be \(\angle B\). Then the sum of the two acute angles \(\angle A + \angle C = 90^\circ\). So, \(\frac{\angle A}{2} + \frac{\angle C}{2} = \frac{1}{2}(\angle A + \angle C) = \frac{1}{2}(90^\circ) = 45^\circ\). Therefore, \(\angle AOC = 180^\circ - 45^\circ = 135^\circ\). A useful shortcut: the angle between bisectors of two angles of a triangle is \(90^\circ + \frac{1}{2}(\text{third angle})\). Here, the third angle (the right angle) is \(90^\circ\). So \(90^\circ + \frac{90^\circ}{2} = 90^\circ + 45^\circ = 135^\circ\).

Answer 1

The Correct Option is C

Concept: This problem involves the properties of angles in a triangle, angle bisectors, and the sum of angles in a triangle. Step 1: Properties of a right triangle Let the right triangle be \(\triangle ABC\), with the right angle at \(B\) (\(\angle B = 90^\circ\)). The other two angles, \(\angle A\) and \(\angle C\), are acute angles. The sum of angles in a triangle is \(180^\circ\), so \(\angle A + \angle B + \angle C = 180^\circ\). Since \(\angle B = 90^\circ\), we have \(\angle A + 90^\circ + \angle C = 180^\circ\), which means \(\angle A + \angle C = 90^\circ\). Step 2: Angle bisectors Let AO be the bisector of \(\angle A\), and CO be the bisector of \(\angle C\). These bisectors meet at point O. This means:
\(\angle OAB = \angle OAC = \frac{1}{2}\angle A\)
\(\angle OCB = \angle OCA = \frac{1}{2}\angle C\) We are interested in the angle \(\angle AOC\), which is the angle at O between the two bisectors. Step 3: Consider the triangle \(\triangle AOC\) The sum of angles in \(\triangle AOC\) is \(180^\circ\): \[ \angle OAC + \angle OCA + \angle AOC = 180^\circ \] Substitute the expressions from the angle bisectors: \[ \left(\frac{1}{2}\angle A\right) + \left(\frac{1}{2}\angle C\right) + \angle AOC = 180^\circ \] \[ \frac{1}{2}(\angle A + \angle C) + \angle AOC = 180^\circ \] Step 4: Use the property \(\angle A + \angle C = 90^\circ\) Substitute \(\angle A + \angle C = 90^\circ\) into the equation from Step 3: \[ \frac{1}{2}(90^\circ) + \angle AOC = 180^\circ \] \[ 45^\circ + \angle AOC = 180^\circ \] Step 5: Solve for \(\angle AOC\) \[ \angle AOC = 180^\circ - 45^\circ \] \[ \angle AOC = 135^\circ \] Thus, the angle at O, between the two bisectors, is \(135^\circ\). General Formula Note: For any triangle \(\triangle ABC\), if the bisectors of \(\angle B\) and \(\angle C\) meet at O, then \(\angle BOC = 90^\circ + \frac{1}{2}\angle A\). In our case, we are looking at the angle formed by bisectors of acute angles A and C, with the third angle B being \(90^\circ\). So the angle at O (i.e., \(\angle AOC\)) can be thought of as \(90^\circ + \frac{1}{2}\angle B\). Since \(\angle B = 90^\circ\), then \(\angle AOC = 90^\circ + \frac{1}{2}(90^\circ) = 90^\circ + 45^\circ = 135^\circ\).