In an isosceles triangle PQR, `/_P = 130^@`. If I is the in-centre of the triangle, what is the value (in degrees) of `/_QIR`?

Options

A. 130
B. 120
C. 155
D. 165

Answer

To find the value of angle \( \angle QIR \) in the isosceles triangle \( PQR \) where \( \angle P = 130^\circ \), we can follow these steps:

### Step 1: Determine the angles \( \angle Q \) and \( \angle R \)

Since triangle \( PQR \) is isosceles with \( \angle P = 130^\circ \), the other two angles \( \angle Q \) and \( \angle R \) must be equal. We know that the sum of angles in a triangle is \( 180^\circ \).

\[ \angle P + \angle Q + \angle R = 180^\circ \]
\[ 130^\circ + \angle Q + \angle Q = 180^\circ \]
\[ 130^\circ + 2\angle Q = 180^\circ \]
\[ 2\angle Q = 180^\circ - 130^\circ \]
\[ 2\angle Q = 50^\circ \]
\[ \angle Q = 25^\circ \]
\[ \angle R = 25^\circ \]

### Step 2: Use the property of the incenter

The incenter \( I \) of a triangle is the point where the angle bisectors of the triangle intersect. The angle \( \angle QIR \) can be calculated using the formula:

\[ \angle QIR = \frac{1}{2} (\angle Q + \angle R) \]

### Step 3: Substitute the values of \( \angle Q \) and \( \angle R \)

Since \( \angle Q = 25^\circ \) and \( \angle R = 25^\circ \):

\[ \angle QIR = \frac{1}{2} (25^\circ + 25^\circ) \]
\[ \angle QIR = \frac{1}{2} (50^\circ) \]
\[ \angle QIR = 25^\circ \]

### Step 4: Calculate \( \angle QIR \)

To find \( \angle QIR \) in terms of \( \angle P \):

Using the property of the incenter, we can also express it as:

\[ \angle QIR = 90^\circ + \frac{\angle P}{2} \]

Substituting \( \angle P = 130^\circ \):

\[ \angle QIR = 90^\circ + \frac{130^\circ}{2} \]
\[ \angle QIR = 90^\circ + 65^\circ \]
\[ \angle QIR = 155^\circ \]

### Final Answer

Thus, the value of \( \angle QIR \) is \( 155^\circ \).

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