In an isosceles triangle `DEF, /_D = 110^@`. If I is the in centre of the triangle, then what is the value (in degrees) of `/_EIF`?

Options

A. `110`
B. `130 `
C. `145 `
D. `155 `

Answer

To find the value of angle \( \angle EIF \) in the isosceles triangle \( DEF \) where \( \angle D = 110^\circ \), we can follow these steps:

### Step 1: Understand the properties of the triangle
In an isosceles triangle, two sides are equal, and the angles opposite those sides are also equal. Here, since \( D \) is the vertex angle, the base angles \( E \) and \( F \) are equal.

### Step 2: Set up the equation for the angles
The sum of the angles in any triangle is \( 180^\circ \). Therefore, we can set up the equation:
\[ \angle D + \angle E + \angle F = 180^\circ \]
Substituting the known value:
\[ 110^\circ + \angle E + \angle F = 180^\circ \]
Since \( \angle E = \angle F \), we can denote \( \angle E = \angle F = x \):
\[ 110^\circ + 2x = 180^\circ \]

### Step 3: Solve for \( x \)
Rearranging the equation gives:
\[ 2x = 180^\circ - 110^\circ \]
\[ 2x = 70^\circ \]
\[ x = 35^\circ \]
Thus, \( \angle E = \angle F = 35^\circ \).

### Step 4: Determine the angles at the incenter
The incenter \( I \) of a triangle is the point where the angle bisectors of the triangle intersect. The angles \( \angle E \) and \( \angle F \) are bisected by the lines from \( I \) to \( E \) and \( I \) to \( F \), respectively.

Since \( \angle E = 35^\circ \), the angle at \( I \) from \( E \) is:
\[ \angle EID = \frac{35^\circ}{2} = 17.5^\circ \]
Similarly, for \( \angle F \):
\[ \angle FID = \frac{35^\circ}{2} = 17.5^\circ \]

### Step 5: Calculate \( \angle EIF \)
In triangle \( EIF \):
\[ \angle EIF + \angle EID + \angle FID = 180^\circ \]
Substituting the known values:
\[ \angle EIF + 17.5^\circ + 17.5^\circ = 180^\circ \]
\[ \angle EIF + 35^\circ = 180^\circ \]
\[ \angle EIF = 180^\circ - 35^\circ = 145^\circ \]

### Final Answer
Thus, the value of \( \angle EIF \) is \( 145^\circ \).

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