Question

1. john is building a doghouse. he decides to use the roof truss design shown. if ( mangle dbf = 35^circ ), what is the measure of the vertex angle of the isosceles triangle?

Explanation:

Step1: Find angle at F

Since \( \angle DBF = 35^\circ \) and \( \angle BDF = 90^\circ \) (right angle), in triangle \( DBF \), \( \angle F = 90^\circ - 35^\circ = 55^\circ \).

Step2: Analyze isosceles triangle

The isosceles triangle here is \( \triangle ABG \) (or the main isosceles triangle with base \( FG \))? Wait, actually, from the diagram, \( AB = AC \) (marked with ticks), and \( \angle F \) and \( \angle G \) should be equal (since the roof truss is symmetric). Wait, no, first, \( \angle F = 55^\circ \), so \( \angle G = 55^\circ \) (symmetry, as the truss is symmetric about the vertical line through \( A \)). Then, the vertex angle at \( A \) (or the vertex angle of the isosceles triangle \( \triangle AFG \)) is \( 180^\circ - 2\times55^\circ = 70^\circ \)? Wait, no, wait the angle at \( B \) and \( C \): Wait, the angle \( \angle FBC \) (the base angle of the isosceles triangle \( \triangle ABC \))? Wait, maybe I misread. Wait, the isosceles triangle is \( \triangle ABC \) with \( AB = AC \), and \( \angle F = 55^\circ \), so \( \angle F = \angle G = 55^\circ \), then the angle at \( B \) ( \( \angle ABC \)) is \( 180^\circ - 55^\circ = 125^\circ \)? No, wait, \( \angle DBF = 35^\circ \), so \( \angle ABF = 90^\circ \) (since \( BD \perp AF \)), so \( \angle ABF = 90^\circ \), so \( \angle ABC = 180^\circ - 90^\circ - 35^\circ \)? No, better: \( \angle F = 55^\circ \), so in \( \triangle AFG \), which is isosceles? Wait, the problem says "the isosceles triangle" – probably \( \triangle ABC \) or \( \triangle AFG \). Wait, let's re-express:

Given \( \angle DBF = 35^\circ \), \( BD \perp AF \), so \( \angle BDF = 90^\circ \), so \( \angle F = 55^\circ \). Since the truss is symmetric, \( \angle G = 55^\circ \). Then the vertex angle of the isosceles triangle ( \( \triangle AFG \)) is \( 180^\circ - 2 \times 55^\circ = 70^\circ \). Wait, but maybe the isosceles triangle is \( \triangle ABC \). Wait, the angle at \( B \) and \( C \): \( \angle ABC = 180^\circ - \angle F - \angle ABF \)? No, \( \angle ABF = 90^\circ \) (because \( BD \) is perpendicular to \( AF \)), so \( \angle ABF = 90^\circ \), so \( \angle ABC = 180^\circ - 90^\circ = 90^\circ \)? No, I'm confused. Wait, let's start over.

1. In right triangle \( DBF \), \( \angle DBF = 35^\circ \), \( \angle BDF = 90^\circ \), so \( \angle F = 90^\circ - 35^\circ = 55^\circ \).
2. The roof truss is symmetric, so \( \angle G = \angle F = 55^\circ \).
3. The isosceles triangle here is \( \triangle AFG \) (with \( AF = AG \)? Wait, no, \( AB = AC \), but the base is \( FG \), and the two equal sides are \( AF \) and \( AG \)? Wait, no, the vertex angle is at \( A \), between \( AF \) and \( AG \). So the sum of angles in \( \triangle AFG \) is \( 180^\circ \), so vertex angle \( \angle A = 180^\circ - 2 \times 55^\circ = 70^\circ \).

Wait, but maybe the isosceles triangle is \( \triangle ABC \). Wait, \( AB = AC \), so \( \angle ABC = \angle ACB \). \( \angle F = 55^\circ \), so \( \angle ABC = 180^\circ - 55^\circ = 125^\circ \)? No, that can't be. Wait, the angle at \( B \): \( \angle FBC \) is a straight line, so \( \angle FBD + \angle DBA + \angle ABC = 180^\circ \). \( \angle FBD = 35^\circ \), \( \angle DBA = 90^\circ \), so \( \angle ABC = 180 - 35 - 90 = 55^\circ \)? No, that's not. Wait, I think I made a mistake. Let's look at the diagram again: \( BD \perp AF \), \( CE \perp AG \), \( AB = AC \), \( FB = CG \) (maybe). So \( \angle F = \angle G \), and \( \angle DBF = 35^\circ \), so \( \angle F = 55^\circ \), so \( \angle G = 55^\circ \). Then the vertex angle of the isosceles triangle ( \( \triangle AFG \)) is \( 180 - 2*55 = 70^\circ \). So the measure of the vertex angle is \( 70^\circ \).

Wait, let's confirm:

- \( \angle DBF = 35^\circ \), \( \angle BDF = 90^\circ \), so \( \angle F = 55^\circ \).
- By symmetry, \( \angle G = \angle F = 55^\circ \).
- In \( \triangle AFG \), which is isosceles ( \( AF = AG \) or \( AB = AC \)? Wait, the problem says "the isosceles triangle" – likely \( \triangle AFG \) with \( AF = AG \), so vertex angle at \( A \) is \( 180 - 2*55 = 70^\circ \).

So the vertex angle is \( 70^\circ \).

Answer:

\( 70^\circ \)