Question

what is the value of v?
h
g
v
62°
f
v = °

Explanation:

Step1: Identify triangle type

The triangle has two equal sides (marked with ticks), so it's isosceles. In an isosceles triangle, the angles opposite the equal sides are equal.

Step2: Determine equal angles

Side \( FH \) and \( FG \) are equal, so angles opposite them (\( \angle G \) and \( \angle H \))? Wait, no, angle at \( F \) is \( 62^\circ \), sides \( FH \) and \( FG \)? Wait, no, the equal sides are \( FH \) (from \( F \) to \( H \)) and \( FG \) (from \( F \) to \( G \))? Wait, no, the marks are on \( FH \) (midway? No, the tick marks: one on \( FH \), one on \( FG \). Wait, maybe \( FH = FG \)? Wait, no, the triangle is \( \triangle FHG \)? Wait, vertices are \( F \), \( H \), \( G \). The side \( FH \) has a tick, side \( FG \) has a tick? Wait, no, the side from \( F \) to \( H \) ( \( FH \)) and from \( G \) to \( F \) ( \( FG \))? Wait, no, the two equal sides: let's see, the angle at \( F \) is \( 62^\circ \), and the sides adjacent to \( F \) ( \( FH \) and \( FG \)) have ticks, so \( FH = FG \). Therefore, the triangle is isosceles with \( FH = FG \), so the base angles (opposite these sides) would be \( \angle G \) and \( \angle H \)? Wait, no, in a triangle, the angles opposite equal sides are equal. So if \( FH = FG \), then the angles opposite them are \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)). Wait, maybe I got the vertices wrong. Let's label the triangle: \( F \) is the vertex with \( 62^\circ \), \( H \) is one vertex, \( G \) is the other. The sides \( FH \) and \( FG \) are equal (ticks), so \( \triangle FHG \) is isosceles with \( FH = FG \). Therefore, angles at \( H \) and \( G \)? Wait, no, angle at \( F \) is \( 62^\circ \), so the other two angles (at \( H \) and \( G \)): wait, no, if \( FH = FG \), then the angles opposite are \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)). Wait, maybe I made a mistake. Wait, the sum of angles in a triangle is \( 180^\circ \). If \( FH = FG \), then \( \angle H = \angle G \)? Wait, no, let's correct: in \( \triangle FHG \), sides \( FH \) and \( FG \) are equal, so the angles opposite them are \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)). Wait, maybe the equal sides are \( FH \) and \( HG \)? Wait, the tick is on \( FH \) (from \( F \) to \( H \)) and on \( HG \) (from \( H \) to \( G \))? Oh! That's probably it. So \( FH = HG \), so the triangle is isosceles with \( FH = HG \), so the angles opposite them are \( \angle G \) (opposite \( FH \)) and \( \angle F \) (opposite \( HG \))? No, angle at \( F \) is \( 62^\circ \), so if \( FH = HG \), then angle at \( G \) (opposite \( FH \)) is equal to angle at \( F \) (opposite \( HG \))? Wait, no, angle at \( F \) is \( 62^\circ \), side \( FH \) and \( HG \) are equal, so angle opposite \( FH \) is \( \angle G \), angle opposite \( HG \) is \( \angle F \). Wait, this is confusing. Let's start over.

In a triangle, the sum of angles is \( 180^\circ \). If two sides are equal (isosceles), the base angles are equal. Let's assume the two equal sides are \( FH \) and \( HG \) (ticks on \( FH \) and \( HG \)). Then the base is \( FG \), and the base angles are \( \angle F \) and \( \angle G \). Wait, angle at \( F \) is \( 62^\circ \), so angle at \( G \) ( \( v \)) would also be \( 62^\circ \)? No, that can't be, because then the third angle would be \( 180 - 62 - 62 = 56^\circ \), but that doesn't match. Wait, maybe the equal sides are \( FH \) and \( FG \), so the base is \( HG \), and the base angles are \( \angle H \) and \( \angle G \). Then angle at \( F \) is \( 62^\circ \), so \( \angle H + \angle G = 180 - 62 = 118^\circ \). Since \( \angle H = \angle G \) (isosceles), then each is \( 118 / 2 = 59^\circ \)? No, that's not right. Wait, maybe I misread the tick marks. Let's look at the diagram again: the tick is on \( FH \) (the side from \( F \) to \( H \)) and on \( FG \) (the side from \( F \) to \( G \))? No, the tick on \( FH \) is closer to \( F \), and the tick on \( FG \) is closer to \( G \)? Wait, no, the two tick marks: one on \( FH \) (the vertical side) and one on \( FG \) (the side from \( F \) to \( G \), the slant side). So \( FH = FG \), so the triangle is isosceles with \( FH = FG \), so angles opposite these sides are \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)). Wait, angle at \( F \) is \( 62^\circ \), so \( \angle H + \angle G + 62^\circ = 180^\circ \). Since \( FH = FG \), \( \angle H = \angle G \). Therefore, \( 2v + 62 = 180 \)? Wait, no, if \( \angle H = \angle G = v \), then \( v + v + 62 = 180 \), so \( 2v = 118 \), \( v = 59 \)? No, that's not matching. Wait, maybe the equal sides are \( FH \) and \( HG \), so \( FH = HG \), then angle at \( F \) ( \( 62^\circ \)) is equal to angle at \( G \) ( \( v \)), so \( v = 62^\circ \). But then the third angle would be \( 180 - 62 - 62 = 56^\circ \), but that's angle at \( H \). But the tick marks are on \( FH \) and \( HG \), so \( FH = HG \), so angles opposite are \( \angle G \) (opposite \( FH \)) and \( \angle F \) (opposite \( HG \)). So \( \angle G = \angle F = 62^\circ \), so \( v = 62^\circ \). Wait, maybe that's it. Let's confirm: in an isosceles triangle, if two sides are equal, the angles opposite them are equal. So if \( FH = HG \), then angle opposite \( FH \) is \( \angle G \), angle opposite \( HG \) is \( \angle F \). Therefore, \( \angle G = \angle F = 62^\circ \), so \( v = 62^\circ \).

Step3: Calculate \( v \)

Wait, no, maybe I had the sides wrong. Let's look at the diagram again: the vertex with \( 62^\circ \) is \( F \), the side from \( F \) to \( H \) ( \( FH \)) has a tick, and the side from \( G \) to \( H \) ( \( GH \)) has a tick. So \( FH = GH \), so the triangle is isosceles with \( FH = GH \), so the angles opposite these sides are \( \angle G \) (opposite \( FH \)) and \( \angle F \) (opposite \( GH \)). Therefore, \( \angle G = \angle F = 62^\circ \), so \( v = 62^\circ \). Wait, but that would make the triangle have angles \( 62^\circ \), \( 62^\circ \), and \( 56^\circ \). But maybe that's correct. Alternatively, if the equal sides are \( FH \) and \( FG \), then angles at \( H \) and \( G \) are equal, so \( 2v + 62 = 180 \), so \( v = (180 - 62)/2 = 59 \). But which is it? Let's check the tick marks: the tick on \( FH \) (from \( F \) to \( H \)) and the tick on \( FG \) (from \( F \) to \( G \)): so \( FH = FG \), so the base is \( HG \), and the base angles are \( \angle H \) and \( \angle G \). Then angle at \( F \) is \( 62^\circ \), so \( \angle H + \angle G = 118^\circ \), and since \( \angle H = \angle G \), each is \( 59^\circ \). But the diagram shows the angle at \( F \) is \( 62^\circ \), and the side \( FH \) and \( FG \) are equal. Wait, maybe the correct approach is: in an isosceles triangle, the angles opposite the equal sides are equal. So if two sides are equal, their opposite angles are equal. Let's label the triangle: \( F \), \( H \), \( G \). \( \angle F = 62^\circ \), \( FH = FG \) (ticks), so opposite angles: \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)) are equal. So \( \angle G = \angle H \). Then \( \angle F + \angle G + \angle H = 180 \), so \( 62 + 2v = 180 \), so \( 2v = 118 \), \( v = 59 \). Wait, that makes sense. So \( v = 59^\circ \).
Wait, I think I made a mistake earlier. Let's do it properly:

1. Identify the isosceles triangle: two sides are equal (ticks), so \( FH = FG \).
2. Therefore, angles opposite these sides: \( \angle G \) (opposite \( FH \)) and \( \angle H \) (opposite \( FG \)) are equal. Let \( \angle G = \angle H = v \).
3. Sum of angles in a triangle: \( 62 + v + v = 180 \).
4. Solve: \( 62 + 2v = 180 \) → \( 2v = 180 - 62 = 118 \) → \( v = 59 \).

Yes, that's correct. So the value of \( v \) is \( 59^\circ \).

Answer:

\( 59 \)