Question

find h.
write your answer as an integer or as a decimal rounded to the ne
h =

Explanation:

Step1: Find angle at G

In triangle \( FGH \), the sum of angles in a triangle is \( 180^\circ \). We know \( \angle F = 57^\circ \) and \( \angle H = 25^\circ \), so \( \angle G=180 - 57 - 25=98^\circ \)

Step2: Apply the Law of Sines

The Law of Sines states that \( \frac{h}{\sin H}=\frac{HG}{\sin F} \). We know \( HG = 12 \), \( \angle H = 25^\circ \), \( \angle F = 57^\circ \). So \( h=\frac{12\times\sin(25^\circ)}{\sin(57^\circ)} \)
Calculate \( \sin(25^\circ)\approx0.4226 \), \( \sin(57^\circ)\approx0.8387 \)
Then \( h=\frac{12\times0.4226}{0.8387}=\frac{5.0712}{0.8387}\approx6.05 \) (rounded to two decimal places, if we round to nearest tenth it's \( 6.1 \), but let's check calculation again)
Wait, maybe I mixed up the sides. Wait, \( HG = 12 \) is opposite to \( \angle F = 57^\circ \), and \( h \) is opposite to \( \angle H = 25^\circ \). So Law of Sines: \( \frac{h}{\sin H}=\frac{HG}{\sin F} \)
So \( h=\frac{HG\times\sin H}{\sin F}=\frac{12\times\sin(25^\circ)}{\sin(57^\circ)} \)
\( \sin(25^\circ)\approx0.4226 \), \( \sin(57^\circ)\approx0.8387 \)
\( 12\times0.4226 = 5.0712 \)
\( 5.0712\div0.8387\approx6.05 \), if we round to nearest tenth, it's \( 6.1 \), but maybe the problem expects more precise or maybe I made a mistake in angle. Wait, wait, maybe the triangle is labeled differently. Wait, \( \angle F = 57^\circ \), \( \angle H = 25^\circ \), so \( \angle G = 180 - 57 - 25 = 98^\circ \). Wait, maybe \( HG = 12 \) is side \( g \) (opposite \( \angle G \))? Wait, no, in standard notation, side \( a \) is opposite \( \angle A \), side \( b \) opposite \( \angle B \), side \( c \) opposite \( \angle C \). So in triangle \( FGH \), side \( f \) is opposite \( \angle F \), side \( g \) opposite \( \angle G \), side \( h \) opposite \( \angle H \). Wait, maybe I got the sides wrong. Wait, the side \( HG \) is between \( H \) and \( G \), so it's side \( f \) (opposite \( \angle F \))? Wait, no, vertex \( F \), \( G \), \( H \). So side \( FG = h \) (between \( F \) and \( G \)), side \( GH = f \) (between \( G \) and \( H \)), side \( FH = g \) (between \( F \) and \( H \)). So \( \angle F = 57^\circ \), \( \angle H = 25^\circ \), \( \angle G = 98^\circ \), side \( GH = 12 \) (which is side \( f \), opposite \( \angle F \)). So Law of Sines: \( \frac{h}{\sin H}=\frac{f}{\sin F} \), so \( h=\frac{f\times\sin H}{\sin F}=\frac{12\times\sin25^\circ}{\sin57^\circ} \)
Calculating again: \( \sin25^\circ\approx0.4226 \), \( \sin57^\circ\approx0.8387 \)
\( 12\times0.4226 = 5.0712 \)
\( 5.0712\div0.8387\approx6.05 \), so approximately \( 6.1 \) (if rounded to nearest tenth) or \( 6.0 \) (if to nearest integer). Wait, maybe the problem has a typo or I misread the angle. Wait, maybe the angle at \( F \) is \( 57^\circ \), angle at \( H \) is \( 25^\circ \), so angle at \( G \) is \( 98^\circ \), side \( GH = 12 \). Alternatively, maybe it's a different approach. Wait, maybe using Law of Sines correctly:

\( \frac{h}{\sin(25^\circ)}=\frac{12}{\sin(57^\circ)} \)

So \( h = \frac{12\times\sin(25^\circ)}{\sin(57^\circ)} \)

\( \sin(25^\circ)\approx0.4226 \), \( \sin(57^\circ)\approx0.8387 \)

\( h\approx\frac{12\times0.4226}{0.8387}\approx\frac{5.0712}{0.8387}\approx6.05 \), so \( h\approx6.1 \) (rounded to nearest tenth) or \( 6 \) (nearest integer). But let's check with calculator:

\( \sin(25) = \sin(25^\circ) \approx 0.4226182617 \)

\( \sin(57) = \sin(57^\circ) \approx 0.8386705679 \)

\( 12 * 0.4226182617 = 5.07141914 \)

\( 5.07141914 / 0.8386705679 ≈ 6.047 \), so approximately \( 6.0 \) (if integer) or \( 6.0 \) (nearest tenth is \( 6.0 \) as \( 6.047 \) is closer to \( 6.0 \) than \( 6.1 \))

Wait, maybe I made a mistake in angle sum. Wait, \( 57 + 25 = 82 \), \( 180 - 82 = 98 \), that's correct. So angle at \( G \) is \( 98^\circ \). Then side \( GH = 12 \) is opposite \( \angle F = 57^\circ \), side \( FG = h \) is opposite \( \angle H = 25^\circ \). So Law of Sines is correct. So the answer is approximately \( 6.0 \) (or \( 6 \))

Answer:

\( \boxed{6.0} \) (or \( 6 \) if integer is preferred)