Question

find the tangent of ∠h.
simplify your answer and write it as a proper fraction, improper fraction, or whole number.
tan(h) =

Explanation:

Step1: Find length of FE

In right triangle FEG, we know hypotenuse \( FG = 68 \), hypotenuse \( FF = 60 \) (wait, correction: \( FG = 68 \), \( FE = 60 \), so by Pythagorean theorem, \( EG=\sqrt{FG^{2}-FE^{2}}=\sqrt{68^{2}-60^{2}} \). Calculate \( 68^{2}=4624 \), \( 60^{2}=3600 \), so \( EG=\sqrt{4624 - 3600}=\sqrt{1024}=32 \). Wait, but in the diagram, \( EG = 32 \) is given? Wait, no, actually, triangles FEG and FDH are similar (since both are right triangles and share angle F). Also, \( ED \) and \( EG \) are both perpendicular, so \( ED \parallel EG \)? Wait, maybe better to find the length of the opposite and adjacent sides for angle H.

Wait, angle H is in triangle FDH (right triangle at D) and triangle FEG (right triangle at E). Since \( \angle F \) is common, triangles FEG and FDH are similar. So \( \tan(H)=\tan(F) \) because \( \angle H + \angle F=90^{\circ} \)? Wait, no, in right triangle, \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For angle H, in triangle FDH (right-angled at D), the opposite side to H would be FD, and adjacent side would be DH? Wait, no, maybe I got the angles wrong. Wait, let's look at the diagram: points F, G, H are colinear (since G is on FH). E is on FD, with FE = 60, FG = 68, EG = 32 (right angle at E). D is on FD, right angle at D (so HD is perpendicular to FD). So angle at H: in triangle DFH, right-angled at D, so angle H and angle F are complementary. In triangle FEG, right-angled at E, \( \tan(F)=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \)? Wait, no, \( FE = 60 \), \( FG = 68 \), so \( EG = 32 \) (as calculated). Wait, but for angle H, in triangle DFH, \( \tan(H)=\frac{FD}{DH} \). But since triangles FEG and FDH are similar (both right-angled, same angle at F), so \( \frac{EG}{FE}=\frac{FD}{FH} \)? Wait, maybe not. Wait, alternatively, since \( \angle H \) and \( \angle F \) are complementary, \( \tan(H)=\cot(F)=\frac{FE}{EG} \)? Wait, no, \( \tan(F)=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \), so \( \tan(H)=\cot(F)=\frac{FE}{EG}=\frac{60}{32}=\frac{15}{8} \)? Wait, no, that can't be. Wait, maybe I mixed up opposite and adjacent.

Wait, let's re-express: in right triangle FEG (right at E), angle at F: opposite side is EG = 32, adjacent side is FE = 60. So \( \tan(F)=\frac{32}{60}=\frac{8}{15} \). In right triangle FDH (right at D), angle at H: opposite side is FD, adjacent side is DH. But since \( \angle F + \angle H = 90^{\circ} \), \( \tan(H)=\cot(F)=\frac{1}{\tan(F)}=\frac{60}{32}=\frac{15}{8} \)? Wait, no, that's not right. Wait, maybe I have the opposite and adjacent wrong for angle H. Let's consider angle H: in triangle DFH, right-angled at D, so the sides: FH is the hypotenuse, FD is one leg, DH is the other leg. Angle at H: the sides adjacent to H is DH, opposite is FD. So \( \tan(H)=\frac{FD}{DH} \). Now, in triangle FEG, right-angled at E, \( FE = 60 \), \( EG = 32 \), \( FG = 68 \). So \( FD = FE + ED \), but ED is equal to DH? Wait, no, EG is parallel to DH (both perpendicular to FD), so EG and DH are both perpendicular to FD, so EG || DH. Therefore, triangles FEG and FDH are similar (by AA similarity: right angles, common angle at F). Therefore, the ratio of corresponding sides is equal. So \( \frac{EG}{DH}=\frac{FE}{FD}=\frac{FG}{FH} \). Wait, \( FG = 68 \), \( FH = FG + GH \), but we don't know GH. Wait, but \( FE = 60 \), \( FG = 68 \), \( EG = 32 \). Let's compute \( FH \): in triangle FEG, \( FG = 68 \), \( FE = 60 \), \( EG = 32 \). Then, in triangle FDH, since EG || DH, \( \frac{EG}{DH}=\frac{FE}{FD}=\frac{FG}{FH} \). But we can also note that \( \tan(H)=\frac{FD}{DH} \), and from similar triangles, \( \frac{EG}{FE}=\frac{DH}{FD} \) (wait, maybe I had the ratio reversed). Let's do it properly:

In similar triangles FEG and FDH:

- Angle at F: common.
- Angle at E and angle at D: both right angles (90°).

Therefore, triangle FEG ~ triangle FDH (AA similarity).

Therefore, corresponding sides:

FE corresponds to FD,

EG corresponds to DH,

FG corresponds to FH.

Therefore, \( \frac{FE}{FD}=\frac{EG}{DH}=\frac{FG}{FH} \).

But we need \( \tan(H) \). In triangle FDH, \( \tan(H)=\frac{FD}{DH} \) (since in right triangle, tangent of angle H is opposite over adjacent: opposite is FD, adjacent is DH).

From the similarity ratio, \( \frac{FE}{EG}=\frac{FD}{DH} \) (since \( \frac{FE}{FD}=\frac{EG}{DH} \) implies \( \frac{FD}{DH}=\frac{FE}{EG} \) when cross-multiplied: \( FE \times DH = FD \times EG \) => \( \frac{FD}{DH}=\frac{FE}{EG} \)).

Wait, \( FE = 60 \), \( EG = 32 \), so \( \frac{FD}{DH}=\frac{60}{32}=\frac{15}{8} \)? But that seems off. Wait, no, in triangle FEG, \( \tan(F)=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \). In triangle FDH, \( \tan(H)=\frac{FD}{DH} \). But since angle F + angle H = 90°, \( \tan(H)=\cot(F)=\frac{1}{\tan(F)}=\frac{15}{8} \). Wait, but that would mean \( \tan(H)=\frac{15}{8} \), but let's check with the sides.

Wait, maybe I made a mistake in identifying the opposite and adjacent sides. Let's take angle H: in triangle DFH, right-angled at D, so the sides are:

- Hypotenuse: FH

- Leg 1: FD (opposite angle H)

- Leg 2: DH (adjacent angle H)

Therefore, \( \tan(H)=\frac{\text{opposite}}{\text{adjacent}}=\frac{FD}{DH} \).

Now, in triangle FEG, right-angled at E:

- Hypotenuse: FG = 68

- Leg 1: FE = 60 (adjacent to angle F)

- Leg 2: EG = 32 (opposite to angle F)

So \( \tan(F)=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \).

Since angle F + angle H = 90°, \( \tan(H)=\cot(F)=\frac{1}{\tan(F)}=\frac{15}{8} \). Wait, but that would be \( \frac{15}{8} \), but let's check with the given lengths. Wait, maybe the diagram is such that angle H is in triangle where the opposite side is 32 and adjacent is 60? No, that doesn't make sense. Wait, maybe I got the angle wrong. Wait, maybe angle H is in triangle where the opposite side is 60 and adjacent is 32? No, that would be \( \frac{60}{32}=\frac{15}{8} \), which is what we have. Wait, but let's confirm with the similar triangles.

Since triangle FEG ~ triangle FDH, then \( \frac{EG}{DH}=\frac{FE}{FD} \), so \( \frac{DH}{FD}=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \), therefore \( \frac{FD}{DH}=\frac{15}{8} \), which is \( \tan(H) \). So yes, \( \tan(H)=\frac{15}{8} \)? Wait, no, wait: \( \tan(H)=\frac{FD}{DH} \), and \( \frac{DH}{FD}=\frac{8}{15} \), so \( \tan(H)=\frac{15}{8} \). Therefore, the tangent of angle H is \( \frac{15}{8} \)? Wait, but let's check the calculation again.

Wait, FE = 60, EG = 32, so in triangle FEG, \( \tan(F)=\frac{EG}{FE}=\frac{32}{60}=\frac{8}{15} \). Then, in triangle FDH, angle H is complementary to angle F, so \( \tan(H)=\cot(F)=\frac{1}{\tan(F)}=\frac{15}{8} \). Yes, that makes sense. So the tangent of angle H is \( \frac{15}{8} \). Wait, but let's check with the sides. If FE = 60, EG = 32, then FD is some length, DH is another, but since they are similar, the ratio holds. So the answer should be \( \frac{15}{8} \)? Wait, no, wait, maybe I mixed up opposite and adjacent. Wait, maybe angle H is in a triangle where the opposite side is 32 and adjacent is 60? No, that would be \( \frac{32}{60}=\frac{8}{15} \), but that's \( \tan(F) \). Wait, I think I made a mistake in the angle. Let's re-express:

In the diagram, points F, G, H are on a straight line (FH). E is on FD, with FE = 60, FG = 68, EG = 32 (right angle at E). D is on FD, right angle at D (so HD is perpendicular to FD). So angle at H: between HD and FH. So in triangle DFH, right-angled at D, angle H is at H, so the sides:

- Opposite to H: FD (vertical side, since HD is horizontal? Wait, no, FD is vertical? Wait, maybe the diagram is such that FD is vertical, FH is horizontal. So E is on FD (vertical), G is on FH (horizontal), EG is vertical? No, the right angle at E means EG is perpendicular to FE, so if FE is horizontal, EG is vertical. Then HD is also vertical (perpendicular to FD, which is horizontal). So FH is horizontal, FD is vertical. Then angle H is at the bottom right, between FH (horizontal) and HD (vertical). So in that case, \( \tan(H)=\frac{\text{opposite (FD)}}{\text{adjacent (DH)}} \). But in triangle FEG, FE is horizontal (length 60), EG is vertical (length 32), FG is the hypotenuse (length 68). Then, in triangle FDH, FD is vertical (length FE + ED = 60 + ED), DH is horizontal (length FG + GH = 68 + GH). But since EG and DH are both vertical (perpendicular to horizontal FH), they are parallel, so triangles FEG and FDH are similar (both right-angled, same angle at F). Therefore, \( \frac{EG}{FE}=\frac{DH}{FD} \), so \( \frac{32}{60}=\frac{DH}{FD} \), so \( \frac{FD}{DH}=\frac{60}{32}=\frac{15}{8} \), which is \( \tan(H) \). Therefore, \( \tan(H)=\frac{15}{8} \). Wait, but that seems correct. Alternatively, maybe the problem is that angle H is in a triangle where the opposite side is 32 and adjacent is 60, but that would be angle F. Wait, no, angle F is at the left, angle H is at the right. So angle H's tangent is opposite over adjacent, where opposite is FD (vertical) and adjacent is DH (horizontal). Since EG is vertical (opposite to angle F) and FE is horizontal (adjacent to angle F), then angle F's tangent is \( \frac{EG}{FE}=\frac{32}{60} \), and angle H's tangent is \( \frac{FD}{DH} \), which is the reciprocal because angle F and angle H are complementary. Therefore, \( \tan(H)=\frac{60}{32}=\frac{15}{8} \).

Step2: Simplify the fraction

We had \( \tan(H)=\frac{60}{32} \), which simplifies by dividing numerator and denominator by 4: \( \frac{60 \div 4}{32 \div 4}=\frac{15}{8} \).

Answer:

\( \frac{15}{8} \)