Question

find hi.

h
g
68°
√13
i

write your answer as an integer or as a decimal rounded to the nearest tenth.
hi =
submit

Explanation:

Step1: Identify trigonometric ratio

In right triangle \( \triangle HIG \), \( \cos(68^\circ) = \frac{GI}{HI} \), so \( HI = \frac{GI}{\cos(68^\circ)} \)

Step2: Substitute values

\( GI = \sqrt{13} \approx 3.6056 \), \( \cos(68^\circ) \approx 0.3746 \)

Step3: Calculate HI

\( HI \approx \frac{3.6056}{0.3746} \approx 9.625 \)? Wait correction: Wait no—wait, angle at G is 68°, right angle at I: so adjacent side to angle G is GI, hypotenuse is HG? Wait no, wait find HI: angle at G is 68°, opposite side to angle G is HI, adjacent is GI. So \( \tan(68^\circ) = \frac{HI}{GI} \). Yes! That's the mistake earlier. So \( HI = GI \times \tan(68^\circ) \)

Step4: Correct calculation

\( \tan(68^\circ) \approx 2.4751 \), \( HI \approx 3.6056 \times 2.4751 \approx 8.92 \)? Wait no—wait \( \sqrt{13} \) is ~3.60555. \( \tan(68) \) is ~2.475087. Multiply: 3.60555 2.475087 ≈ 3.605552.475 ≈ let's compute: 32.475=7.425, 0.605552.475≈1.499, total≈8.924, rounded to nearest tenth is 8.9? Wait wait another check: if angle at G is 68°, right angle at I: sides: GI adjacent to G, HI opposite to G, HG hypotenuse. So yes \( \tan G = opposite/adjacent = HI/GI \). So \( HI = GI \times \tan G = \sqrt{13} \times \tan 68° \). Let's compute precisely: \( \sqrt{13} ≈3.605551275 \), \( \tan68≈2.475086853 \). Product: 3.605551275 2.475086853 ≈ 3.605552.47509≈ let's do 3.605552=7.2111, 3.605550.4=1.44222,3.605550.07=0.2523885,3.605550.00509≈0.01835. Sum:7.2111+1.44222=8.65332+0.2523885=8.9057+0.01835≈8.924. So rounded to nearest tenth is 8.9. Wait but wait did I mix up sides? Let's confirm: angle at G is 68°, vertex G: sides from G are GI (length sqrt13) and GH, with right angle at I. So yes, GI is adjacent to angle G, HI is opposite. So tan(angle G)=HI/GI → HI=GI*tan(angle G). Correct. So final answer is 8.9.

Wait wait I think I messed up the initial ratio but corrected it. So the correct steps are:

Step1: Identify trigonometric ratio

\( \tan(68^\circ) = \frac{HI}{GI} \) → \( HI = GI \times \tan(68^\circ) \)

Step2: Substitute values

\( GI = \sqrt{13} ≈3.6056 \), \( \tan(68^\circ)≈2.4751 \)

Step3: Calculate HI

\( HI≈3.6056×2.4751≈8.9 \)

Answer:

11.2