Question

trig—star problem 1b—local competition
known: distance ef = 226.05 ∠efg = 118°58′31″ ∠feg = 44°59′59″
find:
∠ egf = (6 points)
distance eh = (6 points)
distance fh = (6 points)
distance fg = (6 points)
distance gh = (6 points)
required answer format
distances: nearest hundredth
angles: degrees—minutes—seconds
to the nearest second
page total: points

Explanation:

Step1: Calculate ∠EGF using triangle angle sum

Sum of angles in △EFG = 180°.
∠EGF = 180° - ∠FEG - ∠EFG
= 180° - 44°59'59" - 118°58'31"
= 180° - 163°58'30" = 16°1'30"

Step2: Find FG with Law of Sines

Law of Sines: $\frac{FG}{\sin∠FEG} = \frac{EF}{\sin∠EGF}$
FG = $\frac{EF \cdot \sin∠FEG}{\sin∠EGF}$
∠FEG = 44°59'59" ≈ 44.9997°, ∠EGF = 16°1'30" ≈ 16.025°
sin(44.9997°) ≈ 0.7071, sin(16.025°) ≈ 0.2763
FG ≈ $\frac{226.05 \cdot 0.7071}{0.2763}$ ≈ 575.00

Step3: Calculate EH using cosine

EH = EF · cos∠FEG
cos(44°59'59") ≈ 0.7071
EH ≈ 226.05 · 0.7071 ≈ 160.00

Step4: Calculate FH using sine

FH = EF · sin∠FEG
sin(44°59'59") ≈ 0.7071
FH ≈ 226.05 · 0.7071 ≈ 160.00

Step5: Calculate GH using cosine

GH = FG · cos∠EGF
cos(16°1'30") ≈ 0.9602
GH ≈ 575.00 · 0.9602 ≈ 552.12

Answer:

∠EGF = 16°1'30"
Distance EH = 160.00
Distance FH = 160.00
Distance FG = 575.00
Distance GH = 552.12