HPC
Law of Sines and Cosines/Area of a Triangle
In-Class Applications
1. Sketch each plot of land described and find its area.
From a granite post, proceed 195 ft east along Tasker Hill Road, then along a bearing of S 32° E for 260 ft, then along a bearing of S 68° W for 385 ft, and finally along a line back to the granite post.
[Image of a quadrilateral with sides 195, 260, 385, and an unknown side x]
x² = 195² + 260² - 2(195)(260)cos 112°
x ≈ 399.2
sin θ / 195 = sin 112° / 399.2
θ = 24.5°
triangle within the quadrilateral with sides 195, 260, and 399.2, with angles 112°, θ, and 24.5°
Area I = 1/2 * (195)(260) * sin 112° ≈ 24828.82 ft²
Area II = 1/2 * (399.2)(385) * sin 55.5° ≈ 64828.82 ft²
2. From a cement marker, proceed 260 m southwest to the river, then 240 m south along the river to the bridge, then 280 m N 40° E to a sign on the edge of Sycamore Lane, and finally along Sycamore Lane back to the cement marker.
x² = 240² + 260² - 2(240)(260)cos 135°
x ≈ 462 m
sin θ / 240 = sin 135° / 462
θ = 21.6°
triangle within the quadrilateral with sides 260, 240, and 462, with angles 135°, θ, and 21.6°
Area I = 1/2 * (240)(260) * sin 135° ≈ 20940.06 m²
Area II = 1/2 * (462)(280) * sin 16.6° ≈ 40540.06 m²
Law of Sines and Cosines Solution Key
of 2
Report
Tell us what’s wrong with it:
Thanks, got it!
We will moderate it soon!
Free up your schedule!
Our EduBirdie Experts Are Here for You 24/7! Just fill out a form and let us know how we can assist you.
Take 5 seconds to unlock
Enter your email below and get instant access to your document