Solving for UT Length and Area of Nonrectangular Land Plot | Geometry/Trig Help

  • Thread starter Thread starter ur5pointos2sl
  • Start date Start date
AI Thread Summary
To solve for the length of side UT and the area of a nonrectangular plot of land, the law of sines and law of cosines are essential tools. The law of cosines can be used to relate the sides and angles of the triangle formed by points R, S, and T, while the law of sines can help find the remaining sides and angles. Drawing a diagram can aid in visualizing the relationships between the sides and angles. With the given measurements, applying these laws will lead to the necessary calculations. This approach will effectively address the surveying challenge presented.
ur5pointos2sl
Messages
95
Reaction score
0
A surveyer is required to survey a nonrectangular plot of land but is unable to measure side UT directly. The following data are taken:
RU= 130.0 m
RS= 120.0 m
ST=90.0m
Angle RST= 115 degrees
Angle RUT = 100 degrees

Calculate the length of side UT and area of the plot.

Again I haven't taken any geometry or trig in years and just need help starting the problems(I have a few to post)
 

Attachments

  • pic2.jpg
    pic2.jpg
    5.1 KB · Views: 517
Physics news on Phys.org
hello UR5,
I believe that for this problem, the equation that will help you is either the law of sines or law of cosines equations. This will help you to form oblique triangles and find out the length of the line going through the shape, which will help in finding out the other two sides.

Law of cosines= C2=a2+b2-2ab(cosC)

Law of sines= sinA/a=sinB/a=sinC/a (captial A= angle A, lower case a=side a)

This should definately get you headed in the correct direction. If you need to, draw a shape that resembles a rectangle, just so you can get an idea of where the opposite sides would be and all that.
 
Thanks that worked out well.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
Back
Top