To find the distance from the house at point D to the house at point E, a surveyor measures �ÚDFE, which he finds is 73�‹40'. The distance from the house at point D to point F is 15 ft, and the distance from point F to the house at point E is 28 ft. How far apart are the houses?
this one is just like the one with bearings. Use law of cosines:
DE2 = FD2 + FE2 - 2*FD*FE*cosF
DE2 = 225 + 784 - 2*15*28*cos(73.666666)
DE2 = 773
DE = 28
Thank you so much!
To find the distance between the houses at point D and point E, we can use the Law of Cosines. The Law of Cosines states that in a triangle with sides a, b, and c, and angle C opposite side c, the following formula holds true:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, we have a triangle DFE with D and E as the vertices of interest. The distance from D to F is 15 ft, and the distance from F to E is 28 ft. The angle DFE is given as 73° 40'.
Let's label the sides of the triangle as follows:
- Side DF as side a (15 ft)
- Side DE as side c (what we're trying to find)
- Side FE as side b (28 ft)
Using the Law of Cosines, we can plug in the values we have:
c^2 = (15 ft)^2 + (28 ft)^2 - 2(15 ft)(28 ft) * cos(73° 40')
To compute the cosine of 73° 40', we need to convert it to radians. The formula to convert from degrees to radians is:
radians = degrees * π / 180
Therefore, 73° 40' is equivalent to (73 + 40/60)° or approximately 1.289 rad.
Now we can substitute into the Law of Cosines:
c^2 = (15 ft)^2 + (28 ft)^2 - 2(15 ft)(28 ft) * cos(1.289 rad)
Simplifying the equation:
c^2 = 225 ft^2 + 784 ft^2 - 2(15 ft)(28 ft) * cos(1.289 rad)
Solving for c:
c^2 = 1009 ft^2 - 840 ft * cos(1.289 rad)
Taking the square root of both sides to isolate c:
c = √(1009 ft^2 - 840 ft * cos(1.289 rad))
Now, we can plug in these values into a calculator to find the final answer for c, which is the distance between the houses at point D and point E.