What is the approximate distance from R to T?
To approximate the distance from R and T that are on opposite sides of a lake, a
man walks 600 feet from R to P, then tums 35° to face point T. He then walks 870
feet to point T. What is the approximate distance from R to T?
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I am assuming that man walks form R to P and turns 35 from his original direction to face T, so that RTP triangle is an obtuse-angled triangle with the angle T is an obtuse triangle with the value of (180-35) degrees which is 145 degrees.
To find the distance RT, draw a perpendicular line to extended RP line from T. The pont of intersection will be Q., so that triangle RQT is a right angled triangle so as PQT triangle.
We know RP = 600 and PT = 870.
Therefore from right angled triangle PQT,
`cos(35) = (PQ)/(PT)`
PQ = 870(0.8192) = 712.7 feet.
`RQ = RP+PQ`
Therefore, RQ = 600 + 712.7 = 1312.7 feet.
Again from triangle PQT,
`sin(35) = (QT)/(PT)`
QT = 870(0.5736) = 499 feet.
Now in the triangle RQT, RQ = 1312.7 and QT = 499. From Pythagaras theroem,
`(RT)^2 = (RQ)^2 + (QT)^2`
`(RT)^2 = 1312.7^2 + 499^2`
`(RT)^2 = 1972182.29`
Therefore `RT = sqrt(1972182.29) = 1404.3` feet.
Approximate distance from R to T is 1404.3 feet.
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