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...

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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.**