An ant, located in a square field, is 13 meters from one of the corner posts of the field, 17 meters from the corner post diagonally opposite that one, and 20 meters from a third corner post. Find the area of the field. Assume the land is flat.

Label the vertices of the square A, B, C, D. The ant is at point P, with PB = 13, PC = 20, PD = 17. Rotate triangleCDP 90° counterclockwise about C, so that D goes to B, and P goes to Q. By definition, angleQCP = 90°. Hence, by Pythagoras' Theorem, PQ = 20root 2. Also, since trianglePQC is isosceles, angleCPQ = 45°. Square field, containing an ant, as described above.

Applying the law of cosines (also known as the cosine rule) to triangleBQP 172 = 132 + (20root 2)2 − 2 · 13 · 20root 2 · cos QPB. Simplifying, we find cos QPB = 17root 2 / 26. Then sin2 QPB = 1 − cos2 QPB = 49/338. Hence sin QPB = 7root 2 / 26. (We will need this result below.)

We have angleCPB = angleQPB + angleCPQ = angleQPB + 45°. Hence cos CPB = cos (QPB + 45°) = cos QPB · cos 45° − sin QPB · sin 45°, by trigonometric identity cos(a + b) = cos a · cos b − sin a · sin b = (cos QPB − sin QPB) / root 2 = (10root 2 / 26) / root 2 = 5/13

Applying the law of cosines to triangleCPB BC2 = 202 + 132 − 2 · 20 · 13 · (5/13) = 369.

Label the vertices of the square A, B, C, D. The ant is at point P, with PB = 13, PC = 20, PD = 17.

ReplyDeleteRotate triangleCDP 90° counterclockwise about C, so that D goes to B, and P goes to Q.

By definition, angleQCP = 90°. Hence, by Pythagoras' Theorem, PQ = 20root 2.

Also, since trianglePQC is isosceles, angleCPQ = 45°.

Square field, containing an ant, as described above.

Applying the law of cosines (also known as the cosine rule) to triangleBQP

172 = 132 + (20root 2)2 − 2 · 13 · 20root 2 · cos QPB.

Simplifying, we find cos QPB = 17root 2 / 26.

Then sin2 QPB = 1 − cos2 QPB = 49/338.

Hence sin QPB = 7root 2 / 26. (We will need this result below.)

We have angleCPB = angleQPB + angleCPQ = angleQPB + 45°.

Hence cos CPB = cos (QPB + 45°)

= cos QPB · cos 45° − sin QPB · sin 45°, by trigonometric identity cos(a + b) = cos a · cos b − sin a · sin b

= (cos QPB − sin QPB) / root 2

= (10root 2 / 26) / root 2

= 5/13

Applying the law of cosines to triangleCPB

BC2 = 202 + 132 − 2 · 20 · 13 · (5/13) = 369.

Therefore the area of the field is 369 m2.