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# How to find the distance between these ships?

From an iron post, proceed 500m north east to the brook, then 300m east along the brook to the old mill. then go 200m S 15 degrees E to a post on the edge of the road and finally go along the road back to the iron post.

sketch and find the area

pleeasse halp me

### 2 Answers

- LearnerLv 78 years agoFavorite Answer
Please refer Figure 2 in http://www.flickr.com/photos/46256521@N07/

Solution:

i) North East means it is at 45 deg with North as well with East. BP being an altitude, ABP is a right triangle.

<PAB = 45 deg; so AP = BP = 250√2 m

ii) So area of triangle APB = (1/2)*AP*PB = (1/2)*(250√2)*(250√2) = 62500 sq m

iii) Area of rectangle BPQC = BP x PQ = 250√2 x 300 = 75000√2 sq. m ~ 106066 sq m

iv) <QCD = 15 deg [Since at D it moves S15E]

==> CM = 200*cos(15) = 200(√6 + √2)/4 = 50(√6 + √2) m

So DR = MQ = CQ - CM = 250√2 - 50√2 - 50√6 = 200√2 - 50√6 m ~ 160 m

==> MD = QR = 200*sin(15) = 50(√6 - √2) m = 50√6 - 50√2 m

So area of trapezoid CQRD = (1/2)(QR)(CQ + DR) =

(1/2)(50√6 - 50√2)(250√2 + 200√2 - 50√6) = (25√6 - 25√2)(450√2 - 50√6)

This equals nearly 13301 sq. m

v) From the figure, AR = AP + PQ + QR = 250√2 + 300 + 50√6 - 50√2

= 200√2 + 50√6 + 300 ~ 705 m

So area of triangle ARD = (1/2)(AR)(DR) = (1/2)(705)(160) = 56400 sq. m

vi) Hence of the above, ar ABCD = 62500 + 106066 + 13301 - 56400 = 125467 sq. m

So area is nearly 124567 sq. m

Note: Errors in computation is not ruled out.

Source(s): http://www.flickr.com/photos/46256521@N07/ - 8 years ago
Is there some picture that we could use? 500m North East? How many degrees? Is that X degrees North of East or X degrees East of North. Be specific.