Gret if sum1 could help thansk in advance just a segment (circle) Q

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SUMMARY

The discussion focuses on calculating the grazing area of a goat tethered to a post with a 10-meter rope, restricted by a fence 6 meters away. The total grazing area calculated is 269 m². Participants suggest visualizing the problem by drawing two right triangles, where the hypotenuse represents the rope and one leg is the perpendicular distance to the fence. The angle formed can be determined using the cosine function, specifically cos-1(6/10), and the total angle is double that value.

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A goat is tethered to a post by a rope that is ten meters long. The goat is able to graze over any area that the rope allows it to reach other than that excluded by a straight fence. The perpendicular distance from the post to the fence is 6m. Over wat area can the goat graze- to the nearest meter...ans = 269m^2

If any1 could help me determine the required angle ...that would be awesome...since the Q is straigforward after that! thanks
 
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ku1005 said:
A goat is tethered to a post by a rope that is ten meters long. The goat is able to graze over any area that the rope allows it to reach other than that excluded by a straight fence. The perpendicular distance from the post to the fence is 6m. Over wat area can the goat graze- to the nearest meter...ans = 269m^2

If any1 could help me determine the required angle ...that would be awesome...since the Q is straigforward after that! thanks
Draw a picture! You should see two right triangles where the hypotenuse is the 10 m rope and one leg is the 6 m perpendicular distance. The angle those make is one right triangle is obviously
[tex]cos^{-1}(\frac{6}{10})[/tex]
The total angle is twice that.
 
thanks very much...but it was really stupid wat i was actually doin...i wasn;t including the rest of the circle!i only included the sector which i was caclulating with the angle u define above...but since the answer i got was wrong iassumed the angle was wrong...cheers anyway!