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Ignorance can be fixed. Stupid is forever!
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Answer is 8.
Drop a radius of the arc from the tip of X to the centerline of the large arc, which = 8.
Solve for the shorter leg of the new triangle via tangent of 60 degree & known opposite side of 8.
Calculate hypotenuse; since hypotenuse is common as are angles of the 2 triangles, other legs of the original triangle containing X are the same.
So X = 8.
MM
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Use the 3:4:5 ratio.
3 x 1.333333... = 4
4 x 1.333333...= 5.333333
5 x 1.333333... = 6.666666...
Or you can use proper algebra to come up with the same solution.
Don't be the darkness.
America will perish while those who should be standing guard are satisfying their lusts.
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Never mind. I was working it as though 4 was the length of the sort side of the triangle. I saw it wrong.
MM is right.
I was kinda right...it's a 6:8:10 ratio.
Last edited by RiverRider; 12/03/23.
Don't be the darkness.
America will perish while those who should be standing guard are satisfying their lusts.
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It does not appear that dropping a perpendicular from the apex of the triangle will intersect the center point of the semicircle.
If it did, the problem is fairly simple.
But if it doesn't, I'm stumped.
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Findin dem Xs is easy peasy... Thars b twos o' dem... ![[Linked Image from i.postimg.cc]](https://i.postimg.cc/BnPXZ9B8/Screenshot-2023-12-03-at-19-12-33-Facebookx.png)
If you are not actively engaging EVERY enemy you encounter... you are allowing another to fight for you... and that is cowardice... plain and simple.
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Campfire Ranger
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Circumference of hemisphere is Pi*r
Take half that
Subtract the arc of 60 degree angle
You now have the arc that x subtends.
-OMotS
"If memory serves fails me..."
Quote: ( unnamed) "been prtty deep in the cooler todaay "
Television and radio are most effective when people question little and think even less.
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I may not be smart but I can lift heavy objects
I have a shotgun so I have no need for a 30-06.....
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X= really bad memories of floundering in 10th grade geometry.
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Yeah I'm not thinking that much on Monday morning...or ever.
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X=8
Arc length is 8.373333333
This makes the chord length 8.
Last edited by gregintenn; 12/04/23.
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It does not appear that dropping a perpendicular from the apex of the triangle will intersect the center point of the semicircle.
If it did, the problem is fairly simple.
But if it doesn't, I'm stumped. You're right, it does not intersect the midpoint of the semi circle. And that is why you have to solve for the tangent of 60 degrees with the side opposite being the radian of 8 that you dropped. That gives you the bottom leg of the new triangle. MM
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Are you looking for the length of the triangle side or the area of the space between the triangle leg and the arc?
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I'll make it easy. Get you a piece of paper and mark the distance between the left or right corner of the radius and the center at the dot. We know that is 8. Now compare your 2 marks to the length of the leg of the triangle he has marked as X.
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X=8 Finding the area of the space (inside the circle) in the vicinity of X is a little more complicated X=5.7975
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