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help please
- Thread starter JayJay06
- Start date
Hello, JayJay06!
There should be a theorem which says:
The angle formed by two secants is one-half the difference of the two intercepted arcs.
So we have: \(\displaystyle \L\;25 \:=\:\frac{1}{2}(100\,-\,x)\)
. . . and solve for \(\displaystyle x\).
Refer to the figure. .Find the value of "x".
There should be a theorem which says:
The angle formed by two secants is one-half the difference of the two intercepted arcs.
So we have: \(\displaystyle \L\;25 \:=\:\frac{1}{2}(100\,-\,x)\)
. . . and solve for \(\displaystyle x\).
JayJay06 said:25= 1/2(100-x)
25=50-1/2x
-25= - 1/2x Should be a negative there.
50=x
Is this right?
Thank you
Yes. All you have to do is plug it back in your original equation:
\(\displaystyle \L \;25\,=\frac{1}{2}(100\,-\,(50))\,\Rightarrow\,25\,=\,25\)