Check the picture below.
let's notice that the angle at K is an inscribed angle with an intercepted arc
![\bf \stackrel{\textit{using the inscribed angle theorem}}{K=\cfrac{\widehat{LI}+\widehat{IJ}}{2}}\implies 9x+1=\cfrac{(10x-1)+59}{2} \\\\\\ 9x+1=\cfrac{10x+58}{2}\implies 18x+2=10x+58\implies 8x+2=58 \\\\\\ 8x=56\implies x=\cfrac{56}{8}\implies x=7 \\\\[-0.35em] ~\dotfill\\\\ K=9x+1\implies K=9(7)+1\implies \boxed{K=64}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7Busing%20the%20inscribed%20angle%20theorem%7D%7D%7BK%3D%5Ccfrac%7B%5Cwidehat%7BLI%7D%2B%5Cwidehat%7BIJ%7D%7D%7B2%7D%7D%5Cimplies%209x%2B1%3D%5Ccfrac%7B%2810x-1%29%2B59%7D%7B2%7D%20%5C%5C%5C%5C%5C%5C%209x%2B1%3D%5Ccfrac%7B10x%2B58%7D%7B2%7D%5Cimplies%2018x%2B2%3D10x%2B58%5Cimplies%208x%2B2%3D58%20%5C%5C%5C%5C%5C%5C%208x%3D56%5Cimplies%20x%3D%5Ccfrac%7B56%7D%7B8%7D%5Cimplies%20x%3D7%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20K%3D9x%2B1%5Cimplies%20K%3D9%287%29%2B1%5Cimplies%20%5Cboxed%7BK%3D64%7D)
now, let's notice something again, the angle at L is also an inscribed angle, intercepting and arc of 97 + 59 = 156, so then, by the inscribed angle theorem,
∡L is half that, or 78°.
now, let's take a look at the picture down below, to the inscribed quadrilateral conjecture, since ∡J and ∡I are both supplementary angles, then
∡I = 180 - 64 = 116°.
∡J = 180 - 78 = 102°.
Answer:
Step-by-step explanation:
Surface area is the sum of the areas of all the faces of the solid figure. It's measured in square units. Volume is the number of cubic units that make up a solid figure.
Answer:
18, 20 and 22.
Step-by-step explanation:
Let the lowest number be n, then:
n + n + 2 + n + 4 = 60
3n + 6 = 60
3n = 54
n = 18.
So the other 2 are 20 and 22.