+257 
A plane contains triangle AOB so that line AB is the hypotenuse of right triangle ABC and line CO is perpendicular to the plane. The legs of triangle ABC form angles 45 degrees and 30 degrees with the plane, as shown. If the length of the short leg of triangle ABC is sqaure root 3 how long it line AB?
+129852 Since angle CAO is greater than angle CBO, then AO < BO
And since CO = CO.....then AC < BC
So....AC is the shorter leg of triangle ABC
And since CO is perpendicular to the plane....then triangles AOC and BOC are right
Then
sin 45 = OC / sqrt(3)
1/sqrt(2) * sqrt (3) = OC = sqrt (3/2)
And
sin (30) = OC / BC
(1/2) = sqrt (3/2) / BC
BC = 2sqrt(3/2)
So.....AC^2 = 3 and BC^2 = 6
AB = sqrt [ AC^2 + BC^2 ] = sqrt [ 3 + 6 ] = sqrt (9) = 3
+9479 Let CO = x
△COB is a 30° - 60° - 90° triangle. So BC = 2x
△COA is a 45° - 45° - 90° triangle. So AC = √2 x
And x is positive, so √2 x < 2x . So AC < BC
AC = √3
√2 x = √3
x = √3 / √2 Multiply the numerator and denominator by √2
x = √6 / 2
BC = 2x
BC = 2( √6 / 2 )
BC = √6
By the Pythagorean theorem,
AC2 + BC2 = AB2
(√3)2 + (√6)2 = AB2
3 + 6 = AB2
9 = AB2 AB is a length, so take the positive sqrt of both sides.
3 = AB
