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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? 

 May 10, 2019
 #1
avatar+105415 
+3

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

 

 

cool cool cool

 May 11, 2019
 #2
avatar+8829 
+3

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

 May 11, 2019

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