Answer the following trigo question?

That means BD = x also

Let the angle DAB = y, then from the right triangle ABD

tany = x/29 ............ [1]

Similarly, from the right triangle ABC

tan(y + 20) = 2x/29

==> (tany + tan20)/(1 - tany * tan20) = 2x/29

==> substitute [1]:

==> (x/29 + tan20)/(1 - x/29 * tan20) = 2x/29

==> (x + 29tan20)/(29 - x * tan20) = 2x/29

==> 29(x + 29tan20) = 2x(29 - x * tan20)

==> (2 tan20) x^2 - 29x + 841tan20 = 0

By Quadratic Formula

x = (29 +/- √(841 - 6728tan²20))/(4 tan20)

Since 841 - 6728tan²20 < 0, x is a complex number, but since x represents distance, x cannot be a complex number thus a contradiction, hence there'es no solution for x.

Side note:

For x to be real, angle CAD must be in the range of 0 to 7.1250163 degrees exclusive, if AB is still 29cm

For x to be real, length AB must be in the range of 0 to 0.72429177 cm exclusive, if angle CAD is still 20 degrees.

CD=DB,let the angle BAD =y

tany=x/29

x=tany*29

tan(20+y)=2x/29

2x=tan(y+20)*29

2x=[(tany+tan20)/(1-tanytan20)]*29

2tany*29[1-tanytan20]=(tany+tan20)29

2*29tany-2*29tan^2ytan20=tany*29+29tan20

58tany-21.11tan^2y=29tany+10.56

21.11tan^2y-29tany+10.56=0

It has complex solutions.This means that with the angle 20 degrees AD cannot be a median.

because of the fact this may be a suited triangle, perspective CAB is 40 5 tiers. First, commence via getting the scale of Line advert. considering the fact that all of us understand that perspective CAB is 40 5 tiers, perspective DAB must be 40 5-20=25 tiers. To get the scale of Line advert, use: COS=adjoining/Hypotenuse, it somewhat is COS 25 tiers= 29/Line advert. COS 25=.906. We get .906AD=29; divide out the .906 and you get a length for Line advert as 29/.906= 32 cm. Now that we've the scale of line advert, we are able to be certain the scale of line CD via utilising: Tan=opposite/adjoining. Tan 20 tiers=Line CD/32. We proceed to get .364=Line CD/32. Multiply the two aspects via 32 to be certain the scale of Line CD and it measures .364*32= 11.sixty 5 cm.