How to find out the angle ACD?

Question

I figured that since CD is median that can divide the triangle in two smaller triangle of equal areas, I can use that property.

Area of Triangle CAD = CA*sin(x)*CD/2

Area of Triangle ABC= AB*sin(30)*BC/2

Area of Triangle DCB= DC*sin(15)*CB/2

Area of Triangle ABC = Area of Triangle CAD + Area of Triangle DCB

This approach didn't help. Any idea how to proceed?

Comments

Yes, it does seem to be a good idea at first sight to use the areas of the triangles; but there's a problem deriving the area of triangle ABC, because you don't know the lengths of AC or BC. However, you could use the Sine Rule to work out the length of either of these sides in terms of AD and/or DB and involve x or a related angle (e.g., 135-x). I think it might be a good idea to pursue the solution you started on these grounds.

Answer 1

This can be solved using the Sine Rule. Let y=AD=DB.

Note that CD is common to triangles ACD and BCD. Let z=CD

Also note that CD̂A is the exterior angle to triangle BCD, so ∠B=30°, and sin30=0.5.

sinB/CD=sinDĈB/DB, that is, 0.5/z=sin15/y, from which y=2zsin15.

CÂD=135-x.

sinCÂD/CD=sinAĈD/AD, that is, sin(135-x)/z=sin(x)/y.

ysin(135-x)=zsin(x); but y=2zsin15, so 2zsin15sin(135-x)=zsin(x), and the z's cancel.

2sin15sin(135-x)=sin(x),

2sin15(sin135cos(x)-cos135sin(x))=sin(x), sin135=sin45=√2/2, cos135=-cos45=-√2/2;

√2sin15(cos(x)+sin(x))=sin(x), now divide through by cos(x):

√2sin15(1+tan(x))=tan(x),

√2sin15+√2sin15tan(x)=tan(x),

tan(x)(1-√2sin15)=√2sin15,

tan(x)=√2sin15/(1-√2sin15)=√3/3, making x=30°. (Note that sin15=(√6-√2)/4.)

This may be one of many different ways to solve this problem, particularly because x turned out to be a "nice" angle.

Comments

You can derive sin15 from sin30, because sin30=2sin15cos15=½. I didn't do this, because it's a distraction from the main problem. If you would like to know how I arrived at sin15=(√6-√2)/4, I'd be happy to show you. Just leave a comment after this one.