What would be the value of angle CAD

[Guest]

There is a triangle ABC with point D at the center of AB.
Angle CDA = 45 degree
Angle DCB = 15 degree

Need to find the angle CAD.

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)*AD/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?

[jimbos]

To find the angle CAD in triangle ABC, you can use the Law of Sines. Here's how you can do it:
You know that angle CDA is 45 degrees, and angle DCB is 15 degrees. You want to find angle CAD, so let's call it x.
You can solve this system of equations to find x, which is the angle CAD.
This approach should allow you to calculate the angle CAD in triangle ABC.

[jimbos]

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[Guest]

We are building CH[tex]\bot[/tex]AB (H [tex]\in[/tex]rights AB) ; [tex]\triangle[/tex]HDC -rectangular ,isosceles ; CH=DH= h [tex]\Rightarrow[/tex] CD=h[tex]\sqrt{2}[/tex] (1)
[tex]\angle[/tex]HCD=45[tex]^\circ[/tex] ,[tex]\angle[/tex]DCB=15[tex]^\circ[/tex] [tex]\Rightarrow[/tex] [tex]\angle[/tex]HCB=60[tex]^\circ[/tex] ;[tex]\triangle[/tex]HBC - [tex]\angle[/tex]HBC=30[tex]^\circ[/tex] (2)

From ahgle theorem 30[tex]^\circ[/tex] HC=[tex]\frac{BC}{2}[/tex] ; h=[tex]\frac{BC}{2}[/tex] ; BC=2h (3)
AD=BD=a [tex]\Rightarrow[/tex] AH=DH-AD ; AH=h-a (4)
[tex]\triangle[/tex]HBC -rectangular [tex]HC^{2 }+ HB^{2 }= BC^{2 }[/tex] ; [tex]h^{2 }+( h+a)^{2 } = (2h)^{2 }[/tex]
We find a=h([tex]\sqrt{3}[/tex]-1) (5)

[tex]\triangle[/tex]HAC -rectangular [tex]HC^{2 }+ HA^{2 } = AC^{2 }[/tex] ; [tex]AC^{2 }= h^{2 }+ (h-a)^{2 }[/tex]
[tex]AC^{2 }= h^{2 }+ h^{2 }-2ah+ a^{2 }[/tex] [ see (5) ] [tex]AC^{2 }= 2 h^{2 }[/tex]-2h([tex]\sqrt{3}[/tex]-1)h +[tex]h^{2 } ( \sqrt{3} -1)^{2 }[/tex]
[tex]AC^{2 }=(8-4 \sqrt{3} )h^{2 }[/tex] ; AC=h[tex]\sqrt{8- \sqrt{48} }[/tex] [tex]\Rightarrow[/tex] AC=h([tex]\sqrt{6}- \sqrt{2}[/tex]) (6)

From Guest"s idea is median [tex]\Rightarrow[/tex] [tex]S_{ADC }= S_{DBC }[/tex] ; [tex]\frac{1}{2}[/tex]DC.ACsinACD =[tex]\frac{1}{2}[/tex]BC.DCsin15[tex]^\circ[/tex]

h[tex]\sqrt{2}[/tex].h([tex]\sqrt{6} - \sqrt{2}[/tex])sinACD =2h.h[tex]\sqrt{2}[/tex].[tex]\frac{ \sqrt{6}- \sqrt{2} }{4}[/tex]

sinACD=[tex]\frac{1}{2}[/tex] [tex]\Rightarrow[/tex] [tex]\angle[/tex]ACD=30[tex]^\circ[/tex]

([tex]\triangle[/tex]ADC) [tex]\angle[/tex]CAD+[tex]\angle[/tex]ACD+[tex]\angle[/tex]ADC=180[tex]^\circ[/tex]
[tex]\angle[/tex]CAD+30[tex]^\circ[/tex]+45[tex]^\circ[/tex] =180[tex]^\circ[/tex]
[tex]\angle[/tex]CAD =105[tex]^\circ[/tex]

[Guest]

Thank you very much!