Question

VERY INTERESTING PROBLEM.. ..........​

Answer 1

Heya

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Find DC

In Trangle DCB

Tan 45 = DC/CB

=>

1 = DC/CB

=>

{DC = CB}

=>

Now, in Trangle DCA

Tan 30 = DC/CA

=>

1/√3 = DC/CA

=>

1/√3 = DC/CA

=>

6 + BC = DC √3

{ CA = 6 + BC }

=>

6 + DC = DC√3

=>

DC{√3 - 1 } = 6

=>

DC = 6/(√3 - 1 )

=>

DC = 3{√3 + 1}