Use Pythagoras theorem:
\(AC=\sqrt{25^2-7^2}=24\)
\(CD=DA=12\)(mid point)
Use Pythagoras theorem again:
\(BD=\sqrt{7^2+12^2}=\sqrt{193}\)
Find the sin of angle BCD:
\(\sin \angle BCD =\sin \angle BCA = \dfrac{7}{25}\)(angle BCD and angle BCA refers to the same angle)
Use law of sines on triangle BCD:
\(\dfrac{BD}{\sin\angle BCD}=\dfrac{BC}{\sin\angle BDC}\\ \dfrac{\sqrt{193}}{\frac{7}{25}}=\dfrac{25}{\sin\angle BDC}\\ \sin\angle BDC = \dfrac{25\times \frac{7}{25}}{\sqrt{193}}=\dfrac{7}{\sqrt{193}}\)
Construct a right-angled triangle with
legs = 7, -12(<-- negative because the angle is in the 2nd quadrant), hypotenuse=sqrt193
\(\tan \angle BDC =-\dfrac{7}{12}\)
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