Trigonometry!

As $a=2b \Rightarrow \sin A=2\sin B$ but as $A=B+\frac{\pi}{2}, \sin A=\cos B$

$\cos B= 2 \sin B \Rightarrow \tan B=\frac{1}{2}$

Then $C=\pi-B-B-\frac{\pi}{2}=\frac{\pi}{2}-2B$ and by trigonometric identities $\tan C=\tan(\frac{\pi}{2}-2B)=\cot 2B$ .

$\tan C=\cot 2B=\frac{1}{\tan 2B}=\frac{\cos^2 B- \sin^2 B}{2\sin {B}\cos {B}}$

Drawing a triangle we get:

$\sin B=\frac{1}{\sqrt{5}}$

$\cos B=\frac{2}{\sqrt{5}}$

$\therefore$

$\tan C=\frac{(\frac{2}{\sqrt{5}})^2- (\frac{1}{\sqrt{5}})^2}{2 (\frac{1}{\sqrt{5}})(\frac{2}{\sqrt{5}})}=\boxed{\frac{3}{4}}$

By the Sine Rule , $a=2b\Rightarrow\sin A=2\sin B.$

Now, $A=\frac{\pi}{2}+B$

$\therefore \sin \left(\frac{\pi}{2}+B\right)=\cos B$

$\cos B=2\sin B$

$\tan B = \frac{1}{2}$

Similarly, $B=A-\frac{\pi}{2}$

$\therefore \sin A = \sin \left(A-\frac{\pi}{2}\right)$

$\sin A = -2\cos A$

$\tan A = -2$

Now, we know that $\tan A+\tan B+\tan C = \tan A\tan B\tan C$

Substituting these values, we get $\boxed{\tan C = \frac{3}{4}}$

Let ${\bf BD = 2AC. }$

Using the law of sines ${\bf \implies \frac{sinA}{sinB} = 2 }$ and ${\bf \frac{sinB}{sinA} = \frac{1}{2} \implies sin(A) = 2sinB }$ and ${\bf sinB = \frac{1}{2}sinA }$

${\bf \angle_A - \angle_B = 90 \implies }$

${\bf sinA = sin(90 + B) = cosB }$

${\bf sinB = sin(A - 90) = -cosA }$

${\bf \implies sinA = -2cosA \implies tanA = -2 }$

${\bf sinB = \frac{1}{2}cosB \implies tanB = \frac{1}{2} }$

${\bf \implies }$

${\bf tanC = tan(180 - (A + B)) = -tan(A + B) = -(\frac{tanA + tanB}{1 - tanAtanB}) = \frac{3}{4}. }$

It is given that $A - B = 90^\circ \implies A = 90^\circ + B$ .

$\begin{aligned} \implies C & = 180^\circ - A - B \\ & = 180^\circ - (90^\circ + B) - B \\ & = 90^\circ - 2 B \end{aligned}$

Using sine rule, we have:

$\begin{aligned} \frac{\sin A}{2} & = \frac{\sin B}{1} \\ \implies \sin A & = 2 \sin B \\ \sin (90^\circ + B) & = 2 \sin B \\ \implies \tan B & = \frac{1}{2} \end{aligned}$

Now, we have:

$\begin{aligned} \tan C & = \tan (90^\circ + 2 B) \\ & = \tan (2(45^\circ - B)) \quad \quad \small \color{#3D99F6}{\tan (45^\circ-B) = \frac{1-\tan B}{1+\tan B} = \frac{1-\frac{1}{2}}{1+\frac{1}{2}} = \frac{1}{3}} \\ & = \frac{2\left(\frac{1}{3}\right)}{1-\left(\frac{1}{3}\right)^2} = \frac{6}{8} = \frac{3}{4} = \boxed{0.75} \end{aligned}$

given that , BC = 2AC

=> BC/AC = 2

denoting AB,BC , CA by c,b,a, respectively , we can now write a/b = 2

as we know from Sine law , a/SinA = b/SinB ,

we get , a/b = SinA/SinB = 2 ... Let it be eqn no (1)

again , given that, A-B = pi/2

                           or, A = pi/2+B

so , eqn (1) now becomes ,

                                      Sin(pi/2+B)/SinB = 2

but Sin(pi/2+X) = CosX ,

so , from (1) , CosB/SinB = 2

it implies, CotB = 2

             or,tanB =1/2

             or, B = arctan(1/2) = 26.56505118 degree (aproximately)

again , for any triangle ,

                                  A+ B + C = pi

                  or, pi/2 + B + B +C =pi

                   or,      2B + C = pi/2

                   or,      C = pi/2 - 2B

putting the value of B , we will get C = 36.86989765(aproximately)
now tanC = 3/4