I do not even know how to try to deduce the mass M of the Sirius system in terms of that of the sun, as I do not not how to interpret the given data.

Could you give me a hint?

**EDITED AT THIS POINT**

**1) Estimating the distance to the Sirius binary system.**

I calculated the hypotenuse of the triangle $S E_1 SS$ (Sun - position of the Earth located at the top of a vertical circular orbit - Sirius System):

$$ \theta = 0.378^{\circ} arc \times \frac{1}{3600} \times \frac{2 \pi}{360} = 1.83 \times 10^{-6}arc sec $$

$$D = \frac{D'}{sin(\theta)} = 8.16\times 10^{16} m$$

Where:

D = the distance from the Earth to the Sirius binary system.

D' = Distance from the Sirius system to the Earth ($15 \times 10^{10} m$).

Here's an illustration of what I have done:

**2) Measuring the semi-major axis of the ellipse and estimating the period of the orbit from the figure.**

To compute the semi-major axis I have assumed the major axis is under the scale [0'',12''].

As Sammy Gerbil said :'To measure the semi-major axis from the figure, magnify the diagram to fill your screen, place the edge of a sheet of paper along the diagonal line marked on the ellipse, mark off the ends of this line on the paper, then transfer the paper edge to the scale and read off the length in arc seconds. I get 13.8" for the major axis length, so the semi-major axis is a=6.9". The calculation is then:'

$$a=8.16\times 10^{16}m\times \frac{6.9}{3600}\times \frac{2\pi}{360} = 2.73 \times 10^{12}m$$

Now let's estimate the period of the orbit.

Is this the right thought?

$$\frac{dA}{dt} = SM \times SM \times \frac{d \theta}{2dt}$$

It will not be accurate as 'the two sides of most of the triangles are different'.

$A_1$:

On process...

2) Using Heron's Formula I got an $A_1$ (please see image above) of an order $10^{24}$. Does that make sense for you?