if the angles of triangle are 30 nd 45 and the included side is (root 3+ 1) thne what is the area of triangle?

The angles are given as 45 degree and 30 degree and included side of length (root 3+ 1) unit so we can draw figure as shown below,

Lets draw a perpendicular AD on BC now
BC = $\sqrt{3}+1$ unit
lets BD = x unit so CD = $\sqrt{3}+1-x$  unit
In right angle triangle ABD  ,
$$\begin{gathered} \tan \theta = \frac{AD}{\mathrm{BD}} \\ \tan 30 = \frac{AD}{x} \\ \frac{1}{\sqrt{3}}=\frac{AD}{x}x= AD\sqrt{3} ......(1) \end{gathered}$$
Now taking right angle triangle ADC ,
$\tan 45 = \frac{AD}{\sqrt{3}+1-x}1 = \frac{AD}{\sqrt{3}+1-x}A\mathrm{D} = \sqrt{3}+1-\mathrm{x}....(2)\mathrm{Now} \mathrm{putting} \mathrm{value} \mathrm{of} \mathrm{x} \mathrm{from} \mathrm{equation} (1) \mathrm{we} \mathrm{get}, \mathrm{AD} = \sqrt{3}+1 - \sqrt{3}\mathrm{AD}\mathrm{AD} + \sqrt{3}\mathrm{AD} = \sqrt{3}+1\mathrm{AD}(\sqrt{3}+1 ) = \sqrt{3}+1\mathrm{So} \mathrm{AD} = 1 \mathrm{unit} \mathrm{Now} \mathrm{Area} \mathrm{of} \mathrm{triangle} \mathrm{ABC} = \frac{1}{2} \times \mathrm{base}\times \mathrm{height}=\frac{1}{2} \times (\sqrt{3}+1)\times 1=\frac{(\sqrt{3}+1)}{2}\mathrm{unit} \mathrm{square}$