If from the top of a tower 80 meters high the angles of depression...

Let,& $OA$ is tower and $BC$ is house
Given, $OA = 80$ meters
$∠ACD = {30}^{o}$
$∠ABO = {45}^{o}$
Let, $BC = h$ meters
In $\Delta ABO,$

$\dfrac{OA}{OB} = tan {45}^{o} ⇒ OB = \dfrac{80}{tan {45}^{o}} = 80$ meters
In $\Delta ADC,$

$\dfrac{AD}{DC} = tan {30}^{o}$
$\dfrac{OA - OD}{OB} = \dfrac{1}{\sqrt{3}}\left\{∵ OB = DC\right\}$
$\dfrac{80 - h}{80} = \dfrac{1}{\sqrt{3}}$
$80\sqrt{3} - \sqrt{3}h = 80 ⇒ h = \dfrac{80\left(\sqrt{3} - 1\right)}{\sqrt{3}}$ meters