Let x be the distance of a point P ( lies on the building ) from the top of the building such that AP is perpendicular to the building and y be the distance of the building from point A, ( shown in the below diagram )

Given,

Point A is 8.20 m above level ground,

So, the height of the building = ( x + 8.20 ) meters,

Now, 1 degree = 60 minutes,

⇒ $1\text{ minute } =\frac{1}{60}\text{ degree }$

$20\text{ minutes }=\frac{20}{60}=\frac{1}{3}\text{ degree}$

$50\text{ minutes }=\frac{50}{60}=\frac{5}{6}\text{ degree}$

By the below diagram,

$tan ( 12^{\circ} 50') = \frac{8.20}{y}$

$tan(12+\frac{5}{6})^{\circ}=\frac{8.20}{y}$

$tan (\frac{77}{6})^{\circ}=\frac{8.20}{y}$

$\implies y=\frac{8.20}{tan (\frac{77}{6})^{\circ}}$

Now, again by the below diagram,

$tan (31^{\circ}20')=\frac{x}{y}$

$tan(31+\frac{1}{3})=\frac{x}{y}$

$\implies x=y\times tan(\frac{94}{3})=\frac{8.20}{tan (\frac{77}{6})^{\circ}}\times tan(\frac{94}{3})^{\circ}=21.9142943216\approx 21.91$

Hence, the height of the building = x + 8.20 = 30.11 meters (approx)

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