Answer:
[tex]0.2679[/tex]
Step-by-step explanation:
Consider the given figure.
Let AB denotes the height of a building.
BC = 1 mile
To find: AB
[tex]\tan C[/tex] = side opposite to the angle/side adjacent to the angle
[tex]\tan C=\frac{AB}{BC}[/tex]
Here, [tex]C=15^{\circ}\,,\,BC=1\,\,mile[/tex]
So,
[tex]\tan 15^{\circ}=\frac{AB}{1}\\\Rightarrow AB=\tan 15^{\circ}[/tex]
Here,
[tex]\tan 15^{\circ}=\tan \left (45^{\circ}-30^{\circ} \right )\\=\frac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ}\tan 30^{\circ}}\,\,\left \{ \because \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} \right \}\\=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}\\=\frac{\sqrt{3}-1}{\sqrt{3}+1}\\=\frac{1.73205-1}{1.73205+1}\\=\frac{0.73205}{2.73205}\\=0.2679[/tex]
Therefore,
[tex]AB=0.2679[/tex]
