Answer:
78.64 feet
Explanation:
Refer the attached figure
Height of the tall building = AC
The angle of depression from the top of one building to the foot of a building across the street is 53°. i.e.∠ADC = 53°
The angle of depression to the top of the same building is 19°.i.e.∠AEB = 19°
Height of small building = ED
The two buildings are 80 feet apart i.e. CD =BE= 80 feet
In ΔABC
![Tan \theta = (Perpendicular)/(Base)](https://img.qammunity.org/2020/formulas/mathematics/high-school/exbv4fn58hbl9s7983zbl7bq75j8vuhif8.png)
![Tan 53^(\circ)= (AC)/(CD)](https://img.qammunity.org/2020/formulas/mathematics/high-school/u1x9z3kgz6mh8l5hkdcl1ebehcvnylqid1.png)
![Tan 53^(\circ)= (AC)/(80)](https://img.qammunity.org/2020/formulas/mathematics/high-school/4ir3evhs6msxfn7gzij309y7r504h8atmu.png)
![1.327 * 80= AC](https://img.qammunity.org/2020/formulas/mathematics/high-school/jr88go4k6bwhlg2fqlyatj59m7o60o25vt.png)
![106.16= AC](https://img.qammunity.org/2020/formulas/mathematics/high-school/wubuvf2eaavhnzd2r0k5rpvi6o96ldb2t6.png)
In ΔABE
![Tan \theta = (Perpendicular)/(Base)](https://img.qammunity.org/2020/formulas/mathematics/high-school/exbv4fn58hbl9s7983zbl7bq75j8vuhif8.png)
![Tan 19^(\circ)= (AB)/(BE)](https://img.qammunity.org/2020/formulas/mathematics/high-school/ipzavimxgxhy44wqkuin3z0w0x4ja29hyr.png)
![Tan 19^(\circ)= (AB)/(80)](https://img.qammunity.org/2020/formulas/mathematics/high-school/60xujb7ont02eayd3rpnjx2dmrf5cee33e.png)
![0.344 * 80=AB](https://img.qammunity.org/2020/formulas/mathematics/high-school/k28niwqcvxq221oq5o96bys3vu132810wo.png)
![27.52= AB](https://img.qammunity.org/2020/formulas/mathematics/high-school/zquanw970sbwp6nnlsis8ntd9h5huxyvlo.png)
AC - AB = BC
106.16-27.52=BC
78.64=BC
So, BC = ED = 78.64 feet
Hence the height of the shorter building is 78.64 feet