Answer:
The height of tower is 322.7 feet.
Explanation:
Given
Distance between a building and tower= 250 feet
BCDE is a rectangle .Therefore, we have BC=ED and CD=BE=250 feet
In triangle ABE
![tan\theta=(perpendicula \; side )/(hypotenuse)](https://img.qammunity.org/2020/formulas/mathematics/college/gzb12ikvoxergrsivgkkwy02wpup2ah4i7.png)
![\theta=38^(\circ)](https://img.qammunity.org/2020/formulas/physics/high-school/9kxwykl3u81harl7a7ejfph2irqx6tvshv.png)
![tan38^(\circ)=(AB)/(BE)](https://img.qammunity.org/2020/formulas/mathematics/college/q4wljfou6qkyqm537x3gxdn5ini69hhx39.png)
![(AB)/(250)=0.781](https://img.qammunity.org/2020/formulas/mathematics/college/tvnhjzwlngrwzb46nmhr3xthjqslmbdykd.png)
![AB=0.781*250](https://img.qammunity.org/2020/formulas/mathematics/college/ik3p95lqg26ycqps3dwkc1fdp5l9t4iulg.png)
AB=195.25 feet
In triangle EDC
![\theta=27^(\circ)](https://img.qammunity.org/2020/formulas/mathematics/college/tfpzaw1xjhrpfvlkfz1uk3hbred97tpm62.png)
![tan27^(\circ)=(ED)/(CD)](https://img.qammunity.org/2020/formulas/mathematics/college/578cnlhcwt8ggbpj3x33kexhqv8iczevo4.png)
![(ED)/(250)=0.509](https://img.qammunity.org/2020/formulas/mathematics/college/hom4oy6qhwm5p2xilsiq84eaftd6prmfwl.png)
![ED=250*0.509](https://img.qammunity.org/2020/formulas/mathematics/college/5k8mlux6h4dse9kvrs6smltdb06yn6w6j0.png)
ED=127.25 feet
ED=BC=127.25 feet
The height of tower=AB+BC
The height of tower=195.25+127.25=322.5 feet