Question

please answer these question ​

Answer 1

Answer:

a)Figure attached above

Height of the tower=AB,

Height of the building=AC

Distance between foot of the building to tower=BC=20m

∠CBE=60° , and ∠EAD=45°

b) In ∆EBC,

tan∠CBE=CE/CB

=tan60°=CE/20

=√3=CE/20

CE=20√3

=CE=20×1.73=34.6m

c)In ∆EAD,

tan∠EAD=ED/AD

=tan45°=ED/20

=1=ED/20

ED=20m

Now ,Height of the tower AB=DC

=CE-ED

=34.6m-20m

=14.6m

Step-by-step explanation:

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