Question

Darren designed this triangular shaped entrance, ABC, to a mall.

The figure shows two right triangles ABD and ADC having side AD common. Triangle ABD has length of side AD equal to 15 feet and the measure of angle BAD is 65 degrees. Triangle ADC has angle DAC equal to 25 degrees.

Which expression can be used to find the distance between the points B and C of the entrance?
Answer

15 by cot 65 degrees plus 15 by cot 25 degrees

15 by cot 65 degrees plus 15 by sec 25 degrees

15 (cot 65° + sec 25°)

15(cot 65° + cot 25°)

Answer 1

BD= 15tan(65)
DC=15tan(25)
if you want to use the cofucntion of tan, the cotangent, then cot(90-x)=tanx
BD+DC=BD=15tan(65)+15tan(25)=15cot(90-65)+15cot(90-25)=
15cot25+15cot65

Answer 2

=15(cot65+cot25)

Answer 3

wat dokwerker said

Answer 4

howmany more questions do u hav