Question

Darren designed this triangular shaped entrance, ABC, to a mall. Which expression can be used to find the distance between the points B and C of the entrance?

15/cot65 + 15/sec25

15/cot65 + 15cot25

15/sec65 + 15sec25

15/cot65 + 15/cot25

Answer 1

can u bump it?

Answer 2

thanks calc!!!

Answer 3

Thanks radar

Answer 4

Let's first work on the triangle on the right. You have the angle value and one side value. If:
\[\sec = \frac{hyp.}{adj.}\]\[\cot = \frac{adj.}{opp.}\]which one can you use (you should have one of the values.

Answer 5

BD=15/(Cot 65deg
DC=15Cot 25 degrees
BC=BD+DC

Answer 6

15cos(25

Answer 7

@hang254 I mean out of sec and cot, which one can you use? sec involves the hypotenuse. SInce you don't have the value of the hypotenuse in either triangle, we will be using cot. Do you understand so far?

Answer 8

yes

Answer 9

Ok. Knowing he ratio of cot, let's start with the triangle on the right.
\[\cot 25º = \frac{adj.}{opp.} = \frac{x}{15}\]what would you do here to get x by itself?

Answer 10

cross multiply

Answer 11

is ratio used in division

Answer 12

No. What do you do here to get x by itself:
\[\cot 25º = \frac{x}{15}\]

Answer 13

???

Answer 14

15cot(25 ?

Answer 15

Yup. THerefore, x = CD = 15cot(25º). Keep this measurement in mind. Now let's move onto the next triangle.
\[\cot 65º = \frac{15}{y}\]Get y by itself here.

Answer 16

15/cot65

Answer 17

Ok. y = BD = 15/cot 65º
BD + CD = BC
BC = 15/cot 65º + 15 cot 25º
which one is this answer?

Answer 18

*I mean this is your answer :)

Answer 19

option 2

Answer 20

thank you so much, i was confused on that one

Answer 21

There you go :)

Answer 22

np :)

Answer 23

thanks calc