You don't even need the picture to solve this one.
You said that h = 5 cot(Θ) , and you said that Θ is 30 degrees.
All you need now is to find the cotangent of Θ, plop that into the equation,
and the solution practically jumps off the paper into your lap.
To find the cotangent of 30 degrees, you can use a calculator, look it up
in a book, read it off of a slide rule if you have one, draw a picture of a
30-60-90 right triangle etc. You'll find that the cotangent of 30 degrees
is √3 . That's about 1.732 .
So your equation is h = 5 (1.732) = <em>8.66 </em>(rounded)
Apparently, somebody gave you the equation, and asked you to find 'h'.
Once you had the equation, you didn't even need to know that 'h' has
anything to do with a triangle.
Answer:
1, 2, 3, 6, 47, 94, 141, 282, 1609, 3218, 4827, 9654, 75623, 151246, 226869, 453738
Step-by-step explanation:
I looked it up
If she stopped at 8;05, and she started 20 minutes earlier, just imaging a clock at 8:05 and turn back time 20 minutes.
Five minutes would be 8:00, ten minutes is 7:55, fifteen minutes is 7:50, and twenty minutes is 7:45. This, she started at 7:45
Answer:
15. ∠ABE = 50°
16. ∠EBD = 10n - 19
Step-by-step explanation:
If 
bisects ∠ABD then we have ∠ABE = ∠DBE
So, we will have: (6x + 2)° = (8x - 14)°
⇒ 6x + 2 = 8x - 14
⇒ 8x - 6x = 14 + 2
⇒ 2x =16
⇒ x = 8
∴ ∠ABE = 6(8) + 2
= 48 + 2
= 50°
∴∠ABE = 50°
16. If bisects ∠ABD then ∠ABE + ∠EBD = ∠ABD
⇒(12n - 8) + ∠EBD = (22n - 11)
⇒ ∠EBD = 22n - 11 - 12n + 8
⇒ ∠EBD = 10n - 3
Hence, the answer.
