## MaryJSmith 3 years ago The remedy is experience.. And I don't have any.. Help please ?

1. PaxPolaris

First find the length of BD using △BCD

2. PaxPolaris

using Pythagorean theorem

3. PaxPolaris

$BD^2=18^2+24^2$$\therefore BD=\sqrt{18^2+24^2}$ (you can remove the common factor of 18 and 24 out to make the math easier)$\implies BD = 6\sqrt{3^2+4^2}=6\sqrt{25}=\Large 30 ft.$

4. MaryJSmith

Thanks :)

5. PaxPolaris

Then to find AD you need to know trigonometry ....OR at-least the ratios of the sides of 30-60-90 triangles.

6. MaryJSmith

Do what ?

7. PaxPolaris

|dw:1333147962445:dw| In 30-60-90 triangles the ratios of the sides is always $\Large 1:\sqrt3:2$

8. MaryJSmith

Cool... Now what do I do with that ?

9. PaxPolaris

△ABD is a 30-60-90 triangle.... if you know one side(BD) you can find the others

10. MaryJSmith

so do i do 30*the sqr rt of 3 ?

11. MaryJSmith

That would be 52 (rounded up)

12. PaxPolaris

AD is smaller than BD... ...If the side opposite the 30deg (smallest) angle is length 1(smallest side) ..........then the side opposite the 60deg angle is sqrt3

13. MaryJSmith

I'm confused again..

14. PaxPolaris

$\Large {BD \over AD} = \frac {\sqrt3} 1$

15. MaryJSmith

so would you do 30*the square root of 3 divided by 2 then ? That comes out to be 26 (rounded up)

16. MaryJSmith

Dude , I'm so confused . Can you just give me the answer ? Lol . I'll ask my teacher about all this crap on Monday .

17. PaxPolaris

${ 30 \over AD} = {\sqrt3 \over 1}$ $AD = {30 \over \sqrt 3} = \Large 10\sqrt3\ ft.$

18. PaxPolaris

17ish

19. MaryJSmith

17ish ?

20. AsianDuck

17.3 (3sf)

21. MaryJSmith

thanks (what does 3sf mean ?)

22. AsianDuck

3 significant figures (1, 7 and 3)

23. PaxPolaris

ignore significant figures if you don't know what it means... by seventeen-ish, I meant 10*sqrt3 is around 17

24. AsianDuck

ok. 17.3 (1dp / 1 decimal point)