How do you determine the following? 1. cos 0 2. the length of BD 3. Determine y correct to 2 decimal places 4. calculate the length of EC with out a calculator?

Question

586 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-04T14:58:37

See below

enter image source here

1

#cos(theta)#

Using triangle DBC

#cos(theta)="adjacent"/"hypotenuse"=2/sqrt(7)=color(blue)((2sqrt(7))/7#

2

Using triangle ABD:

#sin(60^@)=(BD)/2=>BD=2sin(60^@)=2(sqrt(3))/2=color(blue)(sqrt(3))#

3

Using triangle DBC:

#sin(y)="opposite"/"hypotenuse"=2/sqrt(7)=(2sqrt(7))/7#

#y=arcsin(sin(y))=arcsin((2sqrt(7))/7)=>y=49.1066053481186^@#

#=color(blue)(49.11^@)color(white)(888)# ( 2 d.p.)

4

First we can find the the length of AC. This will then be the hypotenuse of triangle ACE. We know DC=2. so we need to find AD.

Using triangle ABD:

#cos(60^@)="adjacent"/"hypotenuse"=(AD)/2#

#=>AD=2cos(60^@)=2sqrt(3)/2=sqrt(3)#

So AC = AD + DC = #sqrt(3)+2#

Looking at triangle ACE.

#z=45^@#, we know this because we have angles of #90^@# and #45^@#

#z=180-(90+45)=45#

This means that:

AE = EC.

Letting #AE=EC=x#

By Pythagoras's theorem:

#(AC)^2=x^2+x^2=2x^2#

#2x^2=(AC)^2=(sqrt(3)+2)^2#

Taking square roots:

#xsqrt(2)=sqrt(3)+2#

#x=(sqrt(3)+2)/sqrt(2)#

#x=EC=color(blue)((sqrt(6)+2sqrt(2))/2)#