# Math Help - Lengths of Frame Members

1. ## Lengths of Frame Members

You work in the design department of a company which produces fabricated steel structures and have been given sketches of two frames. For each, work out the lengths of all the members and the angles involved. The roof frame shown below is symmetrical about a centre line axis.

2. ## Re: Little brain teaser :P

You work in the design department of a company which produces fabricated steel structures and have been given sketches of two frames. For each, work out the lengths of all the members and the angles involved. The roof frame shown below is symmetrical about a centre line axis.
so that those interested may see the diagram ...

3. ## Re: Little brain teaser :P

Hello, redbullracer!

You work in the design department of a company which produces
fabricated steel structures and have been given sketches of two frames.
For each, work out the lengths of all the members and the angles involved.
The roof frame shown below is symmetrical about a centre line axis.

Code:
                  B
o
**:**
* * : * *
*20*  :  *  *
6 *   * 5 :   *   *
*    *    :    *    *
A o  *  o  *  o  *  o  *  o C
D     F     E

In $\Delta ABD$, Law of Cosines: . $AD^2 \:=\:5^2 + 6^2 - 2(5)(6)\cos20^o \:=\:4.618442753$

. . Hence: . $AD \:=\:2.149056247 \quad\Rightarrow\quad \boxed{AD \:\approx\:2.15}$

Then: . $\cos A \:=\:\frac{6^2 + 2.15^2 - 5^2}{2(6)(2.15)} \:=\:0.605523256$

. . Hence: . $A \:=\:52.73349563^o \quad\Rightarrow\quad \boxed{A \:\approx\: 52.73^o}$

Then: . $\angle ABF \:=\:90^o - 52.73^o \:=\:37.27^o$

. . Hence: . $\angle DBF \:=\:37.27^o - 20^o \quad\Rightarrow\qyad \boxed{\angle DBF \:=\:17.27^o}$

Then in $\Delta DBF:\,DF \,=\,5\sin12.27^o \:=\:1.484374609 \quad\Rightarrow\quad \boxed{DF \:\approx\:1.48}$

. . and: . $\angle BDF \:=\:90^o - 12.27^o \quad\Rightarrow\quad \boxed{\angle BDF \:=\:77.73^o }$