Roof Structure Question find lengths

**zak_killcity14@hotmail.co** · December 3rd 2006, 05:33 AM

Find the rise of BD?

**ThePerfectHacker** · December 3rd 2006, 05:59 AM

I give you a hint.

The area of Triangle ABD = area ABC +area BCD

Now, call $BD=h$ , the height.

Then you can find,
$CD=\sqrt{4700^2-h^2}$

Thus,
area ABD= $(1/2)h(3300 +\sqrt{4700^2-h^2})$

area ABC can be computed by Heron's Formula

area BCD= $(1/2)h\sqrt{4700^2-h^2}$

You have an equation to solve for!

**Soroban** · December 3rd 2006, 06:03 AM

Hello, Zak!

In $\Delta ABC$ , use the Law of Cosines to find $\angle ACB.$

$\cos(\angle ACB)\:=\:\frac{3300^2 + 4700^2 - 7000^2}{2(3300)(4700)}\:=\:-0.516441006$

Hence: . $\angle ACB \:\approx\:121.1^o$

Then: . $\angle BCD\:=\:180^o - 121.1^o \:=\:58.9^o$

In right triangle $BDC\!:\;\;\sin 58.9^o \:=\:\frac{BD}{4700}$

Therefore: . $BD \:=\:4700\sin58.9^o \:=\:4024.717024 \:\approx\:4025$

**Soroban** · December 3rd 2006, 05:58 PM

Hello, Zak!

An alternate approach . . .

In $\Delta ABC$ , use the Law of Cosines to find angle $A.$

$\cos A\:=\:\frac{7000^2 + 3300^2 - 4700^2}{2(7000)(3300)} \:=\:0.8181818181$

. . Hence: . $A \:\approx\:35.1^o$

In right triangle $BDA\!:\;\sin 35.1^o \:= \:\frac{BD}{7000}\quad\Rightarrow\quad BD \:=\:7000\sin35.1^o \:=\:4025.036764$

. . Therefore: . $BD \:\approx\:4025$

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