• Dec 31st 2008, 08:16 AM
JoanneMac
Ok, here's some problems I hope to have solved...(Headbang)

1. Given triangle ABC with a=12, b=10, and c=8, find the angles in degrees and minutes.

2. Find the area of triangle ABC if a=40ft., b=25ft., and angleC=30degrees.

3. A triangle has sides 18, 27, and 36. What's it's area?

That's it for now. Thanks a bunch!

Joanne
• Dec 31st 2008, 08:22 AM
red_dog
1) $\displaystyle\cos A=\frac{b^2+c^2-a^2}{2bc}$

2) The area is $S=\frac{ab\sin C}{2}$

3) $S=\sqrt{p(p-a)(p-b)(p-c)}$, where $p=\frac{a+b+c}{2}$
• Dec 31st 2008, 09:24 AM
Soroban
Hello, Joanne!

Quote:

1. Given $\Delta ABC$ with: $a=12,\:b=10,\:c=8$
Find the angles in degrees and minutes.

We're expected to know the Law of Cosines and its variations . . .

$\cos A \:=\:\frac{b^2+c^2-a^2}{2bc} \:=\:\frac{10^2+8^2-12^2}{2(10(8)} \:=\:0.125$

. . $A \:=\:\cos^{-1}(0.125) \:\approx\:82.8192^o \quad\Rightarrow\quad\boxed{A\:=\:82^o49'}$

$\cos B \:=\:\frac{a^2+c^2-b^2}{2ac} \:=\:\frac{12^2+8^2-10^2}{2(12)(8)} \:=\:0.5625$

. . $B \:=\:\cos^{-1}(0.5625) \:\approx\:55.7711^o \quad\Rightarrow\quad\boxed{ B\:=\:55^o46'}$

$\cos C \:=\:\frac{a^2+b^2-c^2}{2ab} \:=\:\frac{12^2+10^2-8^2}{2(12)(10)} \:=\:0.75$

. . $C \:=\:\cos^{-1}(0.75) \:\approx\:41.4096^o \quad\Rightarrow\quad\boxed{ C \:=\:41^o25'}$

Quote:

2. Find the area of $\Delta ABC\text{ if }a=40\text{ ft},\; b=25\text{ ft},\;\angle C=30^o$
Area formula: . $A \;=\;\tfrac{1}{2}ab\,\sin C$
. . One-half the product of two sides times the sine of the included angle.

$A \;=\;\tfrac{1}{2}(40)(25)\sin30^o \;=\;\boxed{250\text{ ft}^2}$

Quote:

3. A triangle has sides 18, 27, and 36. Find its area.
We can find angle of the triangle (as we did in part 1).
Then use the area formula in part 2.

$\cos A \:=\:\frac{b^2+c^2-a^2}{2bc} \:=\:\frac{27^2+36^2-18^2}{2(18)(27)} \:=\:0.875$

. . $A \:=\:\cos^{-1}(0.875) \:=\:28.9550... \:\approx\:29^o$

$\text{Area} \;=\;\tfrac{1}{2}bc\,\sin A \:=\:\tfrac{1}{2}(27)(36)\sin29^o \;\approx\;\boxed{235.6 \text{ units}^2}$