# sine rule

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• Nov 19th 2006, 03:39 AM
jim49990
sine rule
using the sine rule solve the following triangles DEF and find their areas
d=17cm f=22cm F=26*
could anyone please show me how to do this example so i could do the others
tha
:confused: nks
• Nov 19th 2006, 04:10 AM
topsquark
Quote:

Originally Posted by jim49990
using the sine rule solve the following triangles DEF and find their areas
d=17cm f=22cm F=26*
could anyone please show me how to do this example so i could do the others
tha
:confused: nks

The Law of Sines is:
$\frac{d}{sin(D)} = \frac{e}{sin(E)} = \frac{f}{sin(F)}$

So for sides d and f:
$\frac{17}{sin(D)} = \frac{22}{sin(26)}$

$sin(D) = \frac{17 \cdot sin(26)}{22} = 0.338741$

So $D = 19.8^o$. Now, D could merely be the reference angle, so it is possible that $D = 180^o - 19.8^o = 160.2^o$ except that the sum of the interior angles in any triangle is 180 degrees. If D were 160.2 degrees then angle D plus angle F accounts for more than 180 degrees not even counting angle E. So angle D must be 19.8 degrees.

So we know two angles in the triangle and as I said above we know that the sum of the interior angles of a triangle is 180 degrees. So for angle E:
$E = 180 - D - F = 180^o - 19.8^o - 26^o = 134.2^o$

Employing the Law of Sines for e and f:
$\frac{e}{sin(134.2)} = \frac{22}{sin(26)}$

$e = \frac{22 \cdot sin(134.2)}{sin(26)} = 35.9788$

So side e is about 36.0 cm in length.

There are various ways to get the area of the triangle. This is Heron's formula:
$A = \sqrt{s(s-d)(s-e)(s-f)}$ where $s = \frac{d+e+f}{2}$ (Called the "semi-perimeter.")

So
$s = \frac{17 + 36 + 22}{2} = 37.4894$ (I am using the unrounded expression for e.)

$A = \sqrt{37.4894 (37.4894 - 17)(37.4894 - 36)(37.4894 - 22)} = 134.063$

So the area of the triangle is about $134.1 \, cm^2$.

-Dan
• Nov 19th 2006, 05:30 AM
Soroban
Hello, Jim!

Quote:

Using the Sine Rule, solve triangle $D{E}F$ and find its area.
. . $d = 17\text{ cm, } f = 22\text{ cm, }F = 26^o$

Dan did an excellent job in solving the triangle.

Here's another way to find its area.
. . $\text{Area } = \:\frac{1}{2}bc\sin A$

One-half the product of two sides and the sine of the included angle.

Since we know: . $d = 17,\:f = 22,\:E = 134.2^o$

. . we have: . $A \:=\:\frac{1}{2}(17)(22)\sin134.2^o \:=\:134.0622836 \:\approx\:134.1$ cm².