1. ## 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
nks

2. 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
nks
The Law of Sines is:
$\displaystyle \frac{d}{sin(D)} = \frac{e}{sin(E)} = \frac{f}{sin(F)}$

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

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

So $\displaystyle D = 19.8^o$. Now, D could merely be the reference angle, so it is possible that $\displaystyle D = 180^o - 19.8^o = 160.2^o$ 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:
$\displaystyle E = 180 - D - F = 180^o - 19.8^o - 26^o = 134.2^o$

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

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

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:
$\displaystyle A = \sqrt{s(s-d)(s-e)(s-f)}$ where $\displaystyle s = \frac{d+e+f}{2}$ (Called the "semi-perimeter.")

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

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

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

-Dan

3. Hello, Jim!

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

Dan did an excellent job in solving the triangle.

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

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

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

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