# Triangle side length

• Jun 4th 2009, 03:55 PM
chengbin
Triangle side length
Given $\displaystyle \triangle$ABC, where AD is perpendicular to BC, B=45$\displaystyle ^{\circ}$, C=60$\displaystyle ^{\circ}$, and AB=10, determine the length of AC.
• Jun 4th 2009, 04:40 PM
Soroban
Hello, chengbin!

Quote:

Given $\displaystyle \Delta ABC$, where $\displaystyle AD \perp BC,\;\;B=45^o,\;\;C=60^o,\;\;AB=10$
Determine the length of $\displaystyle AC$.

Is there a typo?
. . As written, altitude $\displaystyle AD$ is irrelevant.
Code:

                    A                     *                   *  *                 *    *           10 *        *             *          *           *              *         *                *       * 45°            60° *   B * - - - - - - - - - - - * C

Law of Sines: .$\displaystyle \frac{AC}{\sin45^o} \:=\:\frac{10}{\sin60^o} \quad\Rightarrow\quad AC \:=\:\frac{10\sin45^o}{\sin60^o}$

Therefore: .$\displaystyle AC \;=\;8.164965809$

• Jun 4th 2009, 04:56 PM
chengbin
There is no typo.

This was just the first page into the study of triangles.

I didn't get to the law of sine yet. How do you do this without the law of sine?

Can you also check this problem? I didn't get my solution book yet. This problem, and the problem in the first post, are all on the first page. So no advanced stuff yet.

Express the following function in terms of A.

$\displaystyle \sin \frac{B+C}{2}$

These are my steps.

= $\displaystyle \sin (90-\frac {A}{2})$

= $\displaystyle \cos \frac {A}{2}$
• Jun 4th 2009, 05:18 PM
Soroban
Hello again, chengbin!

Okay, so there is no typo and we can't use the Law of Sines.
. . Then that altitude is necessary.
Code:

                    A                     *                   * |*                 *  | *           10 *    |  *             *      |  *           *        |    *         *          |    *       * 45°        |  60° *   B * - - - - - - - * - - - * C                     D
We have to work our way over to side $\displaystyle AC$.

In right triangle $\displaystyle ADB\!:\;\;\sin45^o \:=\:\frac{AD}{10} \quad\Rightarrow\quad AD \:=\:10\sin45^o \:\approx\:7.07$

In right triangle $\displaystyle ADC\!:\;\;\sin60^o \:=\:\frac{AD}{AC} \quad\Rightarrow\quad AC \:=\:\frac{AD}{\sin60^o}$

. . Therefore: .$\displaystyle AC \;=\;\frac{7.07}{\sin60^o} \;\approx\;8.16$