# Math Help - Triangle side length

1. ## Triangle side length

Given $\triangle$ABC, where AD is perpendicular to BC, B=45 $^{\circ}$, C=60 $^{\circ}$, and AB=10, determine the length of AC.

2. Hello, chengbin!

Given $\Delta ABC$, where $AD \perp BC,\;\;B=45^o,\;\;C=60^o,\;\;AB=10$
Determine the length of $AC$.
Is there a typo?
. . As written, altitude $AD$ is irrelevant.
Code:
                    A
*
*  *
*     *
10 *        *
*           *
*              *
*                 *
* 45°            60° *
B * - - - - - - - - - - - * C

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

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

3. 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.

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

These are my steps.

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

= $\cos \frac {A}{2}$

4. 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 $AC$.

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

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

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