Triangle side length

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June 4th 2009, 03:55 PM
chengbin

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.
June 4th 2009, 04:40 PM
Soroban

Hello, chengbin!

Quote:

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

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

These are my steps.

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

= $\cos \frac {A}{2}$
June 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 $AC$ .

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