# Proved that

• Apr 20th 2010, 09:13 AM
dapore
Proved that
• Apr 20th 2010, 01:23 PM
Soroban
Hello, dapore!

Quote:

In any triangle $\displaystyle ABC$ with angles $\displaystyle A,B,C$ and sides $\displaystyle a,b,c\!:$

. . . $\displaystyle b\cos A + a\cos B \:=\:c$

Code:

              C               o             *| *             * |  *           *  |    *         b *  |      * a         *    |        *         *    |          *       *      |            *     A o  *  o  *  *  *  o B           x  D      y       : - - - - - c - - - - - :

Construct altitude $\displaystyle CD$ to side $\displaystyle AB.$
Let $\displaystyle x = AD,\;Y = DB$

In right triangle $\displaystyle CDA\!:\;\;\cos A \,=\,\frac{x}{b} \quad\Rightarrow\quad x \,=\,b\cos A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\cos B \,=\,\frac{y}{a} \quad\Rightarrow\quad y \,=\,a\cos B$ .[2]

Note that: .$\displaystyle x + y \:=\:c$ .[3]

Substitute [1] and [2] into [3]: .$\displaystyle b\cos A + a\cos B \;=\;c$
• Apr 20th 2010, 02:38 PM
dapore
Quote:

Originally Posted by Soroban
Hello, dapore!

Code:

              C               o             *| *             * |  *           *  |    *         b *  |      * a         *    |        *         *    |          *       *      |            *     A o  *  o  *  *  *  o B           x  D      y       : - - - - - c - - - - - :
Construct altitude $\displaystyle CD$ to side $\displaystyle AB.$
Let $\displaystyle x = AD,\;Y = DB$

In right triangle $\displaystyle CDA\!:\;\;\cos A \,=\,\frac{x}{b} \quad\Rightarrow\quad x \,=\,b\cos A$ .[1]

In right triangle $\displaystyle CDB\!:\;\;\cos B \,=\,\frac{y}{a} \quad\Rightarrow\quad y \,=\,a\cos B$ .[2]

Note that: .$\displaystyle x + y \:=\:c$ .[3]

Substitute [1] and [2] into [3]: .$\displaystyle b\cos A + a\cos B \;=\;c$

Thanks Soroban
just for you

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