Basic Trigonometry Proof

[Vertciel]

Hello everyone,

Here is my problem:

In a triangle ABC, we know that b(cosA) = a(cosB). Prove that the triangle is isoceles. (Hint: Use The Cosine Law)

I've tried to isolate b, a, cos A, and cosB from b(cosA) = a(cosB) and substitute them into the law of cosines. This has gotten me nowhere though.

Could anyone please just give a hint?

Thanks.

 

[soroban]

Hello, Vertciel!

n \(\displaystyle \Delta ABC\), we know that: \(\displaystyle \:b\cdot\cos A \:= \:a\cdot\cos B\)
Prove that the triangle is isosceles. .(Hint: Use The Cosine Law)

I haven't found a way to apply the Cosine Law (yet) . . .

            C
            *
           *| *
          * |   *  a
       b *  |     * 
        *   |       *
       *    |         *
    A * * * + * * * * * * B
            D

Draw altitude \(\displaystyle CD\) from vertex \(\displaystyle C\) to side \(\displaystyle AB\).

In right triangle \(\displaystyle CDA:\;\cos A \:= \:\frac{AD}{b}\;\;\Rightarrow\;\;AD \:=\:b\cdot\cos A\)

In right triangle \(\displaystyle CDB:\;\cos B \:=\:\frac{DB}{a}\;\;\Rightarrow\;\;DB \:=\:a\cdot\cos B\)

Since \(\displaystyle \,b\cdot\cos A \:=\:a\cdot\cos B\), then: \(\displaystyle \,AD\:=\B\)


In right triangle \(\displaystyle CDA\) and \(\displaystyle CDB\), we have: \(\displaystyle \:\begin{Bmatrix}AD & = & DB \\ \angle CDA & = & \angle CDB \\ CD & = & CD\end{Bmatrix}\)

Hence: \(\displaystyle \:\Delta CDA\) is congruent to \(\displaystyle \Delta CBD\;\) (s.a.s)

Therefore: \(\displaystyle AC \,=\,BC\) (corresponding parts)

. . and \(\displaystyle \Delta ABC\text{ is isosceles.}\)

 

[pka]

\(\displaystyle \L\begin{array}{rcrr}
\left. {\begin{array}{r}
{a^2 = b^2 + c^2 - 2bc\cos (A)} \\
{b^2 = a^2 + c^2 - 2ac\cos (B)} \\
\end{array}} \right\} &\Rightarrow & \quad \frac{{b^2 + c^2 - a^2 }}{{2c}}& =& \frac{{a^2 + c^2 - b^2 }}{{2c}} \\
&\Rightarrow &2b^2 + c^2 &=& 2a^2 + c^2 \\
&\Rightarrow& b& =& a \\
\end{array}\)

 

[Vertciel]

\(\displaystyle \L\begin{array}{rcrr}
\left. {\begin{array}{r}
{a^2 = b^2 + c^2 - 2bc\cos (A)} \\
{b^2 = a^2 + c^2 - 2ac\cos (B)} \\
\end{array}} \right\} &\Rightarrow & \quad \frac{{b^2 + c^2 - a^2 }}{{2c}}& =& \frac{{a^2 + c^2 - b^2 }}{{2c}} \\\)

Thank you soroban and pka for your assistance.

I have a question for pka's solution:

Shouldn't cos A = (b^2 + c^2 - a^2)/2bc and cos B = (a^2 + c^2 - b^2)/2ac? If so, how did you eliminate the a and b in the denominators? Did you eliminate the a and b on the basis that a = b?

Thanks again.

 

[tkhunny]

Look VERY carefully at your given information.

Do NOT solve for the cosine alone.

 

[pka]

\(\displaystyle \L b\cos (A) = \frac{{b^2 + c^2 - a^2 }}{{2c}}\)

 

[Vertciel]

Thank you very much for your assistance and your reply.