Design a Pennant for the school, using isosceles triangle - Please help?

[Meadow3]

The match club of Meadow High designed a pennant for the school yacht. The pennant was in the shape of an isosceles triangle. Two points, P and Q, are located so that AC=AP=PQ=QB. Find the measure of angle B.

Given: <C=<P1 ;
<B=<P3;
<A2=<Q1;
<A=<A1+<A2
<P1+<P2+<P3=180
<Q1+<Q2=180


I started with <A + <B + <C = 180 with <A=<C therefore <B=180 - 2(<C). From here on, i'm lost. There is no given measurement, I'm going in a circle. Please help.

 

[soroban]

Hello, Meadow3!

Fortunately, I recognized the problem.

The match club of Meadow High designed a pennant for the school yacht.
The pennant was in the shape of an isosceles triangle.
Two points, P and Q, are located so that CA = AP = PQ = QB.
Find the measure of angle B.

             B
             *

            *b\

           *   \

        Q *     \
            *
         /2b  * b\
                *
        /         * P
               * a
       /2b  *      \
         *        a
    A * * * * * * * * C

In isosceles triangle \(\displaystyle APC,\text{ let }a = \angle ACP = \angle APC\)
. . Then:.\(\displaystyle \angle PAC \,=\,180-2a\)

In isosceles triangle \(\displaystyle QBP,\text{ let }b = \angle QBP = \angle QPB\)
Since \(\displaystyle \angle AQP\) is an exterior angle: \(\displaystyle \angle AQP = 2b\)

In isosceles triangle \(\displaystyle APQ,\:\angle QAP = 2b \quad\Rightarrow \angle QPA = 180 - 4b\)

At point \(\displaystyle P\!:\:\angle BPQ + \angle QPA + \angle APC \:=\:180 \)
. . Hence: .\(\displaystyle b + (180-4b) + a \:=\:180 \quad\Rightarrow\quad a - 3b \:=\:0\;\;{\bf[1]}\)

At point \(\displaystyle A\!:\;\angle QAP + \angle PAC \:=\:\angle QAC \:=\:\angle ACP\)
. . Hence: .\(\displaystyle 2b + (180 - 2a) \:=\:a \quad\Rightarrow\quad 3a-2b\:=\:180 \;\;{\bf[2]}\)

. . . \(\displaystyle \begin{array}{ccccccc}&\text{W\!e have:} & 3a - 2b &=& 180 & {\bf[2]} \\ & 3\times [1]\!: & 3a - 9b &=& 0 & {\bf[3]} \end{array}\)

.\(\displaystyle \text{Subtract {\bf[2] - [3]}: }\; 7b \;=\;180\)


Therefore: .\(\displaystyle \angle B \:=\:\dfrac{180}{7} \;=\;25\frac{5}{7}^o\)

 

[Meadow3]

RE: ....isosceles triangle

Thanks soroban, you're good. Hope this won't be on the test. Thanks

 

[soroban]

There is an expansion of this problem.
You may find it fascinating . . . or maybe not.

We have four identical matches.
Join them in a line with "hinges".

. . \(\displaystyle -\!-\!-\!-*-\!-\!-\!-*-\!-\!-\!-*-\!-\!-\!-\)


Using two straightedges, "nudge" the chain into an isosceles triangle
. . as shown in the diagram.

                \ /
[SIZE=3]
             *

            * \

           *   \

          *     \
            *
         /    *  \
                *
        /         *
               *  
       /    *      \
         *         
      * * * * * * * *

     /               \ [/SIZE]

Then the vertex angle is: \(\displaystyle \dfrac{180^o}{7}\)


If we use three matches, we have:

                \ /
[SIZE=3]
             *

            * \

           *   \

          *     \
            *
         /    *  \
                *
        * * * * * *

       /           \ [/SIZE]

And the vertex angle is: \(\displaystyle \dfrac{180^o}{5}\)

If we use five matches, the vertex angle is: \(\displaystyle \dfrac{180^o}{9}\)

Get it?