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In the triangle ABC at the right above, AB=CB, and P and Q are points on the sides CB and AB such that AC=AP=PQ=QB. Find the number of degrees in angle B.

In The Triangle ABC At The Right Above ABCB And P And Q Are Points On The Sides CB And AB Such That ACAPPQQB Find The Number Of Degrees In Angle B class=

Sagot :

Since QP and QB are equal the triangle PQB the angles:

[tex]\begin{gathered} \measuredangle QPB=\measuredangle QBP \\ \measuredangle QPB=x \end{gathered}[/tex]

The last angle can be found by adding all the internal angles and making it equal to 180 degrees.

[tex]\begin{gathered} \measuredangle QPB+\measuredangle QBP+\measuredangle BQP=180 \\ x+x+\measuredangle BQP=180 \\ \measuredangle BQP=180-2x \end{gathered}[/tex]

The angle BQP and the angle AQP are suplementary, this means that their sum is equal to 180 degrees. So we have:

[tex]\begin{gathered} \measuredangle AQP+\measuredangle BQP=180 \\ \measuredangle AQP+180-2x=180 \\ \measuredangle AQP=180-180+2x \\ \measuredangle AQP=2x \end{gathered}[/tex]

Since the sides AP and PQ are equal, then the angle PAQ is equal to AQP.

[tex]\begin{gathered} \measuredangle PAQ=\measuredangle AQP \\ \measuredangle PAQ=2x \end{gathered}[/tex]

To find the last angle on that triangle we can add all the internal angles and make it to 180 degrees.

[tex]\begin{gathered} \measuredangle PAQ+\measuredangle AQP+\measuredangle APQ=180 \\ 2x+2x+\measuredangle APQ=180 \\ \measuredangle APQ=180-4x \end{gathered}[/tex]

The angle APC is suplementary with the sum of the angles APQ and BPQ. So we have:

[tex]\begin{gathered} \measuredangle APC+\measuredangle APQ+\measuredangle BPQ=180 \\ \measuredangle APC+180-4x+x=180 \\ \measuredangle APC=3x \end{gathered}[/tex]

The sides AP and AC are equal, therefore the angles APC and ACP are also equal.

[tex]\begin{gathered} \measuredangle ACP=\measuredangle APC \\ \measuredangle ACP=3x \end{gathered}[/tex]

Then we can find the last angle on that triangle.

[tex]\begin{gathered} \measuredangle CAP+\measuredangle ACP+\measuredangle APC=180 \\ \measuredangle CAP+3x+3x=180 \\ \measuredangle CAP=180-6x \end{gathered}[/tex]

The angle CAB is equal to the sum of CAP and PAQ. So we have:

[tex]\begin{gathered} \measuredangle CAB=\measuredangle CAP+\measuredangle PAQ \\ \measuredangle CAB=180-6x+2x \\ \measuredangle CAB=180-4x \end{gathered}[/tex]

Since the sides AB and BC are equal, then the angles ACB and CAB must also be equal. We can find the value of x with this.

[tex]\begin{gathered} \measuredangle BAC=\measuredangle ACB \\ 180-4x=3x \\ 7x=180 \\ x=\frac{180}{7} \end{gathered}[/tex]

The value of x is 180/7

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