Answered

Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get immediate and reliable solutions to your questions from a knowledgeable community of professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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

We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.