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svet-max [94.6K]
3 years ago
14

I am posting my work for an application of integrals problem. I would love to know if anyone else agrees with my work!

Mathematics
1 answer:
quester [9]3 years ago
8 0
Your work appears to be correct.

The results from a graphing calculator are in agreement.

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In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint : <img src="https://tex.z-dn.net/?f=
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Answer:

See Below.

Step-by-step explanation:

Statements:                                                           Reasons:

\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}      Given

\displaystyle 2) \text{ } m \angle PAB = 60                                                     Definition of equilateral.

3)\text{ } m \angle QAC = 60                                                     Definition of equilateral.

4)\text{ } m\angle PAB = m\angle QAC                                          Substitution

5)\text{ } m\angle PAC=m\angle PAB+m\angle BAC                       Angle Addition

\displaystyle 6)\text{ } m\angle QAB=m\angle QAC+m\angle BAC                       Angle Addition

7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC                       Substitution

\displaystyle 8)\text{ } m\angle PAC=m\angle QAB                                         Substitution

9)\text{ } PA=BA                                                          Definition of equilateral

10)\text{ } AC=AQ                                                        Definition of equilateral

\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ                                            Side-Angle-Side Congruence*

\displaystyle 12)\text{ } PC=BQ                                                        CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

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