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xxMikexx [17]
3 years ago
8

What theorem or postulate can be used to justify that two triangles are congruent

Mathematics
2 answers:
kicyunya [14]3 years ago
4 0
You can use the ASA postulate which means angle side angle. if 2 angles and the including side are congruent to the corresponding parts of another triangle then the triangles are congruent. hope that helps :)
GaryK [48]3 years ago
4 0

Answer:

Following Theorems and postulates used to prove two triangles are congruent:

1). Side-Side-Side ( SSS ) congruence rule,

When three sides of one triangle are equal to three sides of another triangle , then the two triangles are congruent

2). Side-Angle-Side ( SAS ) congruence rule,

When two sides of one triangle are equal to two sides of another triangle and the an angle between two sides of one triangle equal to angle between two sides of another triangle , then the two triangles are congruent

3). Angle-Side-Angle ( ASA ) Congruence Rule,

When two angles of one triangle are equal to two angles of another triangle and side between two angles of one triangle equal to side between two angles of another triangle, then the two triangles are congruent.

4). Right angle - Hypotenuse - side ( RHS ) Congruence Rule,

If in two right triangles the hypotenuse and one side of on triangle are equal tot he hypotenuse and one side of other triangle, then the triangles are congruent.

Note:

there is also Angle-Angle-Side ( AAS ) Congruence Rule,

When two angles of one triangle are equal to two angles of another triangle and any side of one triangle equal to any side of another triangle, then the two triangles are congruent.

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\bf cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{then we can say that}~\hfill }{sin^{-1}\left( -\cfrac{5}{13} \right)\implies \theta }\qquad \qquad \stackrel{\textit{therefore then}~\hfill }{sin(\theta )=\cfrac{\stackrel{opposite}{-5}}{\stackrel{hypotenuse}{13}}}\impliedby \textit{let's find the \underline{adjacent}}

\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies \pm\sqrt{c^2-b^2}=a \qquad \begin{cases} c=hypotenuse\\ a=adjacent\\ b=opposite\\ \end{cases} \\\\\\ \pm\sqrt{13^2-(-5)^2}=a\implies \pm\sqrt{144}=a\implies \pm 12=a \\\\[-0.35em] ~\dotfill\\\\ cos\left[ sin^{-1}\left( -\cfrac{5}{13} \right) \right]\implies cos(\theta )=\cfrac{\stackrel{adjacent}{\pm 12}}{13}

le's bear in mind that the sine is negative on both the III and IV Quadrants, so both angles are feasible for this sine and therefore, for the III Quadrant we'd have a negative cosine, and for the IV Quadrant we'd have a positive cosine.

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77julia77 [94]

Answer:

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Step-by-step explanation:

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