State the third congruence required to prove the congruence of triangles using the indicated postulate.

a)

b)

c)

d)

Jason Farmer
2021-02-21
Answered

State the third congruence required to prove the congruence of triangles using the indicated postulate.

a)

b)

c)

d)

You can still ask an expert for help

estenutC

Answered 2021-02-22
Author has **81** answers

Step 1

Given figure,

we have to state the congruence required to prove the congruence of triangles using the indicated postulate.

Step 2

It is given that both triangles are congruent by criteria of SAS

by criteria of SAS

so, the third congruence is

so, option B is correct.

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Focuses of the hyperbola ${F}_{1}(4,\text{}2),\text{}{F}_{2}(-1,\text{}-10)$ and the tangent equation $3x+4y-5=0$ are given. Find the semiaxes of the hyperbola.

I’ve tried using the definition of hyperbola and the equation for the tangent line to the quadratic curve to form some sort of system of equations, but it didn’t work out.

I’ve tried using the definition of hyperbola and the equation for the tangent line to the quadratic curve to form some sort of system of equations, but it didn’t work out.

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what is the equation of a circle with the center (2, -4) and the radius 4.

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I am trying to figure out what the best method is to go about finding this locus.

$\mathrm{arg}{\displaystyle \frac{z-a}{z-b}}=\theta $

I am aware that it must be part of an arc of a circle that passes through the points $a$ and $b$. The argument means that the vector $(z-a)$ leads the vector $(z-b)$ by $\theta $ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to

$\mathrm{arg}{\displaystyle \frac{z-a}{z-b}}=-\theta $

Would this just be the other part of the same circle as previously or a different circle?

Perhaps you could help by showing my how it would work with the question

$\mathrm{arg}\frac{z-2j}{z+3}=\pi /3$

$\mathrm{arg}{\displaystyle \frac{z-a}{z-b}}=\theta $

I am aware that it must be part of an arc of a circle that passes through the points $a$ and $b$. The argument means that the vector $(z-a)$ leads the vector $(z-b)$ by $\theta $ and so by the argument must lie on an arc of a circle due to the converse of 'angles in the same segment theorem'.

My question is, how do I know what the circle will look like, ie. the centre of the circle and the radius. If I don't need to know this, how will I know what the circle looks like.

Finally, if I changed the locus to

$\mathrm{arg}{\displaystyle \frac{z-a}{z-b}}=-\theta $

Would this just be the other part of the same circle as previously or a different circle?

Perhaps you could help by showing my how it would work with the question

$\mathrm{arg}\frac{z-2j}{z+3}=\pi /3$

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$-2b(3{t}^{2}(s+1)+6t(s+1)+3s+2)-2g(6ts+3t+6s+2)-3t{s}^{2}+6ts+3t-3{s}^{2}+3s+1>0$

for some nonnegative reals $t$,$s$. Here $g$,$b$ are also $\ge 0$. Can it be possible to get a function $f$ such that we get the upper bound on $b$ as $f(g)$ i.e, $b<f(g)$? If one can find $t$,$s$ for which $b$ is maximum, one will get $f(g)$.

$-2b(3{t}^{2}(s+1)+6t(s+1)+3s+2)-2g(6ts+3t+6s+2)-3t{s}^{2}+6ts+3t-3{s}^{2}+3s+1>0$

for some nonnegative reals $t$,$s$. Here $g$,$b$ are also $\ge 0$. Can it be possible to get a function $f$ such that we get the upper bound on $b$ as $f(g)$ i.e, $b<f(g)$? If one can find $t$,$s$ for which $b$ is maximum, one will get $f(g)$.

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Given:

$\mathrm{\u25b3}GRA$

$\mathrm{\angle}GRA={85}^{\circ}$

$\mathrm{\angle}RGA={63}^{\circ}$

$\mathrm{\angle}RAG={?}^{\circ}$