Find the measures of the angles of the triangle whose vertices are A = (-1,0), B = (2,1), C = (1,-2).

opositor5t
2022-08-08
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Deja Navarro

Answered 2022-08-09
Author has **17** answers

$\overline{AB}=\sqrt{(-1-2{)}^{2}+(0-1{)}^{2}}=\sqrt{9+1}=\sqrt{10}$

$\overline{BC}=\sqrt{(1-2{)}^{2}+(-2-1{)}^{2}}=\sqrt{1+9}=\sqrt{10}$

$\overline{AC}=\sqrt{(-1-1{)}^{2}+(0+2{)}^{2}}=\sqrt{4+4}=2\sqrt{2}$

we will use

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}(\theta )$ where $\theta $ is angle A

we have:

$\mathrm{cos}(\mathrm{\angle}ABC)=\frac{\overline{A{B}^{2}}+\overline{B{C}^{2}}-\overline{A{C}^{2}}}{2\ast \overline{B{C}^{2}}\ast \overline{A{B}^{2}}}=\frac{10+10-8}{2\ast \sqrt{10}\ast \sqrt{10}}=\frac{12}{20}=\frac{3}{5}$

$\mathrm{\angle}ABC={\mathrm{cos}}^{-1}(\frac{3}{5})={53.13}^{0}$

$\mathrm{cos}(\mathrm{\angle}ACB)=\frac{(BC{)}^{2}+(AC{)}^{2}-(AB{)}^{2}}{2\ast BC\ast AC}=\frac{10+8-10}{2\ast \sqrt{10}\ast \sqrt{8}}=\frac{8}{2\sqrt{80}}=\frac{4}{4\sqrt{5}}=\frac{\sqrt{5}}{5}$

$\mathrm{\angle}BAC={63.43}^{0}$

$\overline{BC}=\sqrt{(1-2{)}^{2}+(-2-1{)}^{2}}=\sqrt{1+9}=\sqrt{10}$

$\overline{AC}=\sqrt{(-1-1{)}^{2}+(0+2{)}^{2}}=\sqrt{4+4}=2\sqrt{2}$

we will use

${a}^{2}={b}^{2}+{c}^{2}-2bc\mathrm{cos}(\theta )$ where $\theta $ is angle A

we have:

$\mathrm{cos}(\mathrm{\angle}ABC)=\frac{\overline{A{B}^{2}}+\overline{B{C}^{2}}-\overline{A{C}^{2}}}{2\ast \overline{B{C}^{2}}\ast \overline{A{B}^{2}}}=\frac{10+10-8}{2\ast \sqrt{10}\ast \sqrt{10}}=\frac{12}{20}=\frac{3}{5}$

$\mathrm{\angle}ABC={\mathrm{cos}}^{-1}(\frac{3}{5})={53.13}^{0}$

$\mathrm{cos}(\mathrm{\angle}ACB)=\frac{(BC{)}^{2}+(AC{)}^{2}-(AB{)}^{2}}{2\ast BC\ast AC}=\frac{10+8-10}{2\ast \sqrt{10}\ast \sqrt{8}}=\frac{8}{2\sqrt{80}}=\frac{4}{4\sqrt{5}}=\frac{\sqrt{5}}{5}$

$\mathrm{\angle}BAC={63.43}^{0}$

asked 2022-06-25

I'm proving this lemma

If $\mu $ is a finite Borel measure on a metric space $X$ and $\mathcal{A}$ a collection of mutually disjoint Borel sets of $X$, then at most countably many elements of $\mathcal{A}$ have nonzero $\mu $-measure.

This boils down to below result.

Let $({a}_{i}{)}_{i\in I}$ be a collection of non-negative real numbers. Let's define

$\sum _{i\in I}{a}_{i}:=sup\left\{\sum _{i\in J}{a}_{i}\phantom{\rule{thinmathspace}{0ex}}\right|\phantom{\rule{thinmathspace}{0ex}}J\text{is a countable subset of}I\}.$

If $\sum _{i\in I}{a}_{i}<\mathrm{\infty}$, then at most countably many ${a}_{i}$'s are positive.

Could you have a check on my attempt?

Let $J:=\{i\in I\mid {a}_{i}>0\}$ and ${J}_{r}:=\{i\in I\mid {a}_{i}\ge r\}$ for $r\in {\mathbb{Q}}_{>0}$. Then

$J=\bigcup _{r\in {\mathbb{Q}}_{>0}}{J}_{r}.$

It follows from $\sum _{i\in I}{a}_{i}<\mathrm{\infty}$ that ${J}_{r}$ is countable for all $r\in {\mathbb{Q}}_{>0}$. Then it follows from ${\mathbb{Q}}_{>0}$ is countable that $J$ is countable.

If $\mu $ is a finite Borel measure on a metric space $X$ and $\mathcal{A}$ a collection of mutually disjoint Borel sets of $X$, then at most countably many elements of $\mathcal{A}$ have nonzero $\mu $-measure.

This boils down to below result.

Let $({a}_{i}{)}_{i\in I}$ be a collection of non-negative real numbers. Let's define

$\sum _{i\in I}{a}_{i}:=sup\left\{\sum _{i\in J}{a}_{i}\phantom{\rule{thinmathspace}{0ex}}\right|\phantom{\rule{thinmathspace}{0ex}}J\text{is a countable subset of}I\}.$

If $\sum _{i\in I}{a}_{i}<\mathrm{\infty}$, then at most countably many ${a}_{i}$'s are positive.

Could you have a check on my attempt?

Let $J:=\{i\in I\mid {a}_{i}>0\}$ and ${J}_{r}:=\{i\in I\mid {a}_{i}\ge r\}$ for $r\in {\mathbb{Q}}_{>0}$. Then

$J=\bigcup _{r\in {\mathbb{Q}}_{>0}}{J}_{r}.$

It follows from $\sum _{i\in I}{a}_{i}<\mathrm{\infty}$ that ${J}_{r}$ is countable for all $r\in {\mathbb{Q}}_{>0}$. Then it follows from ${\mathbb{Q}}_{>0}$ is countable that $J$ is countable.

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$4\text{lb}=\frac{\u25fb}{\u25fb}=\frac{\u25fb}{\u25fb}=\text{oz}$

$4\text{lb}=\frac{\u25fb}{\u25fb}=\frac{\u25fb}{\u25fb}=\text{oz}$

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Given: Let $X$ be a sample from $P\in \mathcal{P}$, ${\delta}_{0}(X)$ be a decision rule (which may be randomized) in a problem with ${\mathbb{R}}^{k}$ as the action space, and $T$ be a sufficient statistic for $P\in \mathcal{P}$. For any Borel $A\subset {\mathbb{R}}^{k}$, define

${\delta}_{1}(T,A)=E[{\delta}_{0}(X,A)|T]$

Let $L(P,a)$ be a loss function. Show that

$\int L(P,a)d{\delta}_{1}(X,a)=E[\int L(P,a)d{\delta}_{0}(X,a)|T]$

My idea of proving this is to show that this holds for a simple function $L$ and generalize that to non-negative functions $L$ by using the conditional Montone Convergence Theorem. But I can't really show the equality for a simple function $L$. That is, if we take $L=\sum _{i=1}^{n}{c}_{i}{\mathbb{1}}_{{A}_{i}}$ can I show that the given result indeed holds true?

${\delta}_{1}(T,A)=E[{\delta}_{0}(X,A)|T]$

Let $L(P,a)$ be a loss function. Show that

$\int L(P,a)d{\delta}_{1}(X,a)=E[\int L(P,a)d{\delta}_{0}(X,a)|T]$

My idea of proving this is to show that this holds for a simple function $L$ and generalize that to non-negative functions $L$ by using the conditional Montone Convergence Theorem. But I can't really show the equality for a simple function $L$. That is, if we take $L=\sum _{i=1}^{n}{c}_{i}{\mathbb{1}}_{{A}_{i}}$ can I show that the given result indeed holds true?

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