\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}[theorem]{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.6in,right=0.7in,top=0.7in,bottom=0.9in} \pagestyle{fancy} \lhead{\color{blue} \LARGE Lecture Notes} \lfoot{\small \copyright \; Hidegkuti, 2013} \rfoot{\small Last revised: July 6, 2013} \chead{\huge Angles in a Triangle} \rhead{page \thepage} \cfoot{} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{23pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\LARGE Sample Problems\bigskip } \end{center} \begin{enumerate} \item Two angles in a triangle measure $39^{\circ }$ and $58^{\circ }$. \ Compute the measure of the third angle in the triangle. \item One angle in a triangle is $27^{\circ }$. \ The difference between the measures of the other two angles is $41^{\circ }$. \ Find the measure of the missing angles. \item In triangle ABC, the measure of angle $A$ is twice the measure of angle $B$. \ The measure of angle $C$ is $40^{\circ }$ more than the measure of angle $B$. \ Find the measure of the angles in the triangle. \item Consider triangle ABC. \ The measure of angle $B$ is ten degrees less than twice the measure of $A$. \ The measure of angle $C$ is two degrees less than three times the measure of $A$. \ Find the measure of the angles in this triangle. \item Consider the triangle shown on the picture below. \FRAME{dtbpF}{% 2.1318in}{1.9441in}{0pt}{}{}{triangle5.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.1318in;height 1.9441in;depth 0pt;original-width 5.0004in;original-height 4.5558in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../../Desktop/triangle5.bmp';file-properties "XNPEU";}}a) \ Compute the value of $x$. b) \ Compute the measure of the angles in the triangle. \bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Practice Problems \bigskip } \end{center} \begin{enumerate} \item Two angles in a triangle measure $43^{\circ }$ and $70^{\circ }$. \ Compute the measure of the third angle in the triangle. \item One angle in a triangle is $50^{\circ }$. \ The difference between the measures of the other two angles is $38^{\circ }$. \ Find the measure of the missing angles. \item In triangle ABC, the measure of angle $A$ is twice the measure of angle $B$. \ The measure of angle $C$ is $12^{\circ }$ more than the measure of angle $B$. \ Find the measure of the angles in the triangle. \item Consider triangle ABC. \ The measure of angle $B$ is one degree less than twice the measure of $A$. \ The measure of angle $C$ is thirteen degrees more than three times the measure of $A$. \ Find the measure of the angles in this triangle.\pagebreak \item Consider the triangle shown on the picture below. \FRAME{dtbpF}{2.348in% }{1.5714in}{0pt}{}{}{triangle8.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.348in;height 1.5714in;depth 0pt;original-width 5.278in;original-height 3.5137in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../../Desktop/triangle8.bmp';file-properties "XNPEU";}}a) \ Compute the value of $x$. b) \ Compute the measure of the angles in the triangle. {\LARGE \bigskip \bigskip }\bigskip \bigskip \bigskip \bigskip \bigskip \bigskip \end{enumerate} \begin{center} {\LARGE Sample Problems - Answers\bigskip } \end{center} \begin{enumerate} \item $83^{\circ }$ \item $56^{\circ }$ and \ $97^{\circ }$ \item $\angle A=70^{\circ }$, $\angle B=35^{\circ }$, $\angle C=75^{\circ }$ \item $\angle A=32^{\circ }$, \ $\angle B=54^{\circ }$, \ $\angle C=94^{\circ }$ \item a) \ $31^{\circ }\qquad \qquad $b) \ $62^{\circ }$, \ $40^{\circ }$, and $78^{\circ }$ \end{enumerate} \bigskip \bigskip \begin{center} {\LARGE Practice Problems - Answers\bigskip } \end{center} \begin{enumerate} \item $67^{\circ }$ \item $46^{\circ }$ and $84^{\circ }$ \item $\angle A=42^{\circ }$, $\angle B=84^{\circ }$, $\angle C=54^{\circ }$ \item $\angle A=28^{\circ }$, $\angle B=55^{\circ }$, $\angle C=97^{\circ }$ \item a) \ $22^{\circ }\qquad \qquad $b) \ $42^{\circ }$, \ $39^{\circ }$, \ $99^{\circ }$ \pagebreak \end{enumerate} \begin{center} {\LARGE Sample Problems - Solutions\bigskip } \end{center} \begin{enumerate} \item Two angles in a triangle measure $39^{\circ }$ and $58^{\circ }$. \ Compute the measure of the third angle in the triangle. \ Solution: \ Let us denote the missing angle by $x$. \ We know that the three angles add up to $180^{\circ }$.% \begin{eqnarray*} 39^{\circ }+58^{\circ }+x &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 97^{\circ }+x &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ subtract }% 97^{\circ } \\ x &=&83^{\circ } \end{eqnarray*}% So the missing angle measures $83^{\circ }$. \ We check: \begin{equation*} 39^{\circ }+58^{\circ }+83^{\circ }=180^{\circ } \end{equation*}% and so our solution is correct. \item One angle in a triangle is $27^{\circ }$. \ The difference between the measures of the other two angles is $41^{\circ }$. \ Find the measure of the missing angles. Solution: \ Let us denote the smaller unknown angle by $x$. \ Then the other unknown angle measures $x+41^{\circ }$ because these two differ by $% 41^{\circ }$. \ The equation will express the sum of the three triangles. \qquad angle 1: \ $27^{\circ }$ \qquad angle 2: \ $x$ \qquad angle 3: \ $x+41^{\circ }$% \begin{eqnarray*} 27^{\circ }+x+x+41^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 2x+68^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ subtract }68^{\circ } \\ 2x &=&112^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ divide by }2 \\ x &=&56^{\circ } \end{eqnarray*}% Our result, $x=56^{\circ }$ \ means that one unknown angle is $56^{\circ }$, and the other one is $x+41^{\circ }=56^{\circ }+41^{\circ }=97^{\circ }$. \ So the three angles are $27^{\circ }$, $56^{\circ }$ and $97^{\circ }$. \ We check: the sum of the three angles is $27^{\circ }+56^{\circ }+97^{\circ }=180^{\circ }$\ and the difference between the angles we found is $% 97^{\circ }-56^{\circ }=41^{\circ }$. \ Thus our solution is correct. \item In triangle ABC, the measure of angle $A$ is twice the measure of angle $B$. \ The measure of angle $C$ is $40^{\circ }$ more than the measure of angle $B$. \ Find the measure of the angles in the triangle. Solution: \ Notice that angles $A$ and $C$ are both compared to angle $B$. \ So, let us denote angle $B$ by $x$. \ Then we can label the other angles in terms of $x$. \qquad angle $A$: \ $2x$ \qquad angle $B$: \ $x$ \qquad angle $C$: \ $x+40^{\circ }$ The equation will express the sum of the three angles% \begin{eqnarray*} \angle A+\angle B+\angle C &=&180^{\circ } \\ 2x+x+x+40^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 4x+40^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ subtract }% 40^{\circ } \\ 4x &=&140^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ divide by }4 \\ x &=&35^{\circ } \end{eqnarray*} Now that we know the value of $x$, we can compute the measure of the angles: \qquad angle $A$: \ $2x~~~~~\Longrightarrow ~~~~70^{\circ }$ \qquad angle $B$: \ $x~~~~~\Longrightarrow ~~~~35^{\circ }$ \qquad angle $C$: \ $x+40^{\circ }~~~~~\Longrightarrow ~~~~75^{\circ }$ So the three angles are: angle $A$ is $70^{\circ }$, angle $B$ is $35^{\circ }$, and angle $C$ is $75^{\circ }$. \ We check: the sum of the three angles is $70^{\circ }+35^{\circ }+75^{\circ }=180^{\circ }$. \ We also check the connection between the angles: angle $A$ is indeed twice angle $B$: $% 70^{\circ }=2\cdot 35^{\circ }$ and angle $C$ is indeed $40^{\circ }$ greater than angle $B$: $75^{\circ }=35^{\circ }+40^{\circ }$. \item Consider triangle ABC. \ The measure of angle $B$ is ten degrees less than twice the measure of $A$. \ The measure of angle $C$ is two degrees less than three times the measure of $A$. \ Find the measure of the angles in this triangle. Solution: Let us denote the measure of angle $A$ by $x$. \ Then we can label all three angles in terms of $x$. \qquad angle $A$: \ $x$ \qquad angle $B$: \ $2x-10^{\circ }$ \qquad angle $C$: \ $3x-2^{\circ }$% \begin{eqnarray*} \angle A+\angle B+\angle C &=&180^{\circ } \\ x+2x-10^{\circ }+3x-2^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 6x-12^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ add }% 12^{\circ } \\ 6x &=&192^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ \ \ \ divide by }6 \\ x &=&32^{\circ } \end{eqnarray*}% \qquad angle $A$: \ $x=32^{\circ }$ \qquad angle $B$: \ $2x-10^{\circ }=2\cdot 32^{\circ }-10^{\circ }=64^{\circ }-10^{\circ }=54^{\circ }$ \qquad angle $C$: \ $3x-2^{\circ }=3\cdot 32^{\circ }-2^{\circ }=96^{\circ }-2^{\circ }=94^{\circ }$ So the three angles are $A=32^{\circ }$, \ $B=54^{\circ }$, and $C=94^{\circ }$. We check: the sum of the three angles is $32^{\circ }+54^{\circ }+94^{\circ }=180^{\circ }$. \ We also check the connection between the angles: \ the measure of $B$ is indeed ten degrees less than twice the measure of angle $A$% : \ $54^{\circ }=2\cdot 32^{\circ }-10^{\circ }$ and the measure of $C$ is indeed two degrees less than three times the measure of angle $A$: \ $% 94^{\circ }=3\cdot 32^{\circ }-2^{\circ }$. \ Thus our solution is correct.\pagebreak \item Consider the triangle shown on the picture below. \FRAME{dtbpF}{% 2.3921in}{2.1819in}{0pt}{}{}{triangle5.bmp}{\special{language "Scientific Word";type "GRAPHIC";maintain-aspect-ratio TRUE;display "USEDEF";valid_file "F";width 2.3921in;height 2.1819in;depth 0pt;original-width 5.0004in;original-height 4.5558in;cropleft "0";croptop "1";cropright "1";cropbottom "0";filename '../../../../../../Desktop/triangle5.bmp';file-properties "XNPEU";}}a) \ Compute the value of $x$. \ Solution: \ The three angles add up to $180^{\circ }$. \ This will give us an equation we can solve for $x$.% \begin{eqnarray*} 2x+3x-53^{\circ }+x+47^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ combine like terms} \\ 6x-6^{\circ } &=&180^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ add }6^{\circ } \\ 6x &=&186^{\circ }\text{ \ \ \ \ \ \ \ \ \ \ divide by }6 \\ x &=&31^{\circ } \end{eqnarray*}% b) \ Compute the measure of the angles in the triangle. Solution: \ Now that we have the value of $x$, we can compute the measure of all three angles in the triangle. \qquad angle 1: \ $2x=2\cdot 31^{\circ }=62^{\circ }$ \qquad angle 2: \ $3x-53^{\circ }=3\cdot 31^{\circ }-53^{\circ }=40^{\circ }$ \qquad angle 3: \ $x+47^{\circ }=31^{\circ }+47^{\circ }=78^{\circ }$ \pagebreak \end{enumerate} \end{document}