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P and Q are two bases for a mountain climb. PQ is 600 m and QR is a vertical stretch of a rock ace. The angle of elevation of Q from P is 31^{\circ}, and the angle of elevation of R from P is 41^{\circ}. Use the sine law in \triangle PQR to calculate the height of the vertical climb, QR, to the nearest metre.

 

see https://web2.0calc.com/questions/sine-law

 

 

\(\begin{array}{rcll} \frac{ \sin(41^{\circ}-31^{\circ}) } {h} &=& \frac{ \sin(90^{\circ}-41^{\circ}) }{600\ m} \\\\ \frac{ \sin(10^{\circ}) } {h} &=& \frac{ \sin(49^{\circ}) }{600\ m} \\\\ \frac {h} { \sin(10^{\circ}) } &=& \frac{600\ m} { \sin(49^{\circ}) }\\\\ h &=& \frac{\sin(10^{\circ})} { \sin(49^{\circ}) } \cdot 600\ m \\\\ h &=& \frac{0.17364817767} {0.75470958022} \cdot 600\ m \\\\ h &=& 0.23008609168 \cdot 600\ m \\\\ h &=& 138.0516550080\ m \\\\ \mathbf{h} & \mathbf{=} & \mathbf{138\ m}\\ \end{array} \)

 

laugh

heureka  May 25, 2016
edited by heureka  May 25, 2016
edited by heureka  May 25, 2016

Best Answer 

 #3
avatar+18566 
+5

Hallo Melody,

the diagram is in Latex.

My new version:

 

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

 

The source code is:

\(\begin{tikzpicture} % Hilfslinie \coordinate[ ] (D) at (6.14300380421,0); %Punkte \coordinate[label=left:$P$] (P) at (0,0); \draw [dashed,color=red,line width=0.5pt] (P)--(D); \coordinate[label=right:$Q$] (Q) at (31:6); \coordinate[label=right:$R$] (R) at (5.14300380421,4.47074499954); \draw [dashed,color=red,line width=0.5pt] (3.14300380421,4.47074499954)--(R); % Winkel\)
\(\begin{scope}[shift={(P)}] \draw (0,0) -- (0:2.00cm) arc (0:31:2.00cm); \draw (15.5:2.35cm) node {$31^{\circ}$}; \draw[fill=green] (0,0) -- (0:1.25cm) arc (0:41:1.25cm); \draw (20.5:1.55cm) node {$41^{\circ}$}; \end{scope}\)

\(\begin{scope}[shift={(R)}] \draw[fill=green] (0,0) -- (-180:1.25cm) arc (-180:-139:1.25cm); \draw (-160:1.55cm) node {$41^{\circ}$}; \end{scope}\)
\(\draw [color=blue!50,line width=1.5pt] (P)--(Q)--(R)--cycle; \coordinate[label=right:$h$] (h) at ($(R)!.5!(Q)$); \draw ($(P)!.7!(Q)$) node [below,rotate=31]{$600\ m$}; \draw [fill=blue!50] (0,0) circle [radius=1.5pt]; \draw [fill=blue!50] (31:6) circle [radius=1.5pt]; \draw [fill=blue!50] (R) circle [radius=1.5pt]; \end{tikzpicture}\)

 

laugh

heureka  May 26, 2016
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5+0 Answers

 #1
avatar+90227 
0

Very nice Heureka :)

That is a great diagram    laugh

Melody  May 26, 2016
 #2
avatar+18566 
+5

Hallo Melody,

the diagram is in Latex.

My new version:

 

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

 

The source code is:

\begin{tikzpicture}
% Hilfslinie
\coordinate[ ] (D) at (6.14300380421,0);

%Punkte
\coordinate[label=left:$P$]  (P) at (0,0);
\draw [dashed,color=red,line width=0.5pt] (P)--(D);

\coordinate[label=right:$Q$] (Q) at (31:6);
\coordinate[label=right:$R$] (R) at (5.14300380421,4.47074499954);
\draw [dashed,color=red,line width=0.5pt] (3.14300380421,4.47074499954)--(R);

% Winkel
\begin{scope}[shift={(P)}]
   \draw (0,0) -- (0:2.00cm) arc (0:31:2.00cm);
   \draw (15.5:2.35cm) node {$31^{\circ}$};
   \draw[fill=green] (0,0) -- (0:1.25cm) arc (0:41:1.25cm);
   \draw (20.5:1.55cm) node {$41^{\circ}$};
\end{scope}

\begin{scope}[shift={(R)}]
   \draw[fill=green] (0,0) -- (-180:1.25cm) arc (-180:-139:1.25cm);
   \draw (-160:1.55cm) node {$41^{\circ}$};
\end{scope}

\draw [color=blue!50,line width=1.5pt] (P)--(Q)--(R)--cycle;

   \coordinate[label=right:$h$] (h) at ($(R)!.5!(Q)$);
   \draw ($(P)!.7!(Q)$) node [below,rotate=31]{$600\ m$};

\draw [fill=blue!50] (0,0) circle [radius=1.5pt];
\draw [fill=blue!50] (31:6) circle [radius=1.5pt];
\draw [fill=blue!50] (R) circle [radius=1.5pt];

\end{tikzpicture}

 

laugh

heureka  May 26, 2016
 #3
avatar+18566 
+5
Best Answer

Hallo Melody,

the diagram is in Latex.

My new version:

 

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

 

The source code is:

\(\begin{tikzpicture} % Hilfslinie \coordinate[ ] (D) at (6.14300380421,0); %Punkte \coordinate[label=left:$P$] (P) at (0,0); \draw [dashed,color=red,line width=0.5pt] (P)--(D); \coordinate[label=right:$Q$] (Q) at (31:6); \coordinate[label=right:$R$] (R) at (5.14300380421,4.47074499954); \draw [dashed,color=red,line width=0.5pt] (3.14300380421,4.47074499954)--(R); % Winkel\)
\(\begin{scope}[shift={(P)}] \draw (0,0) -- (0:2.00cm) arc (0:31:2.00cm); \draw (15.5:2.35cm) node {$31^{\circ}$}; \draw[fill=green] (0,0) -- (0:1.25cm) arc (0:41:1.25cm); \draw (20.5:1.55cm) node {$41^{\circ}$}; \end{scope}\)

\(\begin{scope}[shift={(R)}] \draw[fill=green] (0,0) -- (-180:1.25cm) arc (-180:-139:1.25cm); \draw (-160:1.55cm) node {$41^{\circ}$}; \end{scope}\)
\(\draw [color=blue!50,line width=1.5pt] (P)--(Q)--(R)--cycle; \coordinate[label=right:$h$] (h) at ($(R)!.5!(Q)$); \draw ($(P)!.7!(Q)$) node [below,rotate=31]{$600\ m$}; \draw [fill=blue!50] (0,0) circle [radius=1.5pt]; \draw [fill=blue!50] (31:6) circle [radius=1.5pt]; \draw [fill=blue!50] (R) circle [radius=1.5pt]; \end{tikzpicture}\)

 

laugh

heureka  May 26, 2016
 #4
avatar+18566 
+5

Hallo Melody,

the diagram is in Latex.

My new version:

 

The Latex package is:

\usepackage{tikz}

\usetikzlibrary{  calc,   positioning,   automata, ... }

 

The source code is:

 

 

laugh

heureka  May 26, 2016
 #5
avatar+90227 
0

Hi Heureka,

I have added the fabulous thread to the Latex sticky thread.

 

http://web2.0calc.com/questions/latex#r72

 

Thanks     laugh laugh laugh

Melody  May 26, 2016

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