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Determine the angle PQR

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anemone

MHB POTW Director
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Feb 14, 2012
3,756
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
 

Plato

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MHB Math Helper
Jan 27, 2012
196
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
From the given you know [tex]m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.[/tex]

You also know [tex]m\left( {\angle PQD} \right)+m\left( {\angle QPD} \right)=135^o[/tex].

Can you finish?
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
From the given you know [tex]m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.[/tex]

You also know [tex]m\left( {\angle PQD} \right)+m\left( {\angle QPD} \right)=135^o[/tex].

Can you finish?
Of course! But only because this is a challenge problem and I am not seeking for help for this problem, Plato!:eek:
 

mente oscura

Well-known member
Nov 29, 2013
172
In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.
Hello.

1º)[tex]\angle DPR=180º-(180º-45º)-30º=15º[/tex]

2º) Draw the height from [tex]Q \ to \ \overline{PR}[/tex], getting the point [tex]O[/tex]

[tex]\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}[/tex]

3º) [tex]\angle PDO=60º-45º=15º[/tex]

4º) For 1º) [tex]\overline{OP}=\overline{OD}=\overline{QO} \rightarrow{} \angle PQO=\angle QPO=45º[/tex]

5º) For 4º): [tex]\angle QPD=45º-15º=30º[/tex]



Regards.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Hello.

1º)[tex]\angle DPR=180º-(180º-45º)-30º=15º[/tex]

2º) Draw the height from [tex]Q \ to \ \overline{PR}[/tex], getting the point [tex]O[/tex]

[tex]\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}[/tex]

3º) [tex]\angle PDO=60º-45º=15º[/tex]

4º) For 1º) [tex]\overline{OP}=\overline{OD}=\overline{QO} \rightarrow{} \angle PQO=\angle QPO=45º[/tex]

5º) For 4º): [tex]\angle QPD=45º-15º=30º[/tex]



Regards.
WOW! That's a brilliant way to draw a line $QO$ such that it is perpendicular to $PR$ and everything immediately becomes obvious.

Bravo, mente oscura!