MHB What is the angle PQR in triangle $PQR$?

[anemone]

In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.

 

[Plato2]

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 m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.

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

Can you finish?

 

[anemone]

From the given you know m\left( {\angle PDR} \right)\;\& \;m\left( {\angle DPR} \right)~.

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

Can you finish?

Of course! But only because this is a challenge problem and I am not seeking for help for this problem, Plato!

 

[mente oscura]

In triangle $PQR$, let $D$ be the midpoint of $QR$. If $\angle PDQ=45^{\circ}$ and $\angle PRD=30^{\circ}$, determine $\angle QPD$.

Hello.

Spoiler

1º)\angle DPR=180º-(180º-45º)-30º=15º

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

\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}

3º) \angle PDO=60º-45º=15º

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

5º) For 4º): \angle QPD=45º-15º=30º

Regards.

 

[anemone]

Hello.

Spoiler

1º)\angle DPR=180º-(180º-45º)-30º=15º

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

\sin 30º=\dfrac{1}{2} \ then \ \overline{QO}=\overline{QD}=\overline{DR}

3º) \angle PDO=60º-45º=15º

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

5º) For 4º): \angle QPD=45º-15º=30º

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!