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!