# Determine the angle PQR

#### anemone

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

#### Plato

##### Well-known member
MHB Math Helper
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

##### MHB POTW Director
Staff member
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

##### Well-known member
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)$$\angle DPR=180º-(180º-45º)-30º=15º$$

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

##### MHB POTW Director
Staff member
Hello.

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!