Given PQS = 45^ and QPS = 45^.

Question

Given PQS = 45^ and QPS = 45^.

Asked on March 27, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-27T16:11:58.762Z

Step 1: Analyze $\triangle PQS$. Given $\angle PQS = 45^\circ$ and $\angle QPS = 45^\circ$. The sum of angles in a triangle is $180^\circ$. $$ \angle PSQ = 180^\circ - \angle PQS - \angle QPS $$ $$ \angle PSQ = 180^\circ - 45^\circ - 45^\circ $$ $$ \angle PSQ = 90^\circ $$ This confirms that $PS$ is perpendicular to $QR$. Step 2: Analyze $\triangle PRS$. Since $PS \perp QR$, we have $\angle PSR = 90^\circ$. We are given $\angle PRT = 30^\circ$. Note that $\angle PRT$ is the same as $\angle PRS$. So, in right-angled $\triangle PRS$: $$ \angle RPS = 180^\circ - \angle PSR - \angle PRS $$ $$ \angle RPS = 180^\circ - 90^\circ - 30^\circ $$ $$ \angle RPS = 60^\circ $$ Step 3: Calculate $\angle SPT$. From the diagram, point $T$ lies on the line segment $SR$. Therefore, $\angle RPS$ is the sum of $\angle RPT$ and $\angle SPT$. $$ \angle RPS = \angle RPT + \angle SPT $$ We know $\angle RPS = 60^\circ$ (from Step 2) and we are given $\angle RPT = 30^\circ$. $$ 60^\circ = 30^\circ + \angle SPT $$ $$ \angle SPT = 60^\circ - 30^\circ $$ $$ \angle SPT = 30^\circ $$ The value $QS = 4.5$ cm is not required to solve for $\angle SPT$. The final answer is $\boxed{\text{30}^\circ}$.

Accepted Answer · 2026-03-27T16:11:58.762Z

Step 1: Analyze $\triangle PQS$. Given $\angle PQS = 45^\circ$ and $\angle QPS = 45^\circ$. The sum of angles in a triangle is $180^\circ$. $$ \angle PSQ = 180^\circ - \angle PQS - \angle QPS $$ $$ \angle PSQ = 180^\circ - 45^\circ - 45^\circ $$ $$ \angle PSQ = 90^\circ $$ This confirms that $PS$ is perpendicular to $QR$. Step 2: Analyze $\triangle PRS$. Since $PS \perp QR$, we have $\angle PSR = 90^\circ$. We are given $\angle PRT = 30^\circ$. Note that $\angle PRT$ is the same as $\angle PRS$. So, in right-angled $\triangle PRS$: $$ \angle RPS = 180^\circ - \angle PSR - \angle PRS $$ $$ \angle RPS = 180^\circ - 90^\circ - 30^\circ $$ $$ \angle RPS = 60^\circ $$ Step 3: Calculate $\angle SPT$. From the diagram, point $T$ lies on the line segment $SR$. Therefore, $\angle RPS$ is the sum of $\angle RPT$ and $\angle SPT$. $$ \angle RPS = \angle RPT + \angle SPT $$ We know $\angle RPS = 60^\circ$ (from Step 2) and we are given $\angle RPT = 30^\circ$. $$ 60^\circ = 30^\circ + \angle SPT $$ $$ \angle SPT = 60^\circ - 30^\circ $$ $$ \angle SPT = 30^\circ $$ The value $QS = 4.5$ cm is not required to solve for $\angle SPT$. The final answer is $\boxed{\text{30}^\circ}$.