Question

quadrilateral abcd is a rhombus and m∠cad = p + 15°. what is the value of p?

Explanation:

Step1: Recall rhombus property

In a rhombus, the diagonals bisect the angles of the rhombus. So, $\angle CAD$ and $\angle BAD$ are related. Also, in triangle $ADC$, $\angle ADC = 62^{\circ}$, and $\triangle ADC$ is isosceles (sides of a rhombus $AD = DC$), so $\angle CAD=\angle ACD$. Using the angle - sum property of a triangle ($\angle ADC+\angle CAD+\angle ACD = 180^{\circ}$), and since $\angle CAD=\angle ACD$, we have $2\angle CAD=180^{\circ}-\angle ADC$.

Step2: Calculate $\angle CAD$

Substitute $\angle ADC = 62^{\circ}$ into the formula $2\angle CAD=180^{\circ}-\angle ADC$. Then $2\angle CAD=180 - 62=118^{\circ}$, so $\angle CAD=\frac{118^{\circ}}{2}=59^{\circ}$.

Step3: Solve for $p$

Given that $\angle CAD=p + 15^{\circ}$, and $\angle CAD = 59^{\circ}$, we set up the equation $p+15^{\circ}=59^{\circ}$. Solving for $p$, we get $p=59^{\circ}-15^{\circ}=44^{\circ}$.

Answer:

$44$