Question

rays da and dc are perpendicular. point b lies in the interior of ∠adc. if m∠adb=(3a + 10)° and m∠bdc = 13a°, find a, m∠adb, and m∠bdc.

Explanation:

Step1: Use angle - addition postulate

Since rays $DA$ and $DC$ are perpendicular, $\angle ADC = 90^{\circ}$. And because point $B$ lies in the interior of $\angle ADC$, we know that $m\angle ADB+m\angle BDC=m\angle ADC$. So, $(3a + 10)+13a=90$.

Step2: Combine like - terms

Combine the terms on the left - hand side of the equation: $3a+13a + 10=90$, which simplifies to $16a+10 = 90$.

Step3: Isolate the variable term

Subtract 10 from both sides of the equation: $16a+10 - 10=90 - 10$, resulting in $16a=80$.

Step4: Solve for a

Divide both sides of the equation by 16: $a=\frac{80}{16}=5$.

Step5: Find $m\angle ADB$

Substitute $a = 5$ into the expression for $m\angle ADB$: $m\angle ADB=3a + 10=3\times5+10=15 + 10=25^{\circ}$.

Step6: Find $m\angle BDC$

Substitute $a = 5$ into the expression for $m\angle BDC$: $m\angle BDC=13a=13\times5 = 65^{\circ}$.

Answer:

$a = 5$, $m\angle ADB=25^{\circ}$, $m\angle BDC=65^{\circ}$