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 - sum property

Since $\overrightarrow{DA}$ and $\overrightarrow{DC}$ are perpendicular, $\angle ADC = 90^{\circ}$. And $\angle ADC=\angle ADB+\angle BDC$. So, $(3a + 10)+13a=90$.

Step2: Combine like - terms

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

Step3: Solve for $a$

Subtract 10 from both sides: $16a=90 - 10=80$. Then divide both sides by 16: $a=\frac{80}{16}=5$.

Step4: Find $\angle ADB$

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

Step5: Find $\angle BDC$

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

Answer:

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