Question

1. in the diagram, m∠bda = 144°. find m∠adc. a. 91° b. 82° c. 62° d. 40°

Explanation:

Step1: Recall angle - sum property

The sum of angles around a point is 360°. Here, assume the angle between \(BD\) and \(CD\) is \(90^{\circ}\), and \(\angle BDA = 144^{\circ}\), and we want to find \(\angle ADC\). Let \(\angle ADC=x\). Then \(90^{\circ}+144^{\circ}+x = 360^{\circ}\).

Step2: Solve for \(x\)

First, simplify the left - hand side of the equation: \(90 + 144+x=234 + x\). Then, solve the equation \(234+x = 360\) for \(x\). Subtract 234 from both sides: \(x=360 - 234\).
\(x = 126\). But this is wrong. Let's assume the non - right angle part of the angle formed by \(BD\) and \(CD\) is \((7x + 14)^{\circ}\) and \(\angle ADC=(9x + 46)^{\circ}\) and \(\angle BDA = 144^{\circ}\). Since the sum of angles around point \(D\) is \(360^{\circ}\), we have \((7x + 14)+(9x + 46)+144=360\).

Step3: Combine like terms

Combine the \(x\) terms and the constant terms: \((7x+9x)+(14 + 46+144)=360\), which gives \(16x+204 = 360\).

Step4: Isolate the variable

Subtract 204 from both sides: \(16x=360 - 204\), so \(16x = 156\), then \(x=\frac{156}{16}=\frac{39}{4}=9.75\).
Then \(\angle ADC=(9x + 46)^{\circ}\), substitute \(x = 9.75\) into it: \(\angle ADC=9\times9.75+46=87.75 + 46=133.75\) (This is wrong. Let's assume the angles are supplementary in a different way).
Since \(\angle BDA+\angle ADC = 180^{\circ}\) (linear - pair of angles, assuming \(B\), \(D\), \(C\) are arranged in a way that they form a straight - line like figure), then \(\angle ADC=180^{\circ}-\angle BDA\).

Step5: Calculate \(\angle ADC\)

Given \(\angle BDA = 144^{\circ}\), then \(\angle ADC=180 - 144\).
\(\angle ADC = 36^{\circ}\) (There is a problem with the given options. But if we assume the angles are related in a different non - shown correct way and we consider the fact that if we assume \(\angle BDA\) and \(\angle ADC\) are supplementary in a correct geometric sense)

If we assume there is an error in the problem setup and we consider the fact that if \(\angle BDA\) and \(\angle ADC\) are supplementary (a common geometric relationship when two angles share a common side and their non - common sides form a straight line), and \(\angle BDA = 144^{\circ}\), then \(\angle ADC=180 - 144=36^{\circ}\). But if we assume the following:
Let's assume the sum of the non - \(\angle BDA\) angles around point \(D\) is \(360-\angle BDA\). If we assume the two non - \(\angle BDA\) angles \((7x + 14)\) and \((9x + 46)\) are part of the angles around \(D\) and we know \(\angle BDA = 144^{\circ}\).
We have \((7x + 14)+(9x + 46)=360 - 144\).
\(16x+60 = 216\).
\(16x=216 - 60=156\).
\(x=\frac{156}{16}=\frac{39}{4}\).
\(\angle ADC=9x + 46=9\times\frac{39}{4}+46=\frac{351}{4}+46=\frac{351 + 184}{4}=\frac{535}{4}=133.75\) (wrong).
If we assume \(\angle BDA\) and \(\angle ADC\) are supplementary (a more likely geometric relationship as there is no other clear indication):

Answer:

\(36^{\circ}\) (None of the given options are correct)