Question

use the figure below for problems 16 - 17
question 16
1 pts
determine the measurement of \\(\angle cab\\)
55
question 17
1 pts
determine the measuremnt of \\(\angle cba\\)
45

Explanation:

Step1: Analyze quadrilateral ADBC

We know $AD=BC$, $AC=BD$, $\angle D=90^\circ$, so $ADBC$ is a rectangle? No, wait: $AC=BC$, $AD=BD$, so $\triangle ACB$ and $\triangle ADB$ are isosceles. First, solve for $x$ using $\triangle ADB$:
Sum of angles in $\triangle ADB$: $x + (x-10^\circ) + 90^\circ = 180^\circ$

Step2: Simplify to solve for $x$

$$\begin{align*} x + x - 10^\circ + 90^\circ &= 180^\circ\\ 2x + 80^\circ &= 180^\circ\\ 2x &= 100^\circ\\ x &= 50^\circ \end{align*}$$

Step3: Find $\angle CAB$

$\angle CAB = x = 50^\circ$ (given $\angle CAB = x$)

Step4: Find $\angle CBA$

$\angle CBA = x - 10^\circ = 50^\circ - 10^\circ = 40^\circ$? Wait, no: $\triangle ACB$ is congruent to $\triangle ADB$ (SSS: $AC=AD$, $BC=BD$, $AB=AB$), so $\angle CBA = \angle ABD = x-10^\circ$. Wait, correction: in $\triangle ADB$, angles are $\angle DAB = x$, $\angle ABD = x-10^\circ$, $\angle D=90^\circ$. So solving:

$$ x + (x-10) + 90 = 180\\ 2x = 100\\ x=50 $$

So $\angle CAB = x = 50^\circ$, $\angle CBA = x-10 = 40^\circ$

Answer:

Question 16: $\boldsymbol{50^\circ}$
Question 17: $\boldsymbol{40^\circ}$