Question

in the diagram below, $overline{ad}perpoverline{dc}$, $mangle c = 55^{circ}$ and $mangle b = 44^{circ}$. find $mangle cab$. step angle reason 1 $mangle c = 55^{circ}$ given 2 $mangle b = 44^{circ}$ given try $manglesquare=square^{circ}$ select a reason

Explanation:

Step1: Find angle CAD in right - triangle ADC

In right - triangle $ADC$ with $\angle D = 90^{\circ}$ and $\angle C=55^{\circ}$, using the angle - sum property of a triangle ($\angle A+\angle B+\angle C = 180^{\circ}$). So, $\angle CAD=180^{\circ}-\angle D-\angle C$.
$\angle CAD = 180^{\circ}-90^{\circ}-55^{\circ}=35^{\circ}$

Step2: Find angle CAB in triangle ABC

In triangle $ABC$, we know $\angle B = 44^{\circ}$. Let $\angle CAB=x$. Using the angle - sum property of a triangle in $\triangle ABC$, we have $x + \angle B+\angle C+\angle CAD=180^{\circ}$. But we can also use the exterior - angle property. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles.
In $\triangle ABC$, consider the exterior angle at $A$ formed by the extension of $CA$. We know that in $\triangle ABD$, we can find $\angle CAB$ using the fact that the sum of angles in $\triangle ABC$.
First, in $\triangle ADC$, $\angle CAD = 35^{\circ}$.
In $\triangle ABC$, using the angle - sum property of a triangle:
Let $\angle CAB = y$. Then $y+44^{\circ}+55^{\circ}+ 90^{\circ}-y=180^{\circ}$ (using the fact that the sum of angles in the whole figure related to the right - angle at $D$).
We can also use the fact that in $\triangle ABC$, $\angle CAB=180^{\circ}-\angle B - \angle ACB$ (where $\angle ACB = 55^{\circ}$ and $\angle B = 44^{\circ}$)
$\angle CAB=180^{\circ}-44^{\circ}-55^{\circ}=81^{\circ}$

Answer:

$81^{\circ}$