Question

in the diagram below we have that $d\widehat{u}v = 73^{circ}$ and $c\widehat{b}t = 61^{circ}$. find: 1. $c\widehat{t}b=square^{circ}$ 2. $a\widehat{b}t = 119^{circ}$ 3. $t\widehat{u}c=square^{circ}$

Explanation:

Step1: Use angle - sum property of a triangle

In $\triangle CBT$, we know that the sum of interior angles of a triangle is $180^{\circ}$. Given $\angle CBT = 61^{\circ}$ and assuming $\angle BCT = 0^{\circ}$ (since $AE\parallel SV$ and we consider the relevant angle - relationships), then $\angle CTB=180^{\circ}-\angle CBT - 0^{\circ}=180 - 61=119^{\circ}$. But if we consider the non - zero angle formed by the intersection of the lines, and since $\angle CBT = 61^{\circ}$ and the sum of angles in $\triangle CBT$ is $180^{\circ}$, and assuming the angles are formed by parallel lines $AE$ and $SV$, $\angle CTB = 61^{\circ}$ (alternate interior angles for parallel lines $AE$ and $SV$).

Step2: Find $\angle ABT$

$\angle ABT$ and $\angle CBT$ are supplementary (linear pair of angles). Since $\angle CBT = 61^{\circ}$, then $\angle ABT=180 - 61=119^{\circ}$.

Step3: Find $\angle TUC$

$\angle TUC$ and $\angle DUV$ are vertical angles. Given $\angle DUV = 73^{\circ}$, so $\angle TUC = 73^{\circ}$ (vertical angles are equal).

Answer:

1. $61$
2. $119$
3. $73$