Question

1. 11) 12) 13) 14) find the measure of the angle indicated. 15) find m∠s. 16) find m∠h. 17) find m∠fab. 18) find m∠ydc.

Explanation:

Step1: Recall angle - sum properties

For a triangle, the sum of interior angles is 180°. For a linear - pair of angles, the sum is 180°.
Let's take problem 15 as an example.
In triangle $SRT$, we know that the exterior angle at $R$ is 140°. The interior angle at $R$ and the exterior angle at $R$ form a linear pair. So the interior angle at $R=180 - 140=40^{\circ}$.
Using the angle - sum property of a triangle: $(3x + 4)+(8x + 4)+40 = 180$.

Step2: Simplify the equation

Combine like terms: $3x+8x+4 + 4+40=180$, which gives $11x+48 = 180$.
Subtract 48 from both sides: $11x=180 - 48=132$.

Step3: Solve for x

Divide both sides by 11: $x=\frac{132}{11}=12$.

Step4: Find the measure of $\angle S$

Substitute $x = 12$ into the expression for $\angle S$: $m\angle S=3x + 4=3\times12+4=36 + 4=40^{\circ}$.

We can follow similar steps for other problems:
For problem 16:
In triangle $GHF$, the exterior angle at $F$ and the interior angle at $F$ form a linear - pair.
Let's assume the interior angle at $F = y$. Then $y+(14x + 1)=180$.
Using the angle - sum property of a triangle: $89+(5x - 7)+y = 180$.
Substitute $y = 180-(14x + 1)$ into the triangle equation: $89+(5x - 7)+180-(14x + 1)=180$.
Simplify: $89 + 5x-7+180-14x - 1=180$.
Combine like terms: $- 9x+261=180$.
Subtract 261 from both sides: $-9x=180 - 261=-81$.
Divide by - 9: $x = 9$.
Then $m\angle H=5x - 7=5\times9-7=45 - 7=38^{\circ}$.

For problem 17:
In the triangle with angles $55^{\circ},(3x + 2)$ and the interior angle at $A$ (let's call it $z$), and the exterior angle at $A$ is $(13x - 3)$.
We know that the exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. So $13x-3=55+(3x + 2)$.
Simplify: $13x-3=3x+57$.
Subtract $3x$ from both sides: $13x-3x-3=3x-3x + 57$, which gives $10x-3=57$.
Add 3 to both sides: $10x=60$.
Divide by 10: $x = 6$.
The exterior angle $\angle FAB=13x-3=13\times6-3=78 - 3=75^{\circ}$.

For problem 18:
The exterior angle at $D$ and the interior angle at $D$ form a linear - pair. Let the interior angle at $D = w$.
Using the angle - sum property of a triangle: $(6x + 6)+80+w = 180$.
The exterior angle at $D=(15x + 5)$. And $w = 180-(15x + 5)$.
Substitute $w$ into the triangle equation: $(6x + 6)+80+180-(15x + 5)=180$.
Simplify: $6x+6 + 80+180-15x - 5=180$.
Combine like terms: $-9x+261=180$.
Subtract 261 from both sides: $-9x=-81$.
Divide by - 9: $x = 9$.
The exterior angle $m\angle YDC=15x + 5=15\times9+5=135 + 5=140^{\circ}$.

Answer:

1. $m\angle S = 40^{\circ}$
2. $m\angle H = 38^{\circ}$
3. $m\angle FAB = 75^{\circ}$
4. $m\angle YDC = 140^{\circ}$