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

Question

Accepted Answer

1. **For problem 14**:
   - **Explanation**:
     - **Step 1: Use the angle - sum property of a triangle**
       - In a triangle, the sum of interior angles is $180^{\circ}$. Let the triangle have angles $40^{\circ}$, $(20x + 5)^{\circ}$, and $(9x-2)^{\circ}$. So, $40+(20x + 5)+(9x-2)=180$.
     - **Step 2: Simplify the left - hand side of the equation**
       - Combine like terms: $40 + 5-2+20x+9x=180$, which gives $43 + 29x=180$.
     - **Step 3: Solve for $x$**
       - Subtract 43 from both sides: $29x=180 - 43=137$, then $x=\frac{137}{29}$.
     - **Step 4: Find the measure of the indicated angle (let's assume we want to find the non - given angles)**
       - If we want to find the angle $(20x + 5)$, substitute $x = \frac{137}{29}$: $20\times\frac{137}{29}+5=\frac{2740}{29}+5=\frac{2740 + 145}{29}=\frac{2885}{29}\approx99.48^{\circ}$. If we want the other non - given angle $(9x - 2)$, substitute $x=\frac{137}{29}$: $9\times\frac{137}{29}-2=\frac{1233}{29}-2=\frac{1233-58}{29}=\frac{1175}{29}\approx40.52^{\circ}$.
2. **For problem 15**:
   - **Explanation**:
     - **Step 1: Use the exterior - angle property of a triangle**
       - The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. Given an exterior angle of $140^{\circ}$ and two interior angles $(8x + 4)^{\circ}$ and $(3x+4)^{\circ}$. So, $(8x + 4)+(3x + 4)=140$.
     - **Step 2: Simplify the left - hand side of the equation**
       - Combine like terms: $8x+3x+4 + 4=140$, which gives $11x+8 = 140$.
     - **Step 3: Solve for $x$**
       - Subtract 8 from both sides: $11x=140 - 8 = 132$, then $x = 12$.
     - **Step 4: Find the measure of $\angle S=(3x + 4)$**
       - Substitute $x = 12$ into $(3x + 4)$: $3\times12+4=36 + 4=40^{\circ}$.
3. **For problem 16**:
   - **Explanation**:
     - **Step 1: Use the angle - sum property of a triangle**
       - In the triangle, we know one angle is $89^{\circ}$, and the other two angles are $(14x + 1)^{\circ}$ and $(5x-7)^{\circ}$. So, $(14x + 1)+(5x-7)+89=180$.
     - **Step 2: Simplify the left - hand side of the equation**
       - Combine like terms: $14x+5x+1-7 + 89=180$, which gives $19x+83=180$.
     - **Step 3: Solve for $x$**
       - Subtract 83 from both sides: $19x=180 - 83 = 97$, then $x=\frac{97}{19}$.
     - **Step 4: Find the measure of $\angle H=(5x-7)$**
       - Substitute $x=\frac{97}{19}$ into $(5x - 7)$: $5\times\frac{97}{19}-7=\frac{485}{19}-7=\frac{485-133}{19}=\frac{352}{19}\approx18.53^{\circ}$.
4. **For problem 17**:
   - **Explanation**:
     - **Step 1: Use the angle - sum property of a triangle**
       - In $\triangle ABC$, we know one angle is $55^{\circ}$, and the other two angles are $(13x-3)^{\circ}$ and $(3x + 2)^{\circ}$. So, $(13x-3)+(3x + 2)+55=180$.
     - **Step 2: Simplify the left - hand side of the equation**
       - Combine like terms: $13x+3x-3 + 2+55=180$, which gives $16x+54=180$.
     - **Step 3: Solve for $x$**
       - Subtract 54 from both sides: $16x=180 - 54 = 126$, then $x=\frac{126}{16}=\frac{63}{8}$.
     - **Step 4: Find the measure of $\angle FAB$ (assuming it is the exterior angle)**
       - The exterior angle $\angle FAB=(3x + 2)+55$. Substitute $x=\frac{63}{8}$: $3\times\frac{63}{8}+2 + 55=\frac{189}{8}+2+55=\frac{189+16+440}{8}=\frac{645}{8}=80.625^{\circ}$.
5. **For problem 18**:
   - **Explanation**:
     - **Step 1: Use the angle - sum property of a triangle**
       - In $\triangle BCD$, we know one angle is $80^{\circ}$, and the other two angles are $(6x + 6)^{\circ}$ and $(15x+5)^{\circ}$. So, $(6x + 6)+(15x+5)+80=180$.
     - **Step 2: Simplify the left - hand side of the equation**
       - Combine like terms: $6x+15x+6 + 5+80=180$, which gives $21x+91=180$.
     - **Step 3: Solve for $x$**
       - Subtract 91 from both sides: $21x=180 - 91 = 89$, then $x=\frac{89}{21}$.
     - **Step 4: Find the measure of $\angle YDC$ (assuming it is the exterior angle)**
       - The exterior angle $\angle YDC=(6x + 6)+80$. Substitute $x=\frac{89}{21}$: $6\times\frac{89}{21}+6 + 80=\frac{534}{21}+6+80=\frac{534+126+1680}{21}=\frac{2340}{21}\approx111.43^{\circ}$.

Since the problems involve finding angle measures in triangles and use properties of triangles such as angle - sum and exterior - angle properties, they fall under the sub - field of Geometry in Mathematics.

# Answer:
14. Angle measures depend on which angle is indicated. If we assume we want the angle $(20x + 5)$ with $x=\frac{137}{29}$, it is $\frac{2885}{29}\approx99.48^{\circ}$; if $(9x - 2)$, it is $\frac{1175}{29}\approx40.52^{\circ}$.
15. $m\angle S = 40^{\circ}$
16. $m\angle H=\frac{352}{19}\approx18.53^{\circ}$
17. $m\angle FAB=\frac{645}{8}=80.625^{\circ}$
18. $m\angle YDC=\frac{2340}{21}\approx111.43^{\circ}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Answer: