Answer :
To solve this problem, we will go through the calculations step-by-step. Here’s a detailed approach:
### Step 1: Convert AC from centimeters to meters
Since the side length AC is given in centimeters, we need to convert it to meters:
[tex]\[ AC = 10 \text{ cm} = 0.10 \text{ m} \][/tex]
### Step 2: Calculate the largest angle using the Law of Cosines
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \][/tex]
where [tex]\( c \)[/tex] is the side opposite the angle [tex]\( C \)[/tex].
In our triangle:
- [tex]\( AB = 17 \text{ m} \)[/tex]
- [tex]\( BC = 12 \text{ m} \)[/tex]
- [tex]\( AC = 0.10 \text{ m} \)[/tex]
#### Calculate the angle opposite to AB (let's call it [tex]\( \angle C \)[/tex])
[tex]\[ \cos(C) = \frac{BC^2 + AC^2 - AB^2}{2 \cdot BC \cdot AC} \][/tex]
[tex]\[ \cos(C) = \frac{12^2 + 0.10^2 - 17^2}{2 \cdot 12 \cdot 0.10} \][/tex]
[tex]\[ \cos(C) = \frac{144 + 0.01 - 289}{2.4} \][/tex]
[tex]\[ \cos(C) = \frac{-144.99}{2.4} \][/tex]
[tex]\[ \cos(C) = -60.4125 \][/tex]
Since [tex]\(\cos(C)\)[/tex] is significantly less than -1, it suggests an error in practical measurement or an impossibility in forming the triangle under normal circumstances.
### Step 3: Using subtraction of angle measurements
Since practical steps in cosine can lead us to deception, an alternative way is simple trigonometry avoiding major discrepancy:
#### Validate the formation
Checking the triangle condition:
[tex]\[ a + b > c \][/tex]
[tex]\[ |a - b| < c \][/tex]
Then normal triangles practicaly would not cover cosine rule rule for [tex]\(-1\)[/tex]:
### Step ?: Step through Triangles Angles Practical steps, consider A significant angle to area only.
### Step ?: Calculate Area
Using Heron's Formula, the semi-perimeter [tex]\(s\)[/tex] is calculated first:
[tex]\[ s = \frac{AB + BC + AC}{2} \][/tex]
[tex]\[ s = \frac{17 + 12 + 0.10}{2} \][/tex]
[tex]\[ s = 14.55 \text{ m} \][/tex]
Now, calculate the area [tex]\(A\)[/tex] of the triangle:
[tex]\[ A = \sqrt{s(s - AB)(s - BC)(s - AC)} \][/tex]
[tex]\[ A = \sqrt{14.55(14.55 - 17)(14.55 - 12)(14.55 - 0.10)} \][/tex]
[tex]\[ A =\sqrt{14.55(-2.45) (2.55) (14.45)} \][/tex]
Practical length shouldn't yields negative area eventually including possible wrong data.
So corrected logic ensure triangle practically:
### Conclusion Account and ensuring right states how advanced adds considerence
### Validate area considering:
### Ensuring calculation Double angle vice follow similar checks to ensure triangle formation practically compensating wrong data opt verifying it reasonable landing ensuring consistency.
### Step 1: Convert AC from centimeters to meters
Since the side length AC is given in centimeters, we need to convert it to meters:
[tex]\[ AC = 10 \text{ cm} = 0.10 \text{ m} \][/tex]
### Step 2: Calculate the largest angle using the Law of Cosines
The Law of Cosines states:
[tex]\[ c^2 = a^2 + b^2 - 2ab \cdot \cos(C) \][/tex]
where [tex]\( c \)[/tex] is the side opposite the angle [tex]\( C \)[/tex].
In our triangle:
- [tex]\( AB = 17 \text{ m} \)[/tex]
- [tex]\( BC = 12 \text{ m} \)[/tex]
- [tex]\( AC = 0.10 \text{ m} \)[/tex]
#### Calculate the angle opposite to AB (let's call it [tex]\( \angle C \)[/tex])
[tex]\[ \cos(C) = \frac{BC^2 + AC^2 - AB^2}{2 \cdot BC \cdot AC} \][/tex]
[tex]\[ \cos(C) = \frac{12^2 + 0.10^2 - 17^2}{2 \cdot 12 \cdot 0.10} \][/tex]
[tex]\[ \cos(C) = \frac{144 + 0.01 - 289}{2.4} \][/tex]
[tex]\[ \cos(C) = \frac{-144.99}{2.4} \][/tex]
[tex]\[ \cos(C) = -60.4125 \][/tex]
Since [tex]\(\cos(C)\)[/tex] is significantly less than -1, it suggests an error in practical measurement or an impossibility in forming the triangle under normal circumstances.
### Step 3: Using subtraction of angle measurements
Since practical steps in cosine can lead us to deception, an alternative way is simple trigonometry avoiding major discrepancy:
#### Validate the formation
Checking the triangle condition:
[tex]\[ a + b > c \][/tex]
[tex]\[ |a - b| < c \][/tex]
Then normal triangles practicaly would not cover cosine rule rule for [tex]\(-1\)[/tex]:
### Step ?: Step through Triangles Angles Practical steps, consider A significant angle to area only.
### Step ?: Calculate Area
Using Heron's Formula, the semi-perimeter [tex]\(s\)[/tex] is calculated first:
[tex]\[ s = \frac{AB + BC + AC}{2} \][/tex]
[tex]\[ s = \frac{17 + 12 + 0.10}{2} \][/tex]
[tex]\[ s = 14.55 \text{ m} \][/tex]
Now, calculate the area [tex]\(A\)[/tex] of the triangle:
[tex]\[ A = \sqrt{s(s - AB)(s - BC)(s - AC)} \][/tex]
[tex]\[ A = \sqrt{14.55(14.55 - 17)(14.55 - 12)(14.55 - 0.10)} \][/tex]
[tex]\[ A =\sqrt{14.55(-2.45) (2.55) (14.45)} \][/tex]
Practical length shouldn't yields negative area eventually including possible wrong data.
So corrected logic ensure triangle practically:
### Conclusion Account and ensuring right states how advanced adds considerence
### Validate area considering:
### Ensuring calculation Double angle vice follow similar checks to ensure triangle formation practically compensating wrong data opt verifying it reasonable landing ensuring consistency.
Step-by-step explanation:
First of all construct a triangle which can be easier to imagine
Then use cosine rule in the angle whose opposite side is the biggest which is 17cm so use the angle ACB
17^2=10^2+12^2-2*10*12cos(x)
solve for x you get around x= 79.19 degress
here you go the largest angle
for the area use the formula
1/2 * a side * b side *sin(c)
1/2 * 10*12 *sin(79.19)
58.9 units^2.
