Question

try this #2: find angle a, side b, and side c.

Explanation:

🆕 New Concept Discovered: Law of Sines
Using ratios to solve non-right triangles[Client Connection Error]

Step 1: Find Angle A

The sum of angles in any triangle is always \( 180^\circ \). We can find the missing angle \( A \) by subtracting the two known angles from \( 180^\circ \).

$$ A = 180^\circ - 141^\circ - 23^\circ $$

$$ A = 16^\circ $$

Step 2: Find Side b

Because this is not a right-angled triangle, we cannot use basic SOH CAH TOA. Instead, we use the Law of Sines, which states that the ratio of the length of a side to the sine of its opposite angle is constant for all three sides:

$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} $$

From the given triangle:

• Side \( a \) (opposite to angle \( A \)) is unknown.
• Side \( b \) (opposite to angle \( B \)) is unknown.
• Side \( c \) (opposite to angle \( C \)) is the side \( AB \), which is unknown.
• We are given side \( BC = 9 \), which is opposite to angle \( A \). Therefore, \( a = 9 \).

Using the ratio for \( a \) and \( b \):

$$ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} $$

$$ \frac{9}{\sin(16^\circ)} = \frac{b}{\sin(141^\circ)} $$

Now, solve for \( b \):

$$ b = \frac{9 \cdot \sin(141^\circ)}{\sin(16^\circ)} $$

$$ b \approx \frac{9 \cdot 0.6293}{0.2756} $$

$$ b \approx 20.55 $$

Step 3: Find Side c

Using the Law of Sines ratio for \( a \) and \( c \):

$$ \frac{a}{\sin(A)} = \frac{c}{\sin(C)} $$

$$ \frac{9}{\sin(16^\circ)} = \frac{c}{\sin(23^\circ)} $$

Now, solve for \( c \):

$$ c = \frac{9 \cdot \sin(23^\circ)}{\sin(16^\circ)} $$

$$ c \approx \frac{9 \cdot 0.3907}{0.2756} $$

$$ c \approx 12.76 $$

Answer:

• Angle A = \( 16^\circ \)
• Side b \approx \( 20.55 \)
• Side c \approx \( 12.76 \)