Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^{\circ} 120''$, find angle $A$. Note that $120'' = 2^{\circ}$, so angle $B = 117^{\circ}$.

1. Problem Description

Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^{\circ} 120''$, find angle $A$. Note that $120'' = 2^{\circ}$, so angle $B = 117^{\circ}$.

2. Solution Steps

We can use the Law of Sines to find the angle $A$. The Law of Sines states that:

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

We are given $a$, $b$, and $B$, so we can use the first two parts of the equation to find $A$:

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

$\frac{7.82}{\sin A} = \frac{14.35}{\sin 117^{\circ}}$

Now we can solve for $\sin A$:

$\sin A = \frac{7.82 \cdot \sin 117^{\circ}}{14.35}$

$\sin A = \frac{7.82 \cdot 0.8910}{14.35}$

$\sin A = \frac{6.96762}{14.35}$

$\sin A = 0.4856$

Now we can find the angle $A$ by taking the inverse sine of $0.4856$:

$A = \arcsin(0.4856)$

$A \approx 29.05^{\circ}$

The handwritten solution has a calculation error. It correctly sets up the equation:

$A = \frac{7.82 \sin 117^{\circ}}{14.35}$

However, the calculation of the numerator is incorrect. Using a calculator:

$7.82 \sin 117 = 6.96762 \dots$

Then, $6.96762/14.35 = 0.4856 \dots$

Taking the inverse sine of that gives $29.05$ degrees. The provided calculation gives $28.1^{\circ}$, likely due to rounding issues or an incorrect intermediate calculation.

The angle B given in the question ($115^\circ 120"$) is equal to $115^\circ + \frac{120}{3600}^\circ = 115^\circ + \frac{1}{30}^\circ = 115.033^\circ$ but it is being simplified to $117^\circ$ by adding $2^\circ$, due to the fact that $1^\circ = 60'$ and $1'=60"$. Thus, $1^\circ = 3600"$, so $120" = 2/60^\circ$, or simply $120" = 2'$. Thus, the correct interpretation is $115^\circ + 2' = 115.033^\circ$, and not $117^\circ$. The provided angle $117^\circ$ seems to be calculated by summing $115^\circ + 2^\circ = 117^\circ$, by interpreting $120"$ as $2^\circ$.

If we follow the given values,

$\frac{7.82}{\sin A} = \frac{14.35}{\sin 115.033}$

$\sin A = \frac{7.82\sin 115.033}{14.35} = \frac{7.82 \cdot 0.90629}{14.35} = \frac{7.0866}{14.35}=0.49397$

$A = \arcsin(0.49397) = 29.614^\circ$

3. Final Answer

$A \approx 29.05^{\circ}$ if we are using $B = 117^{\circ}$.

$A \approx 29.61^{\circ}$ if we are using $B = 115.033^{\circ}$.