Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^\circ + 120''$. We want to find the angle $A$ in degrees. Note that 1 degree = 60 minutes and 1 minute = 60 seconds, so $120'' = 2'$. Therefore, $B = 115^\circ 2'$ which is approximately $117^\circ$.

1. Problem Description

Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^\circ + 120''$. We want to find the angle $A$ in degrees. Note that 1 degree = 60 minutes and 1 minute = 60 seconds, so $120'' = 2'$. Therefore, $B = 115^\circ 2'$ which is approximately $117^\circ$.

2. Solution Steps

First, convert the angle $B$ from degrees and minutes to degrees.

Since $1' = (\frac{1}{60})^\circ$, we have $2' = (\frac{2}{60})^\circ = (\frac{1}{30})^\circ \approx 0.0333^\circ$

Thus, $B = 115^\circ + (\frac{2}{60})^\circ = 115.0333^\circ$

We use the law of sines to find angle $A$:

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

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

$\sin A = \frac{7.82 \sin(115.0333^\circ)}{14.35}$

$A = \arcsin\left(\frac{7.82 \sin(115.0333^\circ)}{14.35}\right)$

Now, we calculate the value of $A$.

$\sin(115.0333^\circ) \approx 0.9062$

$\sin A = \frac{7.82 \times 0.9062}{14.35} \approx \frac{7.0865}{14.35} \approx 0.4938$

$A = \arcsin(0.4938) \approx 29.58^\circ$

The solution in the image calculates:

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

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

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

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

$\sin 117^\circ \approx 0.8910$

$\sin A = \frac{7.82 \times 0.8910}{14.35} = \frac{6.9676}{14.35} \approx 0.4856$

$A = \arcsin(0.4856) = 29.05^\circ$

In the image, the value of $\frac{7.82 \sin 117^\circ}{\sin 14.35}$ is calculated as 7.

2

4

6. $\sin 14.35$ is not being used correctly here.

$A = \arcsin\left(\frac{7.82\sin(117^\circ)}{14.35}\right) \approx \arcsin\left(\frac{7.82 \cdot 0.8910}{14.35}\right) = \arcsin(0.4856) \approx 29.05^\circ$

3. Final Answer

The angle A is approximately $29.05^\circ$.