Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^\circ 20'$, find angle $A$. The given solution starts by converting $115^\circ 20'$ to $117^\circ$, which is incorrect. We should be using $B = 115 + \frac{20}{60} = 115 + \frac{1}{3} = 115.333\ldots$ degrees. The law of sines is being used, which is $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

1. Problem Description

Given a triangle with side $a = 7.82$ cm, side $b = 14.35$ cm, and angle $B = 115^\circ 20'$, find angle $A$. The given solution starts by converting $115^\circ 20'$ to $117^\circ$, which is incorrect. We should be using $B = 115 + \frac{20}{60} = 115 + \frac{1}{3} = 115.333\ldots$ degrees. The law of sines is being used, which is

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

2. Solution Steps

First, we convert $20'$ to degrees:

$20' = \frac{20}{60}^\circ = \frac{1}{3}^\circ \approx 0.333^\circ$

So, $B = 115^\circ 20' = 115.333^\circ$

Using the Law of Sines:

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

$\frac{7.82}{\sin A} = \frac{14.35}{\sin 115.333^\circ}$

$\sin A = \frac{7.82 \sin 115.333^\circ}{14.35}$

$\sin A = \frac{7.82 \times 0.904}{14.35} \approx \frac{7.069}{14.35} \approx 0.4926$

$A = \arcsin(0.4926) \approx 29.51^\circ$

The provided solution uses $117^\circ$ for $B$, and also incorrectly calculates $\sin A$.

The work shown calculates $\frac{7.82\sin 117}{sin 14.35}$, but this is not angle $A$, it is the side $c$ divided by $sin C$.

Using the given values,

$\sin A = \frac{7.82 \sin(117^\circ)}{14.35} = \frac{7.82(0.891)}{14.35} \approx \frac{6.967}{14.35} \approx 0.4855$

$A = \arcsin(0.4855) \approx 29.05^\circ$

3. Final Answer

The value for $A$ is approximately $29.51^\circ$, or $29.05^\circ$ if we use $117^\circ$ for $B$. The value given, 7.246, is not angle A. It appears to be the result of a calculation relating to the side c.

The calculation shown on the image gives $A=arcsin(\frac{7.82 sin(117^\circ)}{14.35})=arcsin(0.4855) \approx 29.05^\circ$