Finding Angle A in a Triangle with Angles in AP

The problem gives us a triangle ABC where the angles are in Arithmetic Progression (AP) and the ratio of sides b and c is $\sqrt{3} : \sqrt{2}$. We need to find the measure of angle A.

Understanding Angles in AP

If the angles A, B, and C of a triangle are in AP, let the common difference be $d$. Then the angles can be represented as $B-d$, $B$, and $B+d$. The sum of angles in any triangle is 180 degrees. So, we have:

\begin{equation*} (B-d) + B + (B+d) = 180^\circ \end{equation*}

This simplifies to:

\begin{equation*} 3B = 180^\circ \end{equation*}

Dividing by 3, we find angle B:

\begin{equation*} B = \frac{180^\circ}{3} = 60^\circ \end{equation*}

So, one of the angles in the triangle is always 60 degrees if the angles are in AP. Let's represent the angles as $A$, $60^\circ$, and $C$. Since they are in AP, $A + C = 180^\circ - 60^\circ = 120^\circ$. We can write $A = 60^\circ - d$ and $C = 60^\circ + d$ (or vice versa) for some value of $d$.

Using the Sine Rule

The Sine Rule for a triangle states that the ratio of a side to the sine of its opposite angle is constant:

\begin{equation*} \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \end{equation*}

We are given the ratio $b : c = \sqrt{3} : \sqrt{2}$. From the Sine Rule, we can write:

\begin{equation*} \frac{b}{c} = \frac{\sin B}{\sin C} \end{equation*}

Substituting Given Values

We know $B = 60^\circ$ and $b/c = \sqrt{3}/\sqrt{2}$. We also know that $C = 180^\circ - A - B = 180^\circ - A - 60^\circ = 120^\circ - A$. Substituting these into the Sine Rule equation:

\begin{equation*} \frac{\sqrt{3}}{\sqrt{2}} = \frac{\sin 60^\circ}{\sin (120^\circ - A)} \end{equation*}

We know that $\sin 60^\circ = \frac{\sqrt{3}}{2}$. Substitute this value:

\begin{equation*} \frac{\sqrt{3}}{\sqrt{2}} = \frac{\frac{\sqrt{3}}{2}}{\sin (120^\circ - A)} \end{equation*}

Now, we can solve for $\sin (120^\circ - A)$. Cancel $\sqrt{3}$ from both sides (assuming $\sqrt{3} \ne 0$, which is true):

\begin{equation*} \frac{1}{\sqrt{2}} = \frac{\frac{1}{2}}{\sin (120^\circ - A)} \end{equation*}

Rearrange the equation to solve for $\sin (120^\circ - A)$:

\begin{equation*} \sin (120^\circ - A) = \frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}} = \frac{1}{2} \times \sqrt{2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \end{equation*}

Finding the Value of Angle A

The general solutions for $\sin \theta = \frac{1}{\sqrt{2}}$ are $\theta = 45^\circ + 360^\circ n$ or $\theta = 135^\circ + 360^\circ n$, where $n$ is an integer. For angles in a triangle, we consider values between $0^\circ$ and $180^\circ$. So, the possible values for $120^\circ - A$ are $45^\circ$ and $135^\circ$.

Case 1:

\begin{equation*} 120^\circ - A = 45^\circ \end{equation*}$$

Solving for A:

\begin{equation*} A = 120^\circ - 45^\circ = 75^\circ \end{equation*}$$

If $A = 75^\circ$, then $C = 120^\circ - A = 120^\circ - 75^\circ = 45^\circ$. The angles are $75^\circ$, $60^\circ$, $45^\circ$. Let's check if these are in AP: $60^\circ - 75^\circ = -15^\circ$ and $45^\circ - 60^\circ = -15^\circ$. Yes, they are in AP with a common difference of $-15^\circ$. This is a valid solution for angle A.

Case 2:

\begin{equation*} 120^\circ - A = 135^\circ \end{equation*}$$

Solving for A:

\begin{equation*} A = 120^\circ - 135^\circ = -15^\circ \end{equation*}$$

An angle in a triangle must be positive (strictly between $0^\circ$ and $180^\circ$). Therefore, this case is not possible for a triangle.

Thus, the only valid measure for angle A is $75^\circ$.

Summary of Angles

With $A = 75^\circ$ and $B = 60^\circ$, $C = 180^\circ - 75^\circ - 60^\circ = 45^\circ$. The angles are $75^\circ, 60^\circ, 45^\circ$, which are in AP.

The measure of angle A is $75^\circ$.