Question

The angles of a triangle ABC are in an arithmetic progression. The larger sides a, b satisfies the relation $\dfrac{{\sqrt 3 }}{2} < \dfrac{b}{a} < 1$, then the possible values of the smallest side are:
A) $\dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{{2a}}$
B) $\dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{{2b}}$
C) $\dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{{2c}}$
D) $\dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{2}$

Answer 1

By using arithmetic progression and the sum of angles of triangle law, find the value of angle B. Then, use the sine rule of the triangle to find the range of the largest angle A. Then use the cosine rule of the triangle to find the value of the shortest side by $\cos B = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}$. It will give a quadratic equation. Now, apply the quadratic formula which will give the possible value of the shortest side.

Complete step-by-step answer:
Given: - The angles of a triangle ABC are in an arithmetic progression.
The larger sides a, b satisfies the relation $\dfrac{{\sqrt 3 }}{2} < \dfrac{b}{a} < 1$.
Let the angles opposite to side a be A, opposite to side b be B, and opposite to side c be C.
As we know that, the sum of the interior angles of a triangle is equal to $180^\circ$. So,
$A + B + C = 180^\circ$ ….. (1)
Since the angles of the triangle, ABC is in arithmetic progression. Then,
$B = \dfrac{{A + C}}{2}$
Multiply both sides by 2,
$2B = A + C$
Substitute the value in equation (1),
$B + 2B = 180^\circ$
Add the terms on the right side,
$3B = 180^\circ$
Divide both sides by 3,
$B = 60^\circ$
The sine rule of the triangle is,
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}$
Take the first 2 terms,
$\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b}$
Since $B = 60^\circ$,
$\dfrac{{\sin A}}{a} = \dfrac{{\sin 60^\circ }}{b}$
Substitute the value of $\sin 60^\circ$,
$\dfrac{{\sin A}}{a} = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{b}$
Interchange the values $b$ and $\sin A$,
$\dfrac{b}{a} = \dfrac{{\sqrt 3 }}{{2\sin A}}$
As $\dfrac{{\sqrt 3 }}{2} < \dfrac{b}{a} < 1$. Then,
$\dfrac{{\sqrt 3 }}{2} < \dfrac{{\sqrt 3 }}{{2\sin A}} < 1$
Divide the terms by $\dfrac{{\sqrt 3 }}{2}$,
$1 < \dfrac{1}{{\sin A}} < \dfrac{2}{{\sqrt 3 }}$
Take the reciprocal,
$1 > \sin A > \dfrac{{\sqrt 3 }}{2}$
Take ${\sin ^{ - 1}}$ on all terms,
${\sin ^{ - 1}}1 > A > {\sin ^{ - 1}}\dfrac{{\sqrt 3 }}{2}$
Substitute the values of ${\sin ^{ - 1}}1$ and ${\sin ^{ - 1}}\dfrac{{\sqrt 3 }}{2}$,
$90^\circ > A > 60^\circ$.
Now, use the cosine formula to find the smallest side,
$\cos B = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}$
Substitute the values of $B$,
$\cos 60^\circ = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}$
Substitute the value of $\cos 60^\circ$,
$\dfrac{1}{2} = \dfrac{{{c^2} + {a^2} - {b^2}}}{{2ac}}$
Cancel out 2 from the denominator of both sides and cross multiply the terms,
${c^2} + {a^2} - {b^2} = ac$
Move $ac$ to the left side,
${c^2} - ac + \left( {{a^2} - {b^2}} \right) = 0$
Find the roots by quadratic formula, i.e. $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$.
$c = \dfrac{{ - \left( { - a} \right) \pm \sqrt {{{\left( { - a} \right)}^2} - 4\left( 1 \right)\left( {{a^2} - {b^2}} \right)} }}{{2 \times 1}}$
Open the brackets and multiply the terms,
$c = \dfrac{{a \pm \sqrt {{a^2} - 4{a^2} + 4{b^2}} }}{2}$
Subtract the like terms,
$c = \dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{2}$
So, the possible value of the smallest side is $\dfrac{{a \pm \sqrt {4{b^2} - 3{a^2}} }}{2}$.

Hence, the option (D) is correct.

Note: This type of problem can be solved by performing the cosine rule, $\cos B = \dfrac{{{a^2} + {c^2} - {b^2}}}{{2ac}}$. Students should be taken care of when adding the like terms, changing the signs. Students must recall the algebraic identity and trigonometry formula when it is required. If anyone of the process or formulae will miss, it will lead to a heavy calculation or wrong answer.
A quadratic equation is a polynomial equation of degree 2. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution, or two imaginary solutions.
There are several methods you can use to solve a quadratic equation:
Factoring
Completing the Square
Quadratic Formula
Graphing
All methods start with setting the equation equal to zero.