Relation between sides and angles, Solutions of triangles Questions in English

(A) We know that $r_1 = \frac{\Delta}{s-a}$,$r_2 = \frac{\Delta}{s-b}$,and $r_3 = \frac{\Delta}{s-c}$.
Then,$r_1 r_2 + r_2 r_3 + r_3 r_1 = \frac{\Delta^2}{(s-a)(s-b)} + \frac{\Delta^2}{(s-b)(s-c)} + \frac{\Delta^2}{(s-c)(s-a)}$.
Taking $\frac{\Delta^2}{(s-a)(s-b)(s-c)}$ as a common factor,we get:
$\frac{\Delta^2}{(s-a)(s-b)(s-c)} [(s-c) + (s-a) + (s-b)]$.
Since $(s-a)(s-b)(s-c) = \frac{\Delta^2}{s}$,the expression becomes:
$\frac{\Delta^2}{\Delta^2/s} [3s - (a+b+c)]$.
Using $a+b+c = 2s$,we have:
$s [3s - 2s] = s^2$.
Since $r = \frac{\Delta}{s}$,we have $s = \frac{\Delta}{r}$,so $s^2 = \frac{\Delta^2}{r^2}$.

(B) Given that $\sin A, \sin B, \sin C$ are in $A.P.$
By the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,where $R$ is the circumradius.
Thus,$a, b, c$ are in $A.P.$
Let $p_1, p_2, p_3$ be the altitudes corresponding to sides $a, b, c$ respectively.
The area of the triangle $\Delta = \frac{1}{2} a p_1 = \frac{1}{2} b p_2 = \frac{1}{2} c p_3$.
This implies $p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}$.
Since $a, b, c$ are in $A.P.$,their reciprocals $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are in $H.P.$
Therefore,$\frac{2\Delta}{a}, \frac{2\Delta}{b}, \frac{2\Delta}{c}$ are in $H.P.$
Hence,the altitudes $p_1, p_2, p_3$ are in $H.P.$

(D) Using the Sine Rule in $\Delta PQR$,we have $\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R} = 2R_{c}$,where $R_{c}$ is the circumradius of the triangle.
Thus,$\sin P = \frac{p}{2R_{c}}$,$\sin Q = \frac{q}{2R_{c}}$,and $\sin R = \frac{r}{2R_{c}}$.
Given the equation $r^{2} \sin P \sin Q = pq$.
Substituting the values of $\sin P$ and $\sin Q$:
$r^{2} \left( \frac{p}{2R_{c}} \right) \left( \frac{q}{2R_{c}} \right) = pq$
$r^{2} \frac{pq}{4R_{c}^{2}} = pq$
Since $p, q \neq 0$,we can divide both sides by $pq$:
$\frac{r^{2}}{4R_{c}^{2}} = 1$
$r^{2} = 4R_{c}^{2}$
$r = 2R_{c}$
Since $r = 2R_{c} \sin R$,we have $2R_{c} \sin R = 2R_{c}$.
$\sin R = 1$
$R = 90^{\circ}$.
Therefore,the triangle is a right-angled triangle.

(B) In $\Delta PQR$,the sum of angles is $P+Q+R = 180^{\circ}$.
Since $P+R = 180^{\circ}-Q$,we substitute this into the expression:
$2pr \sin \left(\frac{P+R-Q}{2}\right) = 2pr \sin \left(\frac{180^{\circ}-Q-Q}{2}\right)$
$= 2pr \sin \left(\frac{180^{\circ}-2Q}{2}\right)$
$= 2pr \sin (90^{\circ}-Q)$
$= 2pr \cos Q$
Using the Law of Cosines,$\cos Q = \frac{p^{2}+r^{2}-q^{2}}{2pr}$.
Substituting this value:
$= 2pr \left(\frac{p^{2}+r^{2}-q^{2}}{2pr}\right)$
$= p^{2}+r^{2}-q^{2}$.

(B) Given that angles $A, B, C$ are in $A$.$P$.,we have $2B = A+C$. Since $A+B+C = 180^{\circ}$,we get $3B = 180^{\circ}$,so $B = 60^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k$,we have $a = k \sin A, b = k \sin B, c = k \sin C$.
Therefore,$\frac{a+c}{b} = \frac{\sin A + \sin C}{\sin B}$.
Using the sum-to-product formula,$\sin A + \sin C = 2 \sin \left(\frac{A+C}{2}\right) \cos \left(\frac{A-C}{2}\right)$.
Since $A+C = 2B$,we have $\frac{A+C}{2} = B$.
Thus,$\frac{a+c}{b} = \frac{2 \sin B \cos \left(\frac{A-C}{2}\right)}{\sin B} = 2 \cos \left(\frac{A-C}{2}\right)$.

(A) Given: $a = 2 \sqrt{2}$,$b = 6$,and $A = 45^{\circ}$.
Using the Law of Sines: $\frac{a}{\sin A} = \frac{b}{\sin B}$.
Substituting the values: $\frac{2 \sqrt{2}}{\sin 45^{\circ}} = \frac{6}{\sin B}$.
$\sin B = \frac{b \sin A}{a} = \frac{6 \times \sin 45^{\circ}}{2 \sqrt{2}}$.
$\sin B = \frac{6 \times \frac{1}{\sqrt{2}}}{2 \sqrt{2}} = \frac{6}{2 \times 2} = \frac{6}{4} = 1.5$.
Since the value of $\sin B$ cannot exceed $1$,$\sin B = 1.5$ is impossible.
Therefore,no triangle is possible.

(C) Given the relation: $\sin A \sin B = \frac{ab}{c^2}$
Using the Sine Rule,we know that $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$.
This implies $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.
Substituting these into the given equation:
$\left(\frac{a}{2R}\right) \left(\frac{b}{2R}\right) = \frac{ab}{c^2}$
$\frac{ab}{4R^2} = \frac{ab}{c^2}$
Since $a, b \neq 0$,we can cancel $ab$ from both sides:
$\frac{1}{4R^2} = \frac{1}{c^2}$ $\Rightarrow c^2 = 4R^2$ $\Rightarrow c = 2R$.
Since $c = 2R$,we have $\frac{c}{\sin C} = 2R \Rightarrow \sin C = \frac{c}{2R} = \frac{2R}{2R} = 1$.
Therefore,$C = 90^{\circ}$.
Thus,the triangle is a right-angled triangle.

(C) Using the Sine Rule in $\triangle ABC$:
$\frac{a}{\sin A} = \frac{b}{\sin B}$
Substituting the given values:
$\frac{2}{2/3} = \frac{3}{\sin B}$
$3 = \frac{3}{\sin B}$
$\sin B = 1$
Since $\sin B = 1$,we have $B = 90^{\circ}$ or $\frac{\pi}{2}$ radians.

(B) Given,$a^{2} \cos^{2} A - b^{2} - c^{2} = 0$
$\Rightarrow a^{2} \cos^{2} A = b^{2} + c^{2}$
Using the Law of Cosines,$\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}$.
Substituting $b^{2} + c^{2} = a^{2} \cos^{2} A$,we get:
$\cos A = \frac{a^{2} \cos^{2} A - a^{2}}{2bc} = \frac{-a^{2}(1 - \cos^{2} A)}{2bc} = \frac{-a^{2} \sin^{2} A}{2bc}$.
Since $a, b, c > 0$ and $\sin^{2} A > 0$ for $0 < A < \pi$,it follows that $\cos A < 0$.
Therefore,$A$ must lie in the second quadrant,i.e.,$\frac{\pi}{2} < A < \pi$.

(B) We know that in a $\triangle ABC$,$A+B+C = \pi$,so $A+C = \pi - B$.
Substituting this into the expression,we get $\frac{A+C-B}{2} = \frac{\pi-B-B}{2} = \frac{\pi}{2} - B$.
Thus,$2ac \sin \left(\frac{A-B+C}{2}\right) = 2ac \sin \left(\frac{\pi}{2}-B\right)$.
Using the identity $\sin \left(\frac{\pi}{2}-\theta\right) = \cos \theta$,we get $2ac \cos B$.
From the Law of Cosines,$\cos B = \frac{a^2+c^2-b^2}{2ac}$.
Substituting this,we get $2ac \left(\frac{a^2+c^2-b^2}{2ac}\right) = a^2+c^2-b^2$.

(D) Let the angles of the triangle be $2x, 3x,$ and $7x$.
Since the sum of angles in a triangle is $180^{\circ}$,we have $2x + 3x + 7x = 180^{\circ}$.
$12x = 180^{\circ} \Rightarrow x = 15^{\circ}$.
Thus,the angles are $30^{\circ}, 45^{\circ},$ and $105^{\circ}$.
The smallest side $a$ is opposite to the smallest angle $30^{\circ}$.
Using the Sine Rule,$\frac{a}{\sin A} = 2R$,where $R = 10 \text{ cm}$.
$\frac{a}{\sin 30^{\circ}} = 2 \times 10$.
$a = 20 \times \frac{1}{2} = 10 \text{ cm}$.