# Theorem on Properties of Triangle

Proof the theorems on properties of triangle $$\frac{p}{sin P}$$ = $$\frac{q}{sin Q}$$ = $$\frac{r}{sin R}$$ = 2K

Proof:

Let O be the circum-centre and K the circum-radius of any triangle PQR.

Since in triangle PQR, three angles are acute in figure (i), then we observe that the triangle PQR is acute-angled in figure (ii), the triangle PQR is obtuse-angled (since its angle P is obtuse) and in figure (iii), the triangle PQR is right-angled (since the angle P is right angle). In figure (i) and figure (ii) we join QO and produce it to meet the circumference at S. Then join RS.

Clearly, QO = circum-radius = K

Therefore, QS   = 2 ∙ QO = 2K and ∠QRS = 90° (being the semi-circular angle).

Now, from figure (i)we get,

∠QSR = ∠QPR = P (being the angles on the same arc QR).

Therefore, from the triangle QRS we have,

QR/QS = sin ∠QSR

⇒ p/2K = sin P

⇒  p/sin P = 2K

Again, from figure (ii) we get,

∠QSR = π - P [Since, ∠QSR + ∠QPR = π]

Therefore, from the triangle QRS we get,

QR/QS = sin ∠QSR

⇒ p/2K = sin (π - P)

⇒ p/2K = sin P

⇒ a/sin P = 2K

Finally, for right-angled triangle, we get from figure (iii),

2K = p = p/sin 90° = p/sin P    [Since, P = 90°]

Therefore, for any triangle PQR (acute-angled, or obtuse-angled or right-angled) we have,

Similarly, if we join PO and produce it to meet the circumference at T then joining RT and QE we can prove

q/sin Q = 2K and  r/sin R = 2K …………………………….. (1)

Therefore, in any triangle PQR we have,

$$\frac{p}{sin P}$$ = $$\frac{q}{sin Q}$$ = $$\frac{r}{sin R}$$ = 2K

Note: (i) The relation $$\frac{p}{sin P}$$ = $$\frac{q}{sin Q}$$ = $$\frac{r}{sin R}$$ is known as Sine Rule.

(ii) Since, p : q : r = sin P : sin Q : sin R

Therefore, in any triangle the lengths of sides are proportional to the sines of opposite angles.

(iii) From (1) we get, p = 2K sin P, q = 2K sin Q and r = 2K sin R. These relations give the sides in terms of sines of angles.

Again, from (1) we get, sin P = p/2K, sin Q = q/2K and sin R = r/2K

These relations give the sines of the angles in terms of the sides of any triangle.

Solved problems using theorem on properties of triangle:

1. In the triangle PQR, if P = 60°, show that,

q + r = 2p cos $$\frac{Q - R}{2}$$

Solution:

We have,

We know that

$$\frac{p}{sin P}$$ = $$\frac{q}{sin Q}$$ = $$\frac{r}{sin R}$$ = 2K.

⇒ p = 2K sin P, q = 2K sin Q and r = 2K sin R.

$$\frac{q + r}{2p}$$ = $$\frac{2K sin Q + 2K sin R}{2 ∙ 2K sin P}$$, [Since, p = 2K sin P, q = 2K sin Q and r = 2K sin R]

= $$\frac{sin Q + sin R}{2 sin P}$$

= $$\frac{2 sin \frac{Q + R}{2} cos \frac{Q - R}{2}}{2 sin 60°}$$

= $$\frac{sin 60° cos \frac{Q - R}{2}}{sin 60°}$$,

[Since, P + Q + R = 180°, and P = 60° Therefore, Q + R = 180° - 60° = 120° ⇒ $$\frac{Q + R}{2}$$ = 60°]

⇒ $$\frac{q + r}{2p}$$ = cos $$\frac{Q - R}{2}$$

Therefore, q + r = 2p cos $$\frac{Q - R}{2}$$        proved.

2. In any triangle PQR, prove that,

(q$$^{2}$$ - r$$^{2}$$) cot P + (r$$^{2}$$ - p$$^{2}$$) cot Q + (p$$^{2}$$ - q$$^{2}$$) cot R = 0.

Solution:

$$\frac{p}{sin P}$$ = $$\frac{q}{sin Q}$$ = $$\frac{r}{sin R}$$ = 2K.

⇒ p = 2K sin P, q = 2K sin Q and r = 2K sin R.

Now, (q$$^{2}$$ - r$$^{2}$$) cot P = (4K$$^{2}$$ sin$$^{2}$$ Q - 4K$$^{2}$$ sin$$^{2}$$ R) cot P

= 2K$$^{2}$$ (2 sin$$^{2}$$ Q - 2 sin$$^{2}$$ R)

= 2K$$^{2}$$ (1 - cos 2Q - 1 + cos 2R) cot P

= 2K$$^{2}$$ [2 sin (Q + R) sin (Q - R)] cot P

=4K$$^{2}$$ sin (π - P) sin (Q - R) cot A, [Since, P + Q + R = π]

= 4K$$^{2}$$ sin P sin (Q - R) $$\frac{cos P}{sin P}$$

= 4K$$^{2}$$ sin (Q - R) cos {π - (Q - R)}

= - 2K$$^{2}$$ ∙ 2sin (Q - R) cos (Q + R)

= - 2K$$^{2}$$ (sin 2Q - sin 2R)

Similarly, (r$$^{2}$$ - p$$^{2}$$) cot Q = -2K$$^{2}$$ (sin 2R - sin 2P)

and (p$$^{2}$$ - q$$^{2}$$) cot R = -2K$$^{2}$$ (sin 2R - sin 2Q)

Now L.H.S. = (q$$^{2}$$ - r$$^{2}$$) cot P + (r$$^{2}$$ - p$$^{2}$$) cot Q + (p$$^{2}$$ - q$$^{2}$$) cot R

= - 2K$$^{2}$$ (sin 2Q - sin 2R) - 2K$$^{2}$$ (sin 2R - sin 2P) - 2K$$^{2}$$(sin 2P - sin 2Q)

= - 2K$$^{2}$$ × 0

= 0 = R.H.S.                        Proved.

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