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.
11 and 12 Grade Math
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