Proof the theorems on properties of triangle p/sin P = q/sin Q = r/sin R = 2K
Proof:
Let O be the circumcentre and K the circumradius of any triangle PQR.
Since in triangle PQR, three angles are acute in figure (i), then we observe that the triangle PQR is acuteangled in figure (ii), the triangle PQR is obtuseangled (since its angle P is obtuse) and in figure (iii), the triangle PQR is rightangled (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 = circumradius = K
Therefore, QS = 2 ∙ QO = 2K and ∠QRS = 90° (being the semicircular 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 rightangled triangle, we get from figure (iii),
2K = p = p/sin 90° = p/sin P [Since, P = 90°]
Therefore, for any triangle PQR (acuteangled, or obtuseangled or rightangled) 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,
p/sin P = q/sin Q = r/sin R = 2K
Note: (i) The relation p/sin P = q/sin Q = 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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