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.

 Properties of Triangles




11 and 12 Grade Math

From Theorem on Properties of Triangle to HOME PAGE




Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.



New! Comments

Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.

Share this page: What’s this?

Recent Articles

  1. Successor and Predecessor | Successor of a Whole Number | Predecessor

    May 24, 24 06:42 PM

    Successor and Predecessor of a Whole Number
    The number that comes just before a number is called the predecessor. So, the predecessor of a given number is 1 less than the given number. Successor of a given number is 1 more than the given number…

    Read More

  2. Counting Natural Numbers | Definition of Natural Numbers | Counting

    May 24, 24 06:23 PM

    Natural numbers are all the numbers from 1 onwards, i.e., 1, 2, 3, 4, 5, …... and are used for counting. We know since our childhood we are using numbers 1, 2, 3, 4, 5, 6, ………..

    Read More

  3. Whole Numbers | Definition of Whole Numbers | Smallest Whole Number

    May 24, 24 06:22 PM

    The whole numbers are the counting numbers including 0. We have seen that the numbers 1, 2, 3, 4, 5, 6……. etc. are natural numbers. These natural numbers along with the number zero

    Read More

  4. Math Questions Answers | Solved Math Questions and Answers | Free Math

    May 24, 24 05:37 PM

    Math Questions Answers
    In math questions answers each questions are solved with explanation. The questions are based from different topics. Care has been taken to solve the questions in such a way that students

    Read More

  5. Estimating Sum and Difference | Reasonable Estimate | Procedure | Math

    May 24, 24 05:09 PM

    Estimating Sum or Difference
    The procedure of estimating sum and difference are in the following examples. Example 1: Estimate the sum 5290 + 17986 by estimating the numbers to their nearest (i) hundreds (ii) thousands.

    Read More