Students get through Maths 1A Important Questions Inter 1st Year Maths 1A Properties of Triangles Important Questions which are most likely to be asked in the exam.

## Intermediate 1st Year Maths 1A Properties of Triangles Important Questions

Question 1.
In ∆ABC, if a = 3, b = 4 and sinA = $$\frac{3}{4}$$, find angle B.
Solution:
By sine Rule $$\frac{a}{\sin A}$$ = $$\frac{b}{\sin B}$$
⇒ sin B = $$\frac{\text { b. } \sin A}{a}$$ = $$\frac{4}{3}$$ ($$\frac{3}{4}$$) = 1
⇒ sin B = 1 ⇒ B = 90°

Question 2.
If the lengths of the sides of a triangle are 3, 4, 5 find the circumradius of the triangle.
Solution:
∴ 32 + 42 = 52
∴ The triangle is right angled and its hypotenuse = 5 = circum diameter.
∴ Circum radius = $$\frac{1}{2}$$ (hypotenu) = $$\frac{5}{2}$$ cms.

Question 3.
If a = 6, b = 5, c = 9, then find angle A.
Solution:
∵ cos A = $$\frac{\mathrm{b}^{2}+c^{2}-a^{2}}{2 b c}$$
= $$\frac{5^{2}+9^{2}-6^{2}}{2(5)(9)}$$ = $$\frac{25+81-36}{2(5)(9)}$$
= $$\frac{70}{90}$$ = $$\frac{7}{9}$$
∴ A = cos-1($$\frac{7}{9}$$)

Question 4.
If ∆ABC, show that ∆ (b + c) cos A = 2s.
Solution:
L.H.S. = (b + c) cos A + (c + a) cos B + (a + b) cos C
= (b cos A + a cos B) + (c cos B + b cos C) + (a cos C + c cos A)
= c + a + b = 2S = R.H.S.

Question 5.
If the sides of a triangle are 13, 14, 15, then find the circum diameter.
Solution:
Let a = 13, b = 14, c = 15
Then 2s = a + b + c = $$\frac{13+14+15}{2}$$ = $$\frac{42}{2}$$
∴ s = 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6
Now ∆ = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{21 \times 8 \times 7 \times 6}$$
= $$\sqrt{21 \times 21 \times 16}$$ = 21 × 4 = 84
∵ ∆ = $$\frac{a b c}{4 R}$$ ⇒ 4R = $$\frac{a b c}{\Delta}$$
= $$\frac{13 \times 14 \times 15}{84}$$ = $$\frac{65}{2}$$
∴ R = $$\frac{65}{8}$$
∴ Circum diameter(2R) = 2 × $$\frac{65}{8}$$ = $$\frac{65}{4}$$cms.

Question 6.
In a ∆A B C, if (a + b + c) (b + c – a) = 3 be, find A.
Solution:
(2s – sa) = 3bc ⇒ $$\frac{s(s-a)}{b c}$$ = $$\frac{3}{4}$$
⇒ cos2 $$\frac{A}{2}$$ = $$\frac{3}{4}$$ ⇒ cos $$\frac{A}{2}$$ = $$\frac{1}{2}$$ = cos 30°
∴ $$\frac{A}{2}$$ = 30° = A = 60°

Question 7.
If a = 4, b = 5, c = 7, find cos $$\frac{B}{2}$$.
Solution:
2s = a + b + c = 4 + 5 + 7 = 16
⇒ s = 8 and s – b = 8 – 5 = 3
Now cos $$\frac{B}{2}$$ = $$\sqrt{\frac{s(s-b)}{a c}}$$ = $$\sqrt{\frac{8 \times 3}{4 \times 7}}$$ = $$\sqrt{\frac{6}{7}}$$

Question 8.
In ∆ ABC, find b cos2 $$\frac{C}{2}$$ + c cos2 $$\frac{B}{2}$$.
Solution:
b cos2 $$\frac{C}{2}$$ + c cos 2 $$\frac{B}{2}$$ = b$\frac{s(s-c)}{a b}$ + c$\frac{s(s-b)}{c a}$
= $$\frac{s(s-c)}{a}$$ + $$\frac{s(s-b)}{a}$$ = $$\frac{\mathrm{s}}{\mathrm{a}}$$[s – c + s – b]
= $$\frac{\mathrm{s}}{\mathrm{a}}$$ • a = s

Question 9.
If tan $$\frac{A}{2}$$ = $$\frac{5}{6}$$ and tan $$\frac{C}{2}$$ = $$\frac{2}{5}$$, determine the relation between a, b, c.    [Mar 05]
Solution:
tan $$\frac{A}{2}$$ • tan $$\frac{C}{2}$$ = $$\frac{5}{6}$$ • $$\frac{2}{5}$$
$$\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \sqrt{\frac{(s-a)(s-b)}{s(s-c)}}$$ = $$\frac{2}{6}$$
⇒ $$\frac{s-b}{s}$$ = $$\frac{1}{3}$$ ⇒ 3s – 3b = s ⇒ 2s = 3b
⇒ a + b + c = 3b ⇒ a + c = 2b Hence a, b, c are in A.P.

Question 10.
If cot $$\frac{A}{2}$$ = $$\frac{b+c}{a}$$, find angle B.
Solution:
cot $$\frac{A}{2}$$ = $$\frac{b+c}{a}$$ ⇒ $$\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}$$ = $$\frac{\cos \left(\frac{B-C}{2}\right)}{\sin \frac{A}{2}}$$ (by Mollweide rule)
⇒ $$\frac{A}{2}$$ = $$\frac{B-C}{2}$$
⇒ A = B – C ⇒ A + C = B ⇒ A + B + C = 2B
∴ 2B = 180° ⇒ B = 90°

Question 11.
If tan ($$\frac{C-A}{2}$$) = k cot $$\frac{B}{2}$$, find k.
Solution:
Comparing with tan ($$\frac{C-A}{2}$$) = $$\frac{c-a}{c+a}$$ cot $$\frac{B}{2}$$, (by tangent law)
we get that k = $$\frac{c-a}{c+a}$$

Question 12.
In ∆ABC, show that $$\frac{b^{2}-c^{2}}{a^{2}}$$ = $$\frac{\sin (B-C)}{\sin (B+C)}$$.
Solution:

Question 13.
Show that a2 cot A + b2 cot B + c2 cot C = $$\frac{a b c}{R}$$.    [Mar 14]
Solution:
L.H.S. = a2 cot A + b2 cot B + c2 cot C
= 4R2 sin2 A. $$\frac{\cos A}{\sin A}$$ + 4R2 sin2B. $$\frac{\cos B}{\sin B}$$ + 4R2 sin2C. $$\frac{\cos C}{\sin C}$$ (by sine rule)
= 2R2 (2 sin A cos A + 2 sin B cos B + 2 sin C cos C)
= 2R2 (sin 2A + sin 2B + sin 2C) .
= 2R2 (4 sin A sin B sin C) (from transformations)
= $$\frac{1}{R}$$ (2R sin A) (2R sin B) (2R sin C)
= $$\frac{abc}{R}$$ = R.H.S

Question 14.
Show that (b – c)2 cos2 $$\frac{A}{2}$$ + (b + c)2 sin2 $$\frac{A}{2}$$ = a2.
Solution:
L.H.S. = (b2 + c2 – 2bc) cos2 $$\frac{A}{2}$$ + (b2 + c2 + 2bc) sin2 $$\frac{A}{2}$$
= (b2 + c2) [cos2 $$\frac{A}{2}$$ + sin2 $$\frac{A}{2}$$] – 2bc (cos2 $$\frac{A}{2}$$ – sin2 $$\frac{A}{2}$$)
= b2 + c2 – 2bc cos A = a2

Question 15.
Prove that a (b cos C – c cos B) = b2 – c2     [Mar 07]
Solution:
L.H.S. = ab cos C – ca cos B
= ($$\frac{a^{2}+b^{2}-c^{2}}{2}$$) – ($$\frac{c^{2}+a^{2}-b^{2}}{2}$$) (by cosine rule)
= $$\frac{1}{2}$$[a2 + b2 – c2 – c2 – a2 + b2]
= b2 – c2 = R.H.S.

Question 16.
Show that $$\frac{c-b \cos A}{b-c \cos A}$$ = $$\frac{\cos B}{\cos C}$$.
Solution:
From projection rule c = a cos B – b cos A and b = c cos A + a cos C

Question 17.
In ∆ABC, if $$\frac{1}{a+c}$$ + $$\frac{1}{b+c}$$ = $$\frac{3}{a+b+c}$$ show that C = 60°.
Solution:
$$\frac{1}{a+c}$$ + $$\frac{1}{b+c}$$ = $$\frac{3}{a+b+c}$$
⇒ $$\frac{b+c+a+c}{(a+c)(b+c)}$$ = $$\frac{3}{a+b+c}$$
⇒ 3(a + c) (b + c) = (a + b + 2c) (a + b + c)
⇒ 3(ab + ac + bc + c2) = (a2 + b2 + 2ab) + 3c(a + b) + 2c2
⇒ ab = a2 + b2 – c2
⇒ 2ab cos C (from cosine rule)
⇒ cos C = $$\frac{1}{2}$$ ⇒ C = 60°

Question 18.
If a = (b – c) sec θ, prove that tan θ = $$\frac{2 \sqrt{b c}}{b-c}$$ sin $$\frac{A}{2}$$.    [Mar 16]
Solution:
a = (b – c) sec θ ⇒ sec θ = $$\frac{a}{b-c}$$
tan2 θ = sec2 θ – 1

Question 19.
In ∆ABC, show that (a + b + c) (tan $$\frac{A}{2}$$ + tan $$\frac{B}{2}$$) = 2c cot $$\frac{C}{2}$$.
Solution:

Question 20.
Show that b2 sin 2C + c2 sin 2B = 2bc sin A.
Solution:
L.H.S. = b2 sin 2C + c2 sin 2B
= 4R2 sin2 B (2 sin C cos C) + 4R2 sin2 C (2 sin B cos B)
= 8R2 sin B sin C (sin B cos C + cos B sin C)
= 8R2 sin B sin C sin (B + C)
= 2(2R sin B) (2R sin C) sin A
= 2bc sin A = R.H.S.

Question 21.
Prove that cot A + cot B + cot C = $$\frac{a^{2}+b^{2}+c^{2}}{4 \Delta}$$    [Mar 15]
Solution:

Question 22.
Show that a cos2 $$\frac{A}{2}$$ + b cos2 $$\frac{B}{2}$$ + c cos 2 $$\frac{C}{2}$$ = s + $$\frac{\Delta}{R}$$.
Solution:
L.H.S. = Σ a cos2 $$\frac{A}{2}$$ = $$\frac{1}{2}$$ Σ a (1 + cos A)
= $$\frac{1}{2}$$ Σ (a + a cos A) = $$\frac{1}{2}$$ (a + b + c) + $$\frac{1}{2}$$ Σ (2R sin A cos A)
= $$\frac{1}{2}$$ (2s) + $$\frac{R}{2}$$ Σ sin 2A
= s + $$\frac{R}{2}$$ (sin 2A + sin 2B + sin 2C)
= s + $$\frac{R}{2}$$ (4 sin A sin B sin C)
= s + $$\frac{1}{R}$$ (2R2 sin A sin B sin C)
= s + $$\frac{\Delta}{R}$$ (∵ ∆ = 2R2 sin A sin B sin C)
= R.H.S.

Question 23.
In ∆ ABC, if a cos A = b cos B, prove that the triangle is either isosceles or right angled.
Solution:
a cos A = b cos B
⇒ 2R sin A cos A = 2R sin B cos B
⇒ sin 2A = sin 2B (or) = sin (180° – 2B)
Hence 2A = 2B or 2A = 180° – 2B
⇒ A = B or A = 90° – B ⇒ A = B or A + B = 90°
⇒ C = 90°
∴ The triangle is isosceles or right angled.

Question 24.
If cot $$\frac{A}{2}$$ : cot $$\frac{B}{2}$$ : cot $$\frac{C}{2}$$ = 3 : 5 : 7, show that a : b : c = 6 : 5 : 4.
Solution:

Then s – a = 3k, s – b = 5k, s – c = 7k
Adding 3s – (a + b + c) = 3k + 5k+ 7k
⇒ 3s – 2s = 15k ⇒ s = 15k
Now a = 12k, b = 10k, c = 8k
∴ a : b : c = 12k : 10k : 8k = 6 : 5 : 4

Question 25.
Prove that a3 cos (B – C) + b3 cos (C – A) + c3 cos(A – B) = 3abc.
Solution:
L.H.S. = Σ a3 cos (B – C)
= Σ a2 (2R sin A) cos (B – C)
= R Σ a2 [2 sin (B + C) cos (B- C)]
= R Σ a2 (sin 2B + sin 2C)
= R Σ a2 (2 sin B cos B + 2 sin C cos C)
= Σ [a2(2R sin B) cos B + a2(2R sin C) cos C] Σ (a2 b cos B + a2c cos C)
= (a2b cos B + a2c cos C) + (b2c cos C + b2 a cos A) + (c2 a cos A + c2b cos B)
= ab (a cos B + b cos A) + bc (b cos C + c cos B) + ca (c cos A + a cos C)
= ab(c) + bc(a) + ca(b) = 3 abc = R.H.S

Question 26.
If p1, p2, p3 are the altitudes of the ∆ A, B, C show that $$\frac{1}{p_{1}^{2}}+\frac{1}{p_{2}^{2}}+\frac{1}{p_{3}^{2}}$$ = $$\frac{\cot A+\cot B+\cot C}{\Delta}$$
Solution:
Since p1, p2, p3 are the altitudes of ∆ ABC,

Question 27.
The angle of elevation of the top point P of the vertical tower PQ of height h from a point A is 45° and from a point B is 60°, where B is a point at a distance 30 meters from the point A measured along the line AB which makes an angle 30° with AQ. Find the height of the tower.
Solution:

PQ = h, ∠PAQ = 45°
∠BAQ = 30° and ∠PBC = 60°
Also AB = 30 mts.
∴ ∠BAP = ∠APB = 15°.
This gives BP = AB = 30 and h = PC + CQ = BP sin 60° + AB sin 30°
= 15 $$\sqrt{3}$$ + 15 = 15($$\sqrt{3}$$ +1) metres.

Question 28.
Two trees A and B are on the same side of a river. From a point C in the river the distances of the trees A and B are 250m and 300m respectively. If the angle C is 45°, find the distance between the trees (use $$\sqrt{2}$$ = 1.414).
Solution:

From the triangle ABC, using the cosine rule
AB2 = 2502 + 3002 – 2(250) (300) cos 45°
= 100 (625 + 900 – 750$$\sqrt{2}$$) = 46450.
∴ AB = 215.5m. (approximately).

Question 29.
In A ABC, prove that $$\frac{1}{r_{1}}$$ + $$\frac{1}{r_{2}}$$ + $$\frac{1}{r_{3}}$$ = $$\frac{1}{r}$$.
Solution:
L.H.S. = $$\frac{1}{r_{1}}$$ + $$\frac{1}{r_{2}}$$ + $$\frac{1}{r_{3}}$$ = $$\frac{s-a}{\Delta}$$ + $$\frac{s-b}{\Delta}$$ + $$\frac{s-c}{\Delta}$$
= $$\frac{3 s-(a+b+c)}{\Delta}$$ = $$\frac{3 s-2 s}{\Delta}$$ = $$\frac{s}{\Delta}$$ = $$\frac{1}{r}$$
= R.H.S

Question 30.
Show that rr1r 2r3 = ∆2
Solution:
L.H.S. = rr1r 2r3 = $$\frac{\Delta}{s} \cdot \frac{\Delta}{s-a} \cdot \frac{\Delta}{s-b} \cdot \frac{\Delta}{s-c}$$
= $$\frac{\Delta^{4}}{\Delta^{2}}$$ = ∆2 = R.H.S

Question 31.
In an equilateral triangle, find the value of $$\frac{r}{R}$$.
Solution:
$$\frac{r}{R}$$ = $$\frac{4 R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}}{R}$$
= 4 sin3 30° (∵ A = B = C = 60°)
= 4 . ($$\frac{1}{2}$$)3 = $$\frac{1}{2}$$

Question 32.
The perimeter of ∆ ABC is 12 cm. and its in radius is 1 cm. Then find the area of the triangle.
Solution:
Given that 2s = 12 ⇒ s = 6 cm.
and r = 1 cm
Area of ∆ ABC = ∆ = rs = (1) (6) = 6 sq.cms.

Question 33.
Show that rr1 = (s – b) (s – c).
Solution:
L.H.S. = rr1
= [(s – b) tan $$\frac{B}{2}$$] [(s – c) cot $$\frac{B}{2}$$]
= (s – b) (s – c) = R.H.S.
ans:

Question 34.
Express $$\frac{a \cos \mathbf{A}+\mathbf{b} \cos \mathbf{B}+\cos \mathbf{C}}{\mathbf{a}+\mathbf{b}+\mathbf{c}}$$ in terms of R and r.
Solution:

Question 35.
In ∆ABC, ∆ = 6sq.cm and s = 1.5 cm, find r.
Solution:
r = $$\frac{\Delta}{s}$$ = $$\frac{6}{1.5}$$ = 4 cm.

Question 36.
Show that rr1 cot$$\frac{A}{2}$$ = ∆.
Solution:
rr1 cot$$\frac{A}{2}$$ = $$\frac{\Delta}{s}$$ (s tan $$\frac{A}{2}$$) cot $$\frac{A}{2}$$ = ∆

Question 37.
If a = 13, b = 14, c = 15, find r1.
Solution:
2s = a + b +c = 42
⇒ s = 21
s – a = 8, s – b = 7, s – c = 6
2 = 21 × 8 × 7 × 6
⇒ ∆ = 7 × 12 = 84 sq. units
∴ r1 = $$\frac{\Delta}{s-b}$$ = $$\frac{84}{8}$$ = 10.5 units

Question 38.
If rr2 = r1r3, then find B.
Solution:

Question 39.
In a ∆ABC, show that the sides a, b, c are in A.P if and only if r1, r2, r3 are in H.P.
Solution:
r1, r2, r3 are in A.P.
⇔ $$\frac{1}{r_{1}}$$, $$\frac{1}{r_{2}}$$, $$\frac{1}{r_{3}}$$ are in A.P.
⇔ $$\frac{s-a}{\Delta}$$, $$\frac{s-b}{\Delta}$$, $$\frac{s-c}{\Delta}$$ are in A.P.
⇔ s – a, s – b, s – c are in A.P.
⇔ -a, -b, -c are in A.P.
⇔ a, b. c, are in A.R

Question 40.
If A = 90°, show that 2(r + R) = b + c.
Solution:
L.H.S = 2r + 2R
= 2(s – a) tan $$\frac{A}{2}$$ + 2R.1
= 2(s – a) tan 45°+ 2RsinA
(∵A = 90°)
= (2s – 2a). 1 + a
= b + c = R.H.S.

Question 41.
If (r2 – r1) (r3 – r1) = 2r2r3, show that A = 90°.
Solution:
(r2 – r1) (r3 – r1) = 2r2r3

⇒ 2(bc – ca – ab + a2) = b2 + c2 + a2 + 2bc – 2ca – 2ab
⇒ 2a2 = b2 + c2 + a2
⇒ b2 + c2 = a2
Hence △ABC is right angled and A = 90°.

Question 42.
Prove that $$\frac{r_{1}\left(r_{2}+r_{3}\right)}{\sqrt{r_{1} r_{2}+r_{2} r_{3}+r_{3} r_{1}}}$$ = a
Solution:

Question 43.
Show that $$\frac{1}{r^{2}}+\frac{1}{r_{1}^{2}}+\frac{1}{r_{2}^{2}}+\frac{1}{r_{3}^{2}}$$ = $$\frac{a^{2}+b^{2}+c^{2}}{\Delta^{2}}$$
Solution:

Question 44.
Prove that Σ(r + r1) tan $$\left(\frac{B-C}{2}\right)$$ = 0
Solution:
r + r1 = 4R sin $$\frac{A}{2}$$ sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$ + 4R sin $$\frac{A}{2}$$ cos $$\frac{B}{2}$$ cos $$\frac{C}{2}$$
= 4R sin$$\frac{A}{2}$$[sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$ + cos $$\frac{B}{2}$$ cos $$\frac{C}{2}$$]

Question 45.
Show that $$\frac{r_{1}}{b c}+\frac{r_{2}}{c a}+\frac{r_{3}}{a b}=\frac{1}{r}-\frac{1}{2 R}$$ .    [May 07]
Solution:

Question 46.
If r: R: r1 = 2 : 5 : 12. then prove that the triangle ¡s right angled at A.
Solution:
r : R : r1 = 2 : 5 : 12 then r = 2k. R = 5k and r1 = 12K
r1 – r = 12k – 2k = 10k = 2(5k) = 2R
⇒ 4R sin $$\frac{A}{2}$$[cos $$\frac{B}{2}$$ cos $$\frac{C}{2}$$ – sin $$\frac{B}{2}$$ sin $$\frac{C}{2}$$] = 2R
⇒ 2 sin$$\frac{A}{2}$$ cos$$\left(\frac{B+C}{2}\right)$$ = 1
⇒ 2 sin $$\frac{A}{2}$$ = $$\frac{1}{2}$$ [cos $$\left(\frac{B+C}{2}\right)$$ = sin $$\frac{A}{2}$$]
⇒ sin $$\frac{A}{2}$$ = $$\frac{1}{\sqrt{2}}$$ = sin 45°
⇒ $$\frac{A}{2}$$ = 45° ⇒ A = 90°
Hence the triangle is right angled at A.

Question 47.
Show that r + r3 + r1 – r2 = 4R cos B.
Solution:
r + r3

Question 48.
If A, A1, A2, A3 are the areas of incircle and excircles of a triangle respectively, then prove that $$\frac{1}{\sqrt{A_{1}}}+\frac{1}{\sqrt{A_{2}}}+\frac{1}{\sqrt{A_{3}}}=\frac{1}{\sqrt{A}}$$ .
Solution:
r, r1, r2, r3 are the in radius and ex-radii of the circles whose areas are A, A1, A2, A3 respectively, then
A = itr2, A1 = πr12, A2 = πr22, A3 = πr32

Question 49.
Show that (r1 + r2) sec2 $$\frac{C}{2}$$ = (r2 + r3) sec2 $$\frac{A}{2}$$ = (r3 + r1) sec2 $$\frac{B}{2}$$.
Solution:
r1 + r2 = 4R cos $$\frac{C}{2}$$ [sin$$\frac{A}{2}$$ cos$$\frac{B}{2}$$ + cos $$\frac{A}{2}$$ sin $$\frac{B}{2}$$]
= 4R cos$$\frac{C}{2}$$ sin$$\left(\frac{A+B}{2}\right)$$
= 4R cos2$$\frac{C}{2}$$ ___________ (1)
⇒ (r1 + r2) sec2$$\frac{C}{2}$$ = 4R
Similarly, we can show that
(r2 + r3) sec2$$\frac{A}{2}$$ = (r3 + r1) sec2$$\frac{B}{2}$$ = 4R
∴ (r1 + r2)sec2$$\frac{C}{2}$$ = (r2 + r3)
sec2$$\frac{A}{2}$$ = (r3 + r1) sec2$$\frac{B}{2}$$ = 4R

Question 50.
In ∆ABC, if AD, BE, CF are the perpendiculars drawn from the vertices A, B, C to the opposite sides, show that
i) $$\frac{1}{A D}+\frac{1}{B E}+\frac{1}{C F}=\frac{1}{r}$$ and
ii) AD. BE. CF = $$\frac{(a b c)^{2}}{8 R^{3}}$$
Solution:

Question 51.
In ∆ ABC, if r1 = 8, r2 = 12, r3 = 24, find a, b, c.
Solution:
∵ $$\frac{1}{r}$$ = $$\frac{1}{r_{1}}$$ + $$\frac{1}{r_{2}}$$ + $$\frac{1}{r_{3}}$$
⇒ $$\frac{1}{r}$$ = $$\frac{1}{8}$$ + $$\frac{1}{12}$$ + $$\frac{1}{24}$$
⇒ $$\frac{1}{r}$$ = $$\frac{3+2+1}{24}$$ ⇒ r = 4
But ∆2 = rr1r2r3 = 4 × 8 × 12 × 24
= (8 × 12)2
⇒ ∆ = 96 sq. cm
∵ r = $$\frac{\Delta}{\mathrm{s}}$$ ⇒ 4 = $$\frac{96}{s}$$ ⇒ s = 24
∵ r1 = $$\frac{\Delta}{s-a}$$
⇒ δ = $$\frac{96}{s-a}$$ ⇒ s – a = $$\frac{96}{8}$$ ⇒ 24 – a
= 12 ⇒ a = 12
Similarly s – b = $$\frac{96}{12}$$ ⇒ 24 – b = 8 ⇒ b = 16
s – c = $$\frac{96}{24}$$ ⇒ 24 – c = 4 ⇒ c = 20

Question 52.
Show that $$\frac{a b-r_{1} r_{2}}{r_{3}}$$ = $$\frac{b c-r_{2} r_{3}}{r_{1}}$$ = $$\frac{c a-r_{3} r_{1}}{r_{2}}$$   [May 08; Mar 08]
Solution: