11 Most FAQ’s of Properties of Triangles Chapter in Inter 1st Year Maths-1A (TS/AP)

LAQ-1 : In a ∆ABC if r₁=8, r₂=12, r₃=24 then find a,b,c

Here, first we find r.

Given r₁​=8, r₂​=12, r₃​=24, then

$$\frac{1}{r} = \frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}$$

$$\Rightarrow \frac{1}{r} = \frac{1}{8} + \frac{1}{12} + \frac{1}{24} = \frac{3 + 2 + 1}{24} = \frac{6}{24} = \frac{1}{4}$$

$$\Rightarrow 1/r = 1/4 \Rightarrow r = 4$$

Now

$$\Delta = \sqrt{(r \cdot r_1 \cdot r_2 \cdot r_3)} = \sqrt{4 \cdot 8 \cdot 12 \cdot 24}$$

$$= \sqrt{4 \cdot 8 \cdot (3 \cdot 4) \cdot (3 \cdot 8)}$$

$$= \sqrt{32 \cdot 4^2 \cdot 8^2} = 3 \cdot 4 \cdot 8 = 96$$

Now, r=Δ/s ⇒ s=Δ/r = 96/4 = 24

(i) $$r_1 = \Delta/(s-a)$$ $$s-a = \Delta/r_1 = 96/8 = 12$$

$$s-a = 12 \Rightarrow 24-a = 12$$

$$\Rightarrow a = 24 – 12 = 12$$

(ii) $$r_2 = \Delta/(s-b)$$ $$s-b = \Delta/r_2 = 96/12 = 8$$

$$s-b = 8 \Rightarrow 24-b = 8$$

$$\Rightarrow b = 24 – 8 = 16$$

(iii) $$r_3 = \Delta/(s-c)$$ $$s-c = \Delta/r_3 = 96/24 = 4$$

$$s-c = 4 \Rightarrow 24-c = 4$$

$$\Rightarrow c = 24 – 4 = 20$$

---

LAQ-2 : If r₁=2, r₂=3, r₃=6, r=1, then prove that a=3, b=4, c=5

Given r_1​=2, r₂​=3, r₃​=6 and r=1, then

$$\Delta = \sqrt{(r \cdot r_1 \cdot r_2 \cdot r_3)} = \sqrt{1 \cdot 2 \cdot 3 \cdot 6} = \sqrt{36} = 6$$

Now, $$r = \Delta/s$$ $$1 = 6/s$$

(i) $$r_1 = \Delta/(s-a)$$ $$s – a = \Delta/r_1 = 6/2 = 3$$

$$s – a = 3$$ $$6 – a = 3$$

$$\Rightarrow a = 6 – 3 = 3$$

(ii) $$r_2 = \Delta/(s-b)$$ $$s – b = \Delta/r_2 = 6/3 = 2$$

$$s – b = 2$$ $$6 – b = 2$$

$$\Rightarrow b = 6 – 2 = 4$$

(iii) $$r_3 = \Delta/(s-c)$$ $$s – c = \Delta/r_3 = 6/6 = 1$$

$$s – c = 1s – c = 1$$ $$6 – c = 1$$

$$\Rightarrow c = 6 – 1 = 5$$

---

LAQ-3 : In a ∆ABC if a=13, b=14, c=15 then show that R=65/8, r=4, r₁=21/2, r₂=12, r₃=14

Given a=13, b=14, c=15, then

$$2s = a + b + c = 13 + 14 + 15 = 42 \Rightarrow s = 21$$

Now

$$\Delta = \sqrt{s(s – a)(s – b)(s – c)} = \sqrt{21(21 – 13)(21 – 14)(21 – 15)}$$

$$= \sqrt{21 \times 8 \times 7 \times 6} = \sqrt{(3 \times 7)(4 \times 2)(7)(3 \times 2)} = \sqrt{3^2 \times 4^2 \times 7^2} = 3 \times 4 \times 7 = 84$$

(i) $$R = \frac{abc}{4\Delta} = \frac{13 \times 14 \times 15}{4 \times 84} = \frac{65}{8}$$

(ii) $$r = \frac{\Delta}{s} = \frac{84}{21} = 4$$

(iii) $$r1 = \frac{\Delta}{s – a} = \frac{84}{21-13} = \frac{84}{8} = \frac{21}{2}$$

(iv) $$r2 = \frac{\Delta}{s-b} = \frac{84}{21-14} = \frac{84}{7} = 12$$

(v) $$r3 = \frac{\Delta}{s-c} = \frac{84}{21-15} = \frac{84}{6} = 14$$

---

LAQ-4 : If a = (b+c)cosθ, then prove that sin θ=(2√bc)/(b+c) cos(A/2)

Given $$a = (b+c)\cos\theta$$ then

$$\cos\theta = \frac{a}{b+c} \Rightarrow \cos^2\theta = \frac{a^2}{(b+c)^2}$$

$$\sin^2\theta = 1 – \cos^2\theta$$

$$= 1 – \frac{a^2}{(b+c)^2} = \frac{(b+c)^2 – a^2}{(b+c)^2} = \frac{b^2 + c^2 + 2bc – a^2}{(b+c)^2}$$

$$\sin^2\theta = 1 – \cos^2\theta$$

$$= 1- \frac{a^2}{(b+c)^2} = \frac{(b+c)^2 – a^2}{(b+c)^2} = \frac{(b^2 + c^2 + 2bc) – a^2}{(b+c)^2}$$

$$= \frac{2bc + (b^2 + c^2 – a^2)}{(b+c)^2} = \frac{2bc + 2bc \cos A}{(b+c)^2}$$

$$= \frac{2bc(1 + \cos A)}{(b+c)^2} = \frac{2bc}{2\cos^2(A/2)}{(b+c)^2} = \frac{4bc \cos^2(A/2)}{(b+c)^2}$$

$$\sin\theta = \frac{2\sqrt{bc}}{b+c} \cos(A/2)$$

---

LAQ-5 : If sin θ=a/(b+c) then show that cosθ=(2√bc)/(b+c) cos(A/2)

Given $$\sin\theta = \frac{a}{b} + c$$

$$\Rightarrow \sin^2\theta = \left(\frac{a}{b+c}\right)^2$$

$$\cos^2\theta = 1 – \sin^2\theta$$

$$= 1 – \left(\frac{a^2}{(b+c)^2}\right) = \frac{(b+c)^2 – a^2}{(b+c)^2}$$

$$= \frac{(b^2 + c^2 + 2bc) – a^2}{(b+c)^2}$$

$$= \frac{2bc + (b^2 + c^2 – a^2)}{(b+c)^2}$$

$$= \frac{2bc + 2bc \cos A}{(b+c)^2}$$

$$= \frac{2bc(1 + \cos A)}{(b+c)^2}$$

$$= \frac{2bc \cdot \cos^2(A/2)}{(b+c)^2}$$

$$= \frac{4bc \cdot \cos^2(A/2)}{(b+c)^2}$$

$$\cos\theta = \frac{2\sqrt{bc}}{b+c} \cdot \cos(A/2)$$

---

LAQ-6 : If a = (b-c)secθ, then prove that tan θ=(2√bc)/(b-c) sin(A/2)

Given $$a = (b-c)\sec\theta$$ $$\sec\theta = \frac{a}{b-c}$$

$$\Rightarrow \sec^2\theta = \left(\frac{a}{b-c}\right)^2$$

$$\tan^2\theta = \sec^2\theta – 1$$

$$= \left(\frac{a^2}{(b – c)^2}\right) – 1 = \frac{a^2 – (b-c)^2}{(b-c)^2}$$

$$= \frac{a^2 – (b^2 – 2bc + c^2)}{(b-c)^2}$$

$$= \frac{2bc – (b^2 + c^2 – a^2)}{(b-c)^2}$$

$$= \frac{2bc – 2bc \cos A}{(b-c)^2}$$

$$= \frac{2bc(1 – \cos A)}{(b-c)^2}$$

$$= \frac{2bc \cdot (2\sin^2(A/2))}{(b-c)^2}$$

$$= \frac{4bc \cdot \sin^2(A/2)}{(b-c)^2}$$

$$\tan\theta = \frac{2\sqrt{bc}}{b-c} \cdot \sin(A/2)$$

---

LAQ-7 : In ∆ABC, show that (a+b+c)²/(a²+b²+c²)=(cot A/2+cot B/2+cot C/2)/(cotA+cotB+cotC)

In the RHS, the numerator is given by:

$$\text{cot} \frac{A}{2} + \text{cot} \frac{B}{2} + \text{cot} \frac{C}{2}$$

$$= \frac{s(s-a)}{\Delta} + \frac{s(s-b)}{\Delta} + \frac{s(s-c)}{\Delta}$$

$$= \frac{s(s-a) + s(s-b) + s(s-c)}{\Delta}$$

$$= \frac{s[3s – (a + b + c)]}{\Delta} = \frac{s[3s – 2s]}{\Delta} = \frac{s^2}{\Delta} \quad \text{(1)}$$

In the RHS, the denominator is given by:

$$\text{cot} A + \text{cot} B + \text{cot} C = \frac{\cos A}{\sin A} + \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C}$$

$$= \frac{b^2 + c^2 – a^2}{2bc(\sin A)} + \frac{c^2 + a^2 – b^2}{2ac(\sin B)} + \frac{a^2 + b^2 – c^2}{2ab(\sin C)},$$

from the cosine rule, we have:

$$= \frac{b^2 + c^2 – a^2}{4\Delta} + \frac{c^2 + a^2 – b^2}{4\Delta} + \frac{a^2 + b^2 – c^2}{4\Delta}$$

$$= \frac{b^2 + c^2 – a^2 + c^2 + a^2 – b^2 + a^2 + b^2 – c^2}{4\Delta} = \frac{a^2 + b^2 + c^2}{4\Delta} \quad \text{(2)}$$

From (1) & (2), RHS is expressed as:

$$\frac{\text{cot} \frac{A}{2} + \text{cot} \frac{B}{2} + \text{cot} \frac{C}{2}}{\text{cot} A + \text{cot} B + \text{cot} C} = \frac{s^2 / \Delta}{(a^2 + b^2 + c^2) / 4\Delta}$$

$$= \frac{4s^2}{a^2 + b^2 + c^2} = \frac{(2s)^2}{a^2 + b^2 + c^2} = \frac{(a + b + c)^2}{a^2 + b^2 + c^2} = \text{LHS}$$

---

LAQ-8 : In ∆ABC prove that acos²  A/2 + bcos²  B/2 + ccos²  C/2 = s + ∆/R

Given the expression for the LHS:

$$\text{LHS} = a\cos^2 \frac{A}{2} + b\cos^2 \frac{B}{2} + c\cos^2 \frac{C}{2}$$

This simplifies to:

$$= a\left(1 + \cos A\right)/2 + b\left(1 + \cos B\right)/2 + c\left(1 + \cos C\right)/2$$

$$= (a/2 + b/2 + c/2) + \left(a \cos A + b \cos B + c \cos C\right)/2$$

$$= (a + b + c)/2 + \left(a \cos A + b \cos B + c \cos C\right)/2$$

$$= s + \left(2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C\right)/2$$

$$= s + R \left(\sin 2A + \sin 2B + \sin 2C\right)/2$$

$$= s + R/2\left(\sin 2A + \sin 2B + \sin 2C\right)$$

$$= s + R/2\left[2\sin(A + B)\cos(A – B) + \sin 2C\right]$$

Since $$A + B + C = \pi$$ $$A + B = \pi – C$$ and thus:

$$= s + R/2\left[2\sin(\pi – C)\cos(A – B) + \sin 2C\right]$$

$$= s + R/2\left(2\sin C\cos(A – B) + 2\sin C\cos C\right)$$

$$= s + R\sin C\left(\cos(A – B) + \cos C\right)$$

Using the trigonometric identity $$\cos(A – B) – \cos(A + B) = -2\sin A\sin B$$

$$= s + R\sin C\left(-2\sin A \sin B\right)$$

$$= s + 2R\sin A\sin B\sin C$$

$$= s + \frac{2R^2 \sin A \sin B \sin C}{R}$$

$$= s + \frac{\Delta}{R}$$

Thus, proving:

$$\text{LHS} = s + \frac{\Delta}{R} = \text{RHS}$$

---

LAQ-9 : In ∆ABC prove that r₁/bc + r₂/ca + r₃/ab = 1/r – 1/2R

Given the expression for the LHS:

$$\text{LHS} = \sum \frac{r_1}{bc}$$

And the transformation involves:

$$= \sum \frac{4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{(2R \sin B)(2R \sin C)}$$

$$= \sum \frac{4R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{4R^2 \sin B \sin C}$$

$$= \sum \frac{\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{R \sin B \sin C}$$

$$= \sum \frac{\sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}}{R(2 \sin \frac{B}{2} \cos \frac{B}{2})(2 \sin \frac{C}{2} \cos \frac{C}{2})}$$

$$= \sum \frac{\sin \frac{A}{2}}{4R \sin \frac{B}{2} \sin \frac{C}{2}}$$

$$= \sum \frac{\sin^2 \frac{A}{2}}{r} = \frac{1}{r} \sum \sin^2 \frac{A}{2}$$

$$= \frac{1}{r}(\sin^2 \frac{A}{2} + \sin^2 \frac{B}{2} + \sin^2 \frac{C}{2})$$

$$= \frac{1}{r}\left((1 – \cos^2 \frac{A}{2}) + (1 – \cos^2 \frac{B}{2}) + (1 – \cos^2 \frac{C}{2})\right)$$

$$= \frac{1}{r}\left(3 – (\cos^2 \frac{A}{2} + \cos^2 \frac{B}{2} + \cos^2 \frac{C}{2})\right)$$

Using the identity $$\cos(A + B) = \cos A \cos B – \sin A \sin B$$ and considering $$\cos 90^\circ = 0$$ simplifies to:

$$= \frac{1}{r}(1 – \sin \frac{C}{2}(\cos \frac{A – B}{2} – \sin \frac{C}{2}))$$

$$= \frac{1}{r}(1 – \sin \frac{C}{2}(\cos \frac{A – B}{2} – \cos \frac{A + B}{2}))$$

Since $$A + B + C = 180^\circ$$ and using trigonometric identities:

$$= \frac{1}{r}(1 – 2 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})$$

$$= \frac{1}{r} – \frac{2(2R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2})}{2R}$$

$$= \frac{1}{r} – \frac{r}{2R} = \frac{1}{r} – \frac{1}{2R} = \text{RHS}$$

---

LAQ-10 : In a ∆ABC if a:b:c = 7:8:9 then show that cosA:cosB:cosC = 14:11:6

Given the ratio $$a:b:c = 7:8:9$$

implying a=7K, b=8K, c=9K, we calculate the cosines of angles A, B, and C using the law of cosines:

For cosA: $$\cos A = \frac{b^2 + c^2 – a^2}{2bc} = \frac{(8K)^2 + (9K)^2 – (7K)^2}{2 \cdot (8K) \cdot (9K)} = \frac{k^2(64 + 81 – 49)}{2 \cdot 8 \cdot 9 \cdot k^2}$$

$$= \frac{96}{144} = \frac{2}{3}$$

For cosB: $$\cos B = \frac{a^2 + c^2 – b^2}{2ac} = \frac{(7K)^2 + (9K)^2 – (8K)^2}{2 \cdot (7K) \cdot (9K)} = \frac{k^2(49 + 81 – 64)}{2 \cdot 7 \cdot 9 \cdot k^2}$$

$$= \frac{66}{126} = \frac{11}{21}$$

For cosC: $$\cos C = \frac{a^2 + b^2 – c^2}{2ab} = \frac{(7K)^2 + (8K)^2 – (9K)^2}{2 \cdot (7K) \cdot (8K)} = \frac{k^2(49 + 64 – 81)}{2 \cdot 7 \cdot 8 \cdot k^2}$$

$$= \frac{32}{112} = \frac{2}{7}$$

To find the ratio cosA:cosB:cosC, let’s simplify:

$$\cos A : \cos B : \cos C = \frac{2}{3} : \frac{11}{21} : \frac{2}{7}$$

$$= 14 : 11 : 6$$

---

LAQ-11 : If p₁,p₂,p₃ are altitudes of a ∆ABC then show that i).1/p₁ + 1/p₂ + 1/p₃ =1/r ii).1/p₁ + 1/p₂ – 1/p₃ =1/r₃ iii).p₁ p₂ p₃= (abc)²/8R³ = 8∆³/abc

(i) Area and Altitudes

Given the area of triangle ABC (Δ) can be expressed as $$\frac{1}{2}\times \text{base} \times \text{height}$$ we have:

For altitude p₁​ (from A to BC):

$$\Delta = \frac{1}{2}a \cdot p_1$$

$$\Rightarrow p_1 = \frac{2\Delta}{a}$$

Similarly, for p₂​ and p₃​:

$$p_2 = \frac{2\Delta}{b}, \quad p_3 = \frac{2\Delta}{c}$$

Hence, the reciprocals are:

$$\frac{1}{p_1} = \frac{a}{2\Delta}, \quad \frac{1}{p_2} = \frac{b}{2\Delta}, \quad \frac{1}{p_3} = \frac{c}{2\Delta}$$

(ii) Sum of Reciprocals of Altitudes

$$\frac{1}{p_1} + \frac{1}{p_2} + \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} + \frac{c}{2\Delta} = \frac{a + b + c}{2\Delta} = \frac{2s}{2\Delta} = \frac{s}{\Delta}$$

$$\Rightarrow \frac{s}{\Delta} = \frac{1}{r}$$

(iii) LHS with Specific Reciprocals

$$\frac{1}{p_1} + \frac{1}{p_2} – \frac{1}{p_3} = \frac{a}{2\Delta} + \frac{b}{2\Delta} – \frac{c}{2\Delta} = \frac{a + b – c}{2\Delta} = \frac{2s – 2c}{2\Delta} = \frac{s – c}{\Delta}$$

$$\Rightarrow \frac{s – c}{\Delta} = \frac{1}{r_3}$$

(iv) Product of Altitudes

$$p_1 \cdot p_2 \cdot p_3 = \frac{2\Delta}{a} \cdot \frac{2\Delta}{b} \cdot \frac{2\Delta}{c} = \frac{8\Delta^3}{abc}$$

$$= \frac{8}{abc} \cdot \left(\frac{abc}{4R}\right)^3 = \frac{8(abc)^2}{64R^3} = \frac{(abc)^2}{8R^3}$$

$$\Rightarrow \frac{(abc)^2}{8R^3} = \text{RHS}$$