Inter 1st Year Maths 1A Properties of Triangles Solutions Ex 10(b)

Practicing the Intermediate 1st Year Maths 1A Textbook Solutions Inter 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b) will help students to clear their doubts quickly.

Intermediate 1st Year Maths 1A Properties of Triangles Solutions Exercise 10(b)

All problems in this exercise have reference to ΔABC.

I.

Question 1.
Express \(\Sigma r_{1} \cot \frac{A}{2}\) in terms of s.
Solution:
\(\Sigma r_{1} \cot \frac{A}{2}\) = \(\Sigma\left(s \tan \frac{A}{2}\right) \cot \frac{A}{2}\)
= Σs
= s + s + s
= 3s

Question 2.
Show that Σa cot A = 2(R + r).
Solution:
L.H.S = Σa . cot A
= Σ2R sin A \(\frac{\cos A}{\sin A}\)
= 2R Σ cos A
= \(2 R\left(1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}\right)\) (From transformants)
= \(2\left(R+4 R \sin \frac{A}{2} \cdot \sin \frac{B}{2} \cdot \sin \frac{C}{2}\right)\)
= 2(R + r)
= R.H.S

Question 3.
In ∆ABC, prove that r₁ + r₂ + r₃ – r = 4R.
Solution:

Question 4.
In ∆ABC, prove that r + r₁ + r₂ – r₃ = 4R cos C.
Solution:

Question 5.
If r + r₁ + r₂ + r₃ then show that C = 90°.
Solution:

II.

Question 1.
Prove that 4(r₁r₂ + r₂r₃ + r₃r₁) = (a + b + c)²
Solution:

Question 2.
Prove that \(\left(\frac{1}{r}-\frac{1}{r_{1}}\right)\left(\frac{1}{r}-\frac{1}{r_{2}}\right)\left(\frac{1}{r}-\frac{1}{r_{3}}\right)=\frac{a b c}{\Delta^{3}}=\frac{4 R}{r^{2} s^{2}}\)
Solution:

Question 3.
Prove that r(r₁ + r₂ + r₃) = ab + bc + ca – s².
Solution:

Question 4.
Show that \(\sum \frac{r_{1}}{(s-b)(s-c)}=\frac{3}{r}\)
Solution:

Question 5.
Show that \(\left(r_{1}+r_{2}\right) \tan \frac{C}{2}=\left(r_{3}-r\right) \cot \frac{C}{2}=c\)
Solution:

Question 6.
Show that r₁r₂r₃ = \(r^{3} \cot ^{2} \frac{A}{2} \cdot \cot ^{2} \frac{B}{2} \cdot \cot ^{2} \frac{C}{2}\)
Solution:

III.

Question 1.
Show that cos A + cos B + cos C = 1 + \(\frac{r}{R}\)
Solution:
L.H.S = cos A + cos B + cos C
= 2 cos(\(\frac{A+B}{2}\)) cos(\(\frac{A-B}{2}\)) + cos C

Question 2.
Show that \(\cos ^{2} \frac{A}{2}+\cos ^{2} \frac{B}{2}+\cos ^{2} \frac{C}{2}=2+\frac{r}{2 R}\)
Solution:

Question 3.
Show that \(\sin ^{2} \frac{A}{2}+\sin ^{2} \frac{B}{2}+\sin ^{2} \frac{C}{2}=1-\frac{r}{2 R}\)
Solution:

Question 4.
Show that
(i) a = (r₂ + r₃) \(\sqrt{\frac{r r_{1}}{r_{2} r_{3}}}\)
(ii) ∆ = r₁r₂ \(\sqrt{\frac{4 R-r_{1}-r_{2}}{r_{1}+r_{2}}}\)
Solution:

Question 5.
Prove that \(r_{1}^{2}+r_{2}^{2}+r_{3}^{2}+r^{2}\) = 16R² – (a² + b² + c²).
Solution:

Question 6.
If p₁, p₂, p₃ are altitudes drawn from vertices A, B, C to the opposite sides of a triangle respectively, then show that
(i) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}+\frac{1}{p_{3}}=\frac{1}{r}\)
(ii) \(\frac{1}{p_{1}}+\frac{1}{p_{2}}-\frac{1}{p_{3}}=\frac{1}{r_{3}}\)
(iii) p₁ . p₂ . p₃ = \(\frac{(a b c)^{2}}{8 R^{3}}=\frac{8 \Delta^{3}}{a b c}\)
Solution:

Question 7.
If a = 13, b = 14, c = 15, show that R = \(\frac{65}{8}\), r = 4, r₁ = \(\frac{21}{2}\), r₂ = 12 and r₃ = 14.
Solution:
a = 13, b = 14, c = 15
s = \(\frac{a+b+c}{2}\)
= \(\frac{13+14+15}{2}\)
= 21
s – a = 21 – 13 = 8
s – b = 21 – 14 = 7
s – c = 21 – 15 = 6

Question 8.
If r₁ = 2, r₂ = 3, r₃ = 6 and r = 1, prove that a = 3, b = 4 and c = 5.
Solution: