Effective S Chand Class 11 Maths Solutions Chapter 7 Properties of Triangle Ex 7 can help bridge the gap between theory and application.
S Chand Class 11 ICSE Maths Solutions Chapter 7 Properties of Triangle Ex 7
Question 1.
In ∆ABC,
(i) If a = 2, b = 3, c = 4, prove that cos A = \(\frac { 7 }{ 8 }\).
(ii) if the sides are 7, 4\(\sqrt{3}\) and \(\sqrt{3}\) cm, prove that the smallest angle is 30°.
(iii) if a = 9, b = 8, c = 4, prove that 6 cos C = 4 + 3 cos B.
(iv) The sines of the angles of a triangle are in the ratio of 4 : 5 : 6 ; prove that the cosines of the angles are 12 : 9 : 2.
(v) If the two angles of a triangle are 30° and 45° and the included side is (\(\sqrt{3}\) + 1) cm, find the area of the triangle.
(vi) If in a ∆ABC, a = 6, b = 3 and cos (A – B) = \(\frac { 4 }{ 5 }\), find its area.
(vii) In a triangle ABC, ∠C = 60° and ∠A = 75°. If D is a point on AC such that the area of the ∆ABAD is \(\sqrt{3}\) times the area of the ∆ABCD, find the ∠ABD.
Solution:
(i) Given a = 2, b = 2, c = 4
By cosine formula, we have
cos A = \(\frac{b^2+c^2-a^2}{2 b c}=\frac{3^2+4^2-2^2}{2 \times 3 \times 4}=\frac{9+16-4}{24}=\frac{21}{24}=\frac{7}{8}\)
(ii) Let a = 7 ; b = 4\(\sqrt{3}\); c = \(\sqrt{13}\)
Since smallest side has smallest angle opposite to it.
Here a > b > c
Hence the smallest angle of ∆ABC be \(\frac { π }{ 6 }\).
(iii) Given a = 9, b = 8, c = 4
By using cosine’s formula, we have
∴ from eqn. (1) and eqn. (2); we have
6 cos C = 4 + 3 cos B
(iv) Using sine formula, we have
\(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}\) … (1)
Since sin A : sin B : sin C = 4 : 5 : 6
Thus using eqn. (1); we have
(v) Since the sum of angles of ∆ABC is 180°.
∴ ∠A + ∠B + ∠C = 180°
⇒ ∠A + 30° + 45° = 180°
⇒ ∠A = 180° – 75° = 105°
Using sine formula, we have
(vi) Given a = 6, b = 3 and cos (A – B) = \(\frac { 4 }{ 5 }\)
(vii) Since the sum of angles of ∆ABC is 180°.
Thus, ∠A = 180°- 75° – 60° = 45°
Let ∠BAD = θ ∴ ∠CBD = 45° – θ
Now, area of ∆BAD = \(\frac { 1 }{ 2 }\) x BA x BD sin θ
and area of ∆BCD = \(\frac { 1 }{ 2 }\) x BC x BD sin (45° – θ)
according to given condition, we have
In any ∆ABC, Prove that
Question 2.
\(\frac{\sin A}{\sin (A+B)}=\frac{a}{c}\)
Solution:
Question 3.
\(\frac{a-b}{a+b}=\frac{\tan \frac{1}{2}(A-B)}{\tan \frac{1}{2}(A+B)}\)
Solution:
Question 4.
ac cos B – bc cos A = a² – b²
Solution:
Question 5.
\(\frac{\sin (A-C)}{\sin (A+B)}=\frac{a^2-b^2}{c^2}\)
Solution:
Question 6.
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0
Solution:
Using sine formula, we have
\(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}\)
Then sin A = ak; sin B = bk and sin C = ck
L.H.S = a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)
= a (bk – ck) + b (ck – ak) + c (ak – bk)
= k[ab – ac + bc – ba + ca – bc] = k x 0
= 0 = R.H.S.
Question 7.
a cos (A + B + C) – b cos (B + A) – c cos (A + C) = 0
Solution:
L.H.S. = a cos (A + B + C) – b cos (B + A) – c cos (A + C)
= a cos π – b cos (π – C) – c cos (π – B)
= – a – b (- cos C) – c (- cos B)
= – a + b cos C + c cos B
= – a + a = 0 = R.H.S.
[using projection formula, b cos C + c cos B = a]
Question 8.
a (cos C – cos B) = 2 (b – c) cos² \(\frac { 1 }{ 2 }\) A
Solution:
Question 9.
a sin\(\frac { 1 }{ 2 }\) (B – C) = (b – c) cos \(\frac { 1 }{ 2 }\)A
Solution:
Question 10.
a sin(\(\frac { A }{ 2 }\) + B) = (b + c)sin\(\frac { A }{ 2 }\)
Solution:
Question 11.
c² = (a – b)² cos² \(\frac { 1 }{ 2 }\) C + (a + b)² sin² \(\frac { 1 }{ 2 }\)C
Solution:
Question 12.
a sin (B – C) + b sin (C – A) + c sin (A – B) = 0
Solution:
Using sine formula, we have \(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k ≠ 0 (say)
Then a = k sin A; b – k sin B ; c = k sin C
L.H.S. = a sin (B – C) + b sin (C – A) + c sin (A – B)
= k [sin A sin (B – C) + sin B sin (C-A) + sin C sin (A – B)]
= k [sin (π – \(\overline{B+C}\)) sin(B – C) + sin(π – \(\overline{A+C}\)) sin (C – A) + sin (π – \(\overline{A+B}\)) sin (A – B)] [∵ A + B + C = π]
= k [sin (B + C) sin (B – C) + sin (C + A) sin (C – A) + sin (A + B) sin (A – B)]
= k [sin² B – sin² C + sin² C – sin² A + sin² A – sin² B]
= k x 0 = 0 = R.H.S.
Question 13.
\(\frac{\cos 2 A}{a^2}-\frac{\cos 2 B}{b^2}=\frac{1}{a^2}-\frac{1}{b^2}\)
Solution:
L.H.S = \(\frac{\cos 2 \mathrm{~A}}{a^2}-\frac{\cos 2 \mathrm{~B}}{b^2}=\frac{1-2 \sin ^2 \mathrm{~A}}{a^2}-\frac{1-2 \sin ^2 \mathrm{~B}}{b^2}\)
Using sine formula, \(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}\) = k ≠ 0 (say)
Then sin A = ak; sin B = bk; sin C = ck
= \(\frac{1-2 a^2 k^2}{a^2}-\frac{1-2 b^2 k^2}{b^2}=\frac{1}{a^2}-2 k^2-\frac{1}{b^2}+2 k^2=\frac{1}{a^2}-\frac{1}{b^2}\) = R.H.S
Question 14.
\(\frac{1+\cos (A-B) \cos C}{1+\cos (A-C) \cos B}=\frac{a^2+b^2}{a^2+c^2}\)
Solution:
Question 15.
(b² – c²) cot A + (c² – a²) cot B + (a² – b²) cot C = 0
Solution:
Question 16.
a³ sin (B – C) cosec² A + b³ sin (C – A) cosec² B + c³ sin (A – B) cosec² C = 0
Solution:
L.H.S = a³ sin (B – C) cosec² A + b³ sin (C – A) cosec² B + c³ sin (A – B) cosec² C
= \(\frac{a^3}{\sin ^2 \mathrm{~A}} \sin (\mathrm{B}-\mathrm{C})+\frac{b^3}{\sin ^2 \mathrm{~B}} \sin (\mathrm{C}-\mathrm{A})+\frac{c^3}{\sin ^2 \mathrm{C}} \sin (\mathrm{A}-\mathrm{B})\)
[using sine formula, \(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k ≠ 0]
= [k³ sin A sin (B – C) + k³ sin B sin (C – A) + k³ sin C sin (A – B)]
= k³ [sin (π – \(\overline{B+C}\)) sin (B – C) + sin (π – \(\overline{A+C}\)) sin (C – A) + sin (π – \(\overline{A+B}\)) sin (A – B)]
= k³ [sin (B + C) sin (B – C) + sin (A + C) sin (C-A) + sin (A + B) sin (A – B)] [∵ A + B + C = π]
= k³ [sin² B – sin² C + sin² C – sin² A + sin² A – sin² B] = k³ x 0 = 0 = R.H.S.
Question 17.
a³ cos (B – C) + b³ cos (C – A) + c³ cos (A – B) = 3abc.
Solution:
L.H.S. = a³ cos (B – C) + b³ cos (C – A) + c³ cos (A – B)
= a² [k sin A cos (B – C)] + b² [k sin B cos (C – A)] + c² [k sin C cos (A – B)] [Using sine formula, \(\frac{\sin \mathrm{A}}{a}=\frac{\sin \mathrm{B}}{b}=\frac{\sin \mathrm{C}}{c}=\frac{1}{k}\)]
= ka² sin (π – \(\overline{B+C}\)) cos (B – C) + kb² sin (π – \(\overline{A+C}\)) cos (C – A) + kc² sin (π – \(\overline{A+B}\)) cos (A – B)
= ka² sin (B + C) cos (B – C) + kb² sin (C + A) cos (C – A) + kc² sin (A + B) cos (A – B)
= \(\frac{k a^2}{2}\) (sin 2B + sin 2C) + \(\frac { 1 }{ 2 }\)kb² (sin 2C + sin 2A) + \(\frac { 1 }{ 2 }\)kc² (sin 2A + sin 2B)
= \(\frac{k a^2}{2}\) [2 sin B cos B + 2 sin C cos C] + \(\frac{k b^2}{2}\) [2 sin C cos C + 2 sin A cos A] + \(\frac{k c^2}{2}\) [2 sin A cos A + 2 sin B cos B]
= a² (b cos B + c cos C) + b² (c cos C + a cos A) + c² (a cos A + b cos B) [using sine formula]
= ab (a cos B + b cos A) + be (b cos C + c cos B) + ac (a cos C + c cos A)
= abc + bca + acb = 3abc = R.H.S. [using projection formulae]
Question 18.
(i) In a ∆ABC, if \(\frac{2 \cos A}{a}+\frac{\cos B}{b}+\frac{2 \cos C}{c}=\frac{a}{b c}+\frac{b}{c a}\), prove that ∠A = 90°
(ii) In a ∆ABC, AD is the altitude from A. Given b > c, ∠C = 23° and AD = \(\frac{a b c}{\left(b^2-c^2\right)}\), find ∠B.
Solution:
(i) Given
Thus ∆ABC is right angled at A ∴ ∠A = 90°
(ii) Given ∠B = 23° and AD = \(\frac{a b c}{b^2-c^2}\) … (1)
Also from right angled ∆ADC, \(\frac { AD }{ AC }\) = sin 23°
⇒ AD = b sin 23° … (2)
∴ from eqn. (1) and eqn. (2); we have
Question 19.
In ∆ABC, \(\frac{a^2+b^2}{a^2-b^2}=\frac{\sin (A+B)}{\sin (A-B)}\), Prove the triangle is isosceles or right triangle.
Solution:
Question 20.
If \(\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}\), prove that a², b², c² are in A.P.
Solution:
Question 21.
If sin 2A + sin 2B = sin 2C, prove that A = 90° or B = 90°
Solution:
Question 22.
If C = 60°, prove that \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)
Solution:
Let \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\) be true.
⇒ \(\frac{b+c+a+c}{(a+c)(b+c)}=\frac{3}{a+b+c}\)
⇒ (b + a + 2c) (a + b + c) = 3 (a + c) (b + c)
⇒ a² + b² + 2ab + 3c(a + b) + 2c² = 3ab + 3ac + 3bc + 3c²
⇒ a² + b² – c² = ab
⇒ \(\frac{a^2+b^2-c^2}{2 a b}=\frac{1}{2}\) ⇒ cos C = \(\frac { 1 }{ 2 }\) [using cosine formula]
∴ ∠C = \(\frac { π }{ 3 }\) or 60° which is true
Thus \(\frac{1}{a+c}+\frac{1}{b+c}=\frac{3}{a+b+c}\)
Question 23.
(i) In a ∆ABC the angles A, B, C are in A.P. show that 2 cos \(\frac{\mathbf{A}-\mathbf{C}}{2}=\frac{a+c}{\sqrt{\left(a^2-a c+c^2\right)}}\)
(ii) If the angles A, B, C of ∆ABC are in A.P. and b : c = \(\sqrt{3}\) : \(\sqrt{2}\) show that A = 75°.
(iii) If A = 45° and B = 75°, show that a + c\(\sqrt{2}\) = 2b.
Solution:
(i) Since angles A, B, C of ∆ABC are in A.P.
∴ 2B = A + C
Since sum of all angles of ∆ABC is 180°.
(ii) since A, B, C are the angles of AABC forms A.P.
∴ 2B = A + C … (1)
Also sum of all angles of ∆ABC be 180°.
∴ A + B + C = 180°
⇒ 2B + B = 180°
⇒ B = 60°
given b : c = \(\sqrt{3}\) : \(\sqrt{2}\)
⇒ \(\frac{b}{c}=\frac{\sqrt{3}}{\sqrt{2}}\)
⇒ \(\frac{\sin B}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}\) [using sine formulae]
⇒ \(\frac{\sin 60^{\circ}}{\sin C}=\frac{\sqrt{3}}{\sqrt{2}}\)
⇒ \(\frac{\sqrt{3}}{2 \sin C}=\frac{\sqrt{3}}{\sqrt{2}}\)
⇒ sin C = \(\frac{1}{\sqrt{2}}\)
⇒ C = 45°
∴ A = 2B – C = 2 x 60° – 45° = 120° – 45° = 75
(iii) Since sum of all angles of ∆ABC is 180°.
∴ A + B + C= 180° ⇒ 45° + 75° + C = 180°
⇒ C = 180° – 120° = 60°
L.H.S = a + c\(\sqrt{2}\) = k sin A + \(\sqrt{2}\)k sin C [using sine formulae, \(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k]
= k sin 45° + \(\sqrt{2}\)k sin 60° = k\(\left[\frac{1}{\sqrt{2}}+\sqrt{2} \times \frac{\sqrt{3}}{2}\right]=k\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)\)
R.H.S = 2b = 2k sin B = 2 k sin 75° = 2k sin (45° + 30°)
= 2k [sin 45° cos 30° + cos 45° sin 30°]
= 2k \(\left[\frac{1}{\sqrt{2}}+\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \cdot \frac{1}{2}\right]=\frac{k(\sqrt{3}+1)}{\sqrt{2}}\)
∴ L.H.S = R.H.S
Question 24.
The angles A, B, C of a triangle are in the ratio 3:5:4, prove that a + c\(\sqrt{2}\) = 2b.
Solution:
Since the angles A, B and C of a triangle are in the ratio 3 : 5 : 4.
Let the angles of ∆ABC are, A = 3k; B = 5k; C = 4k
Since the sum of angles of ∆ABC be 180°.
∴ A + B + C = 180° ⇒ 3k+ 5k + 4k = 180° ⇒ 12k = 180° ⇒ k = 15°
∴ A = 3k = 45°; B = 5k = 75° and C = 4k = 60°
∴ LH.S = a + c\(\sqrt{2}\) = k[sin A + \(\sqrt{2}\) sinC] = k [sin 45° + \(\sqrt{2}\) sin 60°]
= k \(\left[\frac{1}{\sqrt{2}}+\sqrt{2} \times \frac{\sqrt{3}}{2}\right]=k\left(\frac{1+\sqrt{3}}{\sqrt{2}}\right)\)
R.H.S = 2b = 2k sin B = 2k sin 75° = 2k sin (45° + 30°)
= 2k [sin 45° cos 30° + cos 45° sin 30°]
= 2k \(\left[\frac{1}{\sqrt{2}} \times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}} \times \frac{1}{2}\right]=\frac{k(1+\sqrt{3})}{\sqrt{2}}\)
∴ L.H.S = R.H.S
Question 25.
In a ∆ABC, if cos A = \(\frac { 17 }{ 22 }\), cos C = \(\frac { 1 }{ 14 }\), prove that the ratio of the sides is 7 : 9 :11.
Solution:
Question 26.
The angle of a triangle are in the ratio 1:2:7, prove that the ratio of the greatest side to the least side is (75 +1): (75 -1).
Solution:
Since the angles of triangle are given to be in the ratio 1 : 2 : 7.
Let A = x; B = 2x and C = 7x
since the sum of the angles of AABC is 180°.
A + B + C = 180° ⇒ x + 2x + 7x = 180° ⇒ 10x = 180° ⇒ x = 18°
∴ A = 18° ; B = 2 x 18° = 36° and C = 7 x 18°= 126°
Thus greatest side : least side = sin 126° : sin 18°
= sin (180° – 54°): sin 18° = sin 54° : sin 18°
= sin (90° – 36°): sin 18° – cos 36°: sin 18°
= \(\frac{\sqrt{5}+1}{4}: \frac{\sqrt{5}-1}{4} \text { i.e. } \sqrt{5}+1: \sqrt{5}-1\)
Question 27.
If the sides of ∆ABC are in the ratio 4 : 5 : 6, prove that one angle it twice that of the other.
Solution:
Since the sides of AABC are in the ratio 4 : 5 : 6
Let BC = 4k; AC = 5k; AB = 6k
Using cosine formula, we have
Hence one angle of ∆ABC is twice the other.
Question 28.
Two sides and included angles of a triangle are respectively 3 + \(\sqrt{3}\), 3 – \(\sqrt{3}\) and 60°. Show that the remaining elements of the triangle are 105°, 15°, 3\(\sqrt{2}\).
Solution:
Let a = BC ; b = CA = 3 – \(\sqrt{3}\); c = AB = 3 + \(\sqrt{3}\) and ∠A = 60°
Question 29.
In a ∆ABC, if B = 3C, prove that
(i) cos C = \(\sqrt{\left(\frac{b+c}{4 c}\right)}\)
(ii) sin \(\frac { A }{ 2 }\) = \(\frac{b-c}{2 c}\).
Solution:
(i) Given B = 3C
Using sine formula,
\(\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}\) = k ≠ 0 (say)
⇒ a = k sin A; b = k sin B and c = k sin C
since the sum of angles of ∆ABC be 180°.
∴ A + B + C = 180°
⇒ A + 3C + C = 180°
⇒ A + 4C = 180°
⇒ 4C = 180° – A
⇒ 2C = \(\frac{180^{\circ}-\mathrm{A}}{2}\) = 90° – \(\frac { A }{ 2 }\)
∴ from (1) ; we have
\(\frac{b-c}{2 c}=\cos \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)=\sin \frac{\mathrm{A}}{2}\)