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
$$\frac{b-c}{2 c}=\cos \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)=\sin \frac{\mathrm{A}}{2}$$