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ML Aggarwal Class 11 Maths Solutions Section A Chapter 3 Trigonometry Ex 3.9

Very short answer/objective questions (1 to 4) :

Question 1.
In a ∆ABC, if a = 4, b = 5 and c = 6, find cos C.
Solution:
Given a = 4, b = 5 and c = 6 using cosine formula, we have
cos C = $$\frac{a^2+b^2-c^2}{2 a b}$$
= $$\frac{4^2+5^2-6^2}{2 \times 4 \times 5}$$
= $$\frac{16+25-36}{40}$$
= $$\frac{5}{40}=\frac{1}{8}$$

Question 2.
In a ∆ABC, if a = 7, c = 8 and ∠B = 45°, find area of ∆ABC.
Solution:
Given a = 7 ; c = 8 ; ∠B = 45°
∴ area of ∆ABC = $$\frac{1}{2}$$ × a × c
= $$\frac{1}{2}$$ × 7 × 8 × sin 45°
= $$\frac{28}{\sqrt{2}}$$
= 14√2 sq. units

Question 3.
In a ∆ABC, if a = 4, b = 6 and c = 8, find the value of 4 cos B + 3 cos C.
Solution:
Given a = 4, b = 6, c = 8
∴ cos B = $$\frac{a^2+c^2-b^2}{2 a c}$$
= $$\frac{4^2+8^2-6^2}{2 \times 4 \times 8}$$
= $$\frac{80-36}{64}$$
= $$\frac{44}{64}=\frac{11}{16}$$
and cos C = $$\frac{a^2+b^2-c^2}{2 a b}$$
= $$\frac{4^2+6^2-8^2}{2 \times 4 \times 6}$$
= $$-\frac{12}{48}=-\frac{1}{4}$$
Thus 4 cos B + 3 cos C = $$4 \times \frac{11}{16}+3\left(-\frac{1}{4}\right)$$
= $$\frac{11-3}{4}$$
= 2

Question 4.
In a ∆ABC, if a = 3, b = 5 and c = 6, then verify that c = a cos B + b cos A.
Solution:
Given a = 3, b = 5 and c = 6
R.H.S = a cos B + b cos A
= $$a\left(\frac{a^2+c^2-b^2}{2 a c}\right)+b\left(\frac{b^2+c^2-a^2}{2 b c}\right)$$
= $$\frac{1}{2 c}$$ [a2 + c2 – b2 + b2 + c2 – a2]
= $$\frac{2 c^2}{2 c}$$
= c = L.H.S

Short and long answer questions (5 to 25) :

Question 5.
2(b ccos A + ca cos B + ab cos C) = a2 + b2 + c2.
Solution:
LH.S. = 2 (bc cos A + ca cos B + ah cos C)
= 2 $$\left[b c\left(\frac{b^2+c^2-a^2}{2 b c}\right)+c a\left(\frac{c^2+a^2-b^2}{2 c a}\right)+a b\left(\frac{a^2+b^2-c^2}{2 a b}\right)\right]$$
[using cosine’s formula]
= $$\frac{2}{2}$$ [b2 + c2 – a2 – c2 + a2 – b2 + a2 + b2 – c2]
= a2 + b2 + c2
= R.H.S.

Question 6.
$$\frac{\cos \mathrm{A}}{a}+\frac{\cos \mathrm{B}}{b}+\frac{\cos \mathrm{C}}{c}=\frac{a^2+b^2+c^2}{2 a b c}$$
Solution:
L.H.S.

Question 7.
(a + b) sin $$\frac{C}{2}$$ = c cos $$\frac{A-B}{2}$$
Solution:
In any ∆ABC, using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C

Question 8.
(a – b) cos $$\frac{C}{2}$$ = c sin $$\frac{A-B}{2}$$
Solution:
In any ∆ABC, using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C

Question 9.
(c – b)2 cos2 $$\frac{A}{2}$$ + (c + b)2 sin2 $$\frac{A}{2}$$ = a2.
Solution:
L.H.S. = (c – b)2 cos2 $$\frac{A}{2}$$ + (c + b)2 sin2 $$\frac{A}{2}$$
= (b2 + c2) $$\left(\cos ^2 \frac{\mathrm{A}}{2}+\sin ^2 \frac{\mathrm{A}}{2}\right)$$ – 2 bc $$\left(\cos ^2 \frac{\mathrm{A}}{2}-\sin ^2 \frac{\mathrm{A}}{2}\right)$$
= (b2 + c2) × 1 – 2bc cos A
= (b2 + c2) – 2bc $$\left(\frac{b^2+c^2-a^2}{2 b c}\right)$$
= b2 + c2 – b2 – c2 + a2
= a2
= R.H.S.

Question 10.
$$\frac{a^2 \sin (B-C)}{\sin B+\sin C}+\frac{b^2 \sin (C-A)}{\sin C+\sin A}+\frac{c^2 \sin (A-B)}{\sin A+\sin B}$$ = 0
Solution:
L.H.S. = $$\frac{a^2 \sin (B-C)}{\sin B+\sin C}+\frac{b^2 \sin (C-A)}{\sin C+\sin A}+\frac{c^2 \sin (A-B)}{\sin A+\sin B}$$

= k2 sin A (sin B – sin C) + k2 sin B (sin C – sin A) + k2 sin C (sin A – sin B)
= k2 [sin A sin B – sin A sin C + sin B sin C – sin A sin B + sin C sin A – sin B sin C]
= k2 × 0
= 0 = R.H.S.

Question 11.
$$\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}$$ = 0.
Solution:
L.H.S. = $$\frac{a^2-b^2}{\cos A+\cos B}+\frac{b^2-c^2}{\cos B+\cos C}+\frac{c^2-a^2}{\cos C+\cos A}$$
In any ∆ ABC, bby sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k (say)
⇒ a = k sin A ;
b = k sin B
and c = k sin C
∴ L.H.S. = $$\frac{k^2\left[\sin ^2 \mathrm{~A}-\sin ^2 \mathrm{~B}\right]}{\cos \mathrm{A}+\cos \mathrm{B}}+\frac{k^2\left[\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}\right]}{\cos \mathrm{B}+\cos \mathrm{C}}+\frac{k^2\left[\sin ^2 \mathrm{C}-\sin ^2 \mathrm{~A}\right]}{\cos \mathrm{C}+\cos \mathrm{A}}$$
= $$\frac{k^2\left[\cos ^2 \mathrm{~B}-\cos ^2 \mathrm{~A}\right]}{\cos \mathrm{B}+\cos \mathrm{A}}+\frac{k^2\left[\cos ^2 \mathrm{C}-\cos ^2 \mathrm{~B}\right]}{\cos \mathrm{C}+\cos \mathrm{B}}+\frac{k^2\left[\cos ^2 \mathrm{~A}-\cos ^2 \mathrm{C}\right]}{\cos \mathrm{A}+\cos \mathrm{C}}$$
[∵ sin2 θ + cos2 θ = 1
sin2 θ = 1 – cos2 θ]
= k2 [cos B – cos A + cos C – cos B + cos A – cos C]
= k2 × 0
= 0 = R.H.S.

Question 12.
b2 sin 2C + c2 sin 2B = 2bc sin A
Solution:
L.H.S. = b2 sin 2C + c2 sin 2B
= 2b2 sin C cos C + 2c2 sin B cos B
By sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
∴ L.H.S. = 2b2 × $$\frac{c}{k}\left(\frac{b^2+a^2-c^2}{2 a b}\right)$$ + 2c2 × $$\frac{b}{k} \frac{\left(c^2+a^2-b^2\right)}{2 a c}$$
= $$\frac{b c}{a k}$$ (b2 + a2 – c2) + $$\frac{b c}{a k}$$ (c2 + a2 – b2)
= $$\frac{b c}{a k}$$ [b2 + a2 – c2 + c2 + a2 – b2]
= $$\frac{b c}{a k}$$ × 2a2
= $$\frac{2 a b c}{k}$$
= 2bc $$\left(\frac{a}{k}\right)$$
= 2bc sin A [using sine formula]
= R.H.S.

Question 13.
$$\frac{b^2-c^2}{a} \cos \mathrm{A}+\frac{c^2-a^2}{b} \cos \mathrm{B}+\frac{a^2-b^2}{c} \cos \mathrm{C}$$ = 0
Solution:
L.H.S. = $$\frac{b^2-c^2}{a} \cos \mathrm{A}+\frac{c^2-a^2}{b} \cos \mathrm{B}+\frac{a^2-b^2}{c} \cos \mathrm{C}$$
= $$\frac{b^2-c^2}{a}\left(\frac{b^2+c^2-a^2}{2 b c}\right)+\frac{c^2-a^2}{b}\left(\frac{c^2+a^2-b^2}{2 a c}\right)+\frac{a^2-b^2}{c}\left(\frac{a^2+b^2-c^2}{2 a b}\right)$$
[using cosine’s formula]
= $$\frac{1}{2 a b c}$$ [(b2 – c2) (b2 + c2 – a2) + (c2 – a2) (c2 + a2 – b2) + (a2 – b2) (a2 + b2 – c2)]
= $$\frac{1}{2 a b c}$$ [b4 – c4 – a2 (b2 – c2) + c4 – a4 – b2 (c2 – a2) + a4 – b4 – c2 (a2 – b2)]
= $$\frac{1}{2 a b c}$$ [- a2b2 + a2c2 – b2c2 + a2b2 – c2a2 + b2c2]
= $$\frac{1}{2 a b c}$$ × 0
= 0 = R.H.S.

Question 14.
b2 cos 2A – a2 cos 2B = b2 – a2.
Solution:
Using sine formula
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
L.H.S. = b2 cos 2A – a2 cos 2B
= b2 (1 – 2 sin2 A) – a2 (1 – 2 sin2 B)
= b2 – a2 – 2b2 $$\left(\frac{a}{k}\right)^2$$ + 2a2 $$\left(\frac{b}{k}\right)^2$$
= b2 – a2 – $$\frac{2 b^2 a^2}{k^2}+\frac{2 a^2 b^2}{k^2}$$
= b2 – a2
= R.H.S.

Question 15.
(c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C.
Solution:
Using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
⇒ a = k sin A ;
b = k sin B
and c = k sin C
Now (c2 – a2 + b2) tan A

Thus,
(c2 – a2 + b2) tan A = (a2 – b2 + c2) tan B = (b2 – c2 + a2) tan C
[using (1), (2) and (3)].

Question 16.
2 [a sin2 $$\frac{C}{2}$$ + c sin2 $$\frac{A}{2}$$] = c + a – b
Solution:
LH.S. = $$\left[a \sin ^2 \frac{\mathrm{C}}{2}+c \sin ^2 \frac{\mathrm{A}}{2}\right]$$
= 2 $$\left[a\left(\frac{1-\cos \mathrm{C}}{2}\right)+c\left(\frac{1-\cos \mathrm{A}}{2}\right)\right]$$
= a + c – a cos C – c cos A
= a + c – b [using projection formula)
= R.H.S.

Question 17.
If in a triangle ABC, sin 2A + sin 2B = sin 2C, prove that either A = 90° or B = 90°.
Solution:
Given sin 2A + sin 2B = sin 2C
⇒ 2 sin (A + B) cos (A – B) = 2 sin C cos C
⇒ sin [π – C] cos (A – B) = sin C cos C
[∵ A + B + C = π
⇒ A + B = π – C]
⇒ sin C [cos (A – B) – cos C] = 0
⇒ sin C [cos (A – B) – cos (π – $$\overline{A+B}$$)] = 0
⇒ sin C [cos(A – B) cos (A + B)] = 0
⇒ sin C [2 cos A cos B] = 0
either sin C = 0 or cos A = 0 or cos B = 0
C = 0 or A = 90° or B = 90°
since A, B and C are the angles of ∆ABC
∴ neither of the angles A, B and C is 0.
Hence either A = 90° or B = 90°.

Question 18.
If in a triangle ABC, sin2 A + sin2 B = sin2 C. prove that ¿ABC is right angled.
Solution:
Given sin2 A + sin2 B = sin2 C ………………..(1)
using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k
⇒ a = k sin A ;
b = k sin B
and c = k sin C
Thus from (1) ; we have
$$\frac{a^2}{k^2}+\frac{b^2}{k^2}=\frac{c^2}{k^2}$$
⇒ a2 + b2 = c2
So the triangles ABC is right angled ∆ at C.
since pythagoras theorem holds.

Question 19.
If in a triangle ABC, $$\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}$$, show that the triangle is isosceles.
Solution:
Given, $$\frac{\cos \mathrm{A}}{a}=\frac{\cos \mathrm{B}}{b}$$
⇒ b cos A = a cos B ………………..(1)
In any ABC, using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$ = k
∴ from (1) ;
k sin B cos A = k sin A cos B
⇒ 2 sin B cosA = 2 sin A cos B
⇒ sin (A + B) – sin (A – B) = sin (A + B) + sin (A – B)
⇒ 2 sin (A – B) = 0
⇒ A – B = 0
⇒ A = B
Hence the given triangle is isosceless.

Question 20.
In triangle ABC, if a = 18, b = 24 and c = 30 then find
(i) cos A, cos B, cos C
(ii) sin A sin B, sin C.
Solution:
(i) Given a = 18, b = 24 ; c = 30
Using cosine formula, we have

(ii) Using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$
⇒ $$\frac{18}{\sin A}=\frac{24}{\sin B}=\frac{30}{\sin \frac{\pi}{2}}$$
⇒ sin A = $$=\frac{18}{30}=\frac{3}{5}$$ ;
sin B = $$\frac{24}{30}=\frac{4}{5}$$
and sin C = 1

Question 21.
If a = 7 cm, b = 5 cm and c = 3 cm, prove that the triangle ha 14ij angle.
Solution:
Given a = 7 cm ;
b = 5 cm
and c = 3 cm
∴ cos A = $$\frac{b^2+c^2-a^2}{2 b c}$$
= $$\frac{5^2+3^2-7^2}{2 \times 5 \times 3}$$
= $$\frac{34-49}{30}$$
= $$-\frac{15}{30}=-\frac{1}{2}$$
cos A = – cos $$\frac{\pi}{3}$$
= cos (π – $$\frac{\pi}{3}$$)
= cos $$\frac{2 \pi}{3}$$
⇒ A = $$\frac{2 \pi}{3}$$
Clearly one of the angle of given ∆ be 120° > 90°.
Hence one of the angle ofi be obtuse angle.

Question 22.
If the angles of a triangle are in the ratio 1 : 2 : 3, prove that the corresponding sides of the triangle are in ratio 1 : √3 : 2.
Solution:
Let the angles of triangle be x°, 2x° and 3x°
since, x° + 2x° + 3x° = 180°
⇒ 6x° = 180°
⇒ x = 330°
Hence the angles of a triangle are 30°, 60° and 90°.
Using sine formula, we have
$$\frac{a}{\sin \mathrm{A}}=\frac{b}{\sin \mathrm{B}}=\frac{c}{\sin \mathrm{C}}$$
⇒ $$\frac{a}{\sin 30^{\circ}}=\frac{b}{\sin 60^{\circ}}=\frac{c}{\sin 90^{\circ}}$$
⇒ $$\frac{a}{\frac{1}{2}}=\frac{b}{\frac{\sqrt{3}}{2}}=\frac{c}{1}$$
⇒ $$\frac{a}{1}=\frac{\dot{b}}{\sqrt{3}}=\frac{c}{2}$$
Hence a : b : c :: 1 : √3 : 2
Thus, the corresponding sides of the triangle are in the ratio 1 : √3 : 2.

Question 23.
In a ABC, a = 1, b = √3 and C = $$\frac{\pi}{6}$$. Find the other two angles and the third side.
Solution:
Given a = 1 ;
b = √3
and C = $$\frac{\pi}{6}$$ or 30°
In only ∆ABC, using sine formula, we have

⇒ 3 = 4 – c2
⇒ c2 = 1
⇒ c = 1 [∵ c > 0]
∴ from (1) ;
$$\frac{1}{\sin A}=\frac{\sqrt{3}}{\sin B}=\frac{1}{\frac{1}{2}}$$
⇒ sin A = $$\frac{1}{2}$$
⇒ A = $$\frac{\pi}{6}$$
Since A. B and C are the angles of ∆ABC.
∴ A + B + C = 180°
⇒ B = 180° – 30° – 30° = 120°

Question 24.
Two boats leave a place at the same time. One travels 56 km in the direction N 40° E, while the other travels 48 km in the direction S 80° E. What is the distance between the boats?
Solution:
Clearly ∠POQ = 180° – 40° – 80° = 60°
using cosine’s formula, we have
cos 60° = $$\frac{\mathrm{OP}^2+\mathrm{OQ}^2-\mathrm{PQ}^2}{2 \mathrm{OP} \cdot \mathrm{OQ}}$$
⇒ $$\frac{1}{2}=\frac{56^2+48^2-\mathrm{PQ}^2}{2 \times 56 \times 48}$$

⇒ 56 × 48 = 562 + 482 – PQ2
⇒ PQ2 = 3136 + 2304 – 2688 = 2752
⇒ PQ = $$\sqrt{2752}$$ cm = 52.5 km

Question 25.
Two trees A and B are on the same side of the river. From a point C in the river the distance of the trees A and B is 250 m and 300 m respectively. 1f the angle C is 45, find the distance between the trees. Use √2 = 1.44.
Solution:
Using cosine formula, we have
cos C = $$\frac{a^2+b^2-c^2}{2 a b}$$
= $$\frac{\mathrm{CA}^2+\mathrm{CB}^2-\mathrm{AB}^2}{2 \mathrm{CA} \cdot \mathrm{CB}}$$
⇒ cos 45° = $$\frac{(250)^2+(300)^2-\mathrm{AB}^2}{2 \times 250 \times 300}$$

⇒ $$\frac{1}{\sqrt{2}}$$ × 2 × 250 × 300 = 62500 + 90000 – AB2
⇒ AB2 = 152500 – 75000 √2
⇒ AB2 = 152500 – 106050 = 46450
[∵ √2 = 1.414]
⇒ AB = $$\sqrt{46450}$$ = 215.5 m
Hence, the required distance between trees be 215.5 metre.