Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 3 Trigonometry Ex 3.9 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 3 Trigonometry Ex 3.9

Question 1.
In a ∆ ABC, if $$\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$$ prove that a2, b2, C2 are in Arithmetic progression.
$$\frac{\sin A}{\sin C}=\frac{\sin (A-B)}{\sin (B-C)}$$
sin A . sin (B – C) = sin C . sin (A – B)
sin (180° – (B + C)) . sin (B – C) = sin (180° – (A + B)) . sin (A – B)
sin (B + C) sin (B – C) = sin (A + B) sin (A – B) ——— (1)
sin(B + C) . sin(B – C) = (sin B cos C + cos B sin C) × (sin B cos C – cos B sin C)
= (sin B cos C)2 – (cos B sin C)2
= sin2 B cos2 C – cos2 B sin2 C
= sin2 B (1 – sin2 C) – (1 – sin2 B) sin2 C
= sin2 B – sin2 B sin2 C – sin2 C + sin2 B sin2 C
sin ( B + C) . sin ( B – C) = sin2 B – sin2 C
Similarly,
sin (A + B ) . sin (A – B) = sin2 A – sin2 B
(1) ⇒ sin2 B – sin2 C = sin2 A – sin2 B
sin2 B + sin2 B = sin2 A + sin2 C
2 sin2 B = sin2 A + sin2 C ——— (2)

Question 2.
The angles of a triangle A B C, are in Arithmetic Progression and if b : c = √3 : √2 find ∠A.
Given that the angles A, B, C are in A. P.
∴ 2B = A + C
Also A + B + C = 180°
B + (A + C) = 180°
B + 2B = 180°
3B = 180° ⇒ B = 60°
A + C = 2B = 2 × 60° = 120°

A + 45° = 120°
A = 120° – 45° = 75°
A = 75°

Question 3.
In a ∆ ABC, if cos c = $$\frac{\sin \mathbf{A}}{2 \sin B}$$ show that the triangle is isosceles.

a2 + b2 – c2 = a2
b2 – c2 = 0
b2 = c2 ⇒ b = c
Two sides of is ∆ ABC are equal.
∴ ∆ ABC is an isosceles triangle.

Question 4.
In a ∆ ABC, prove that

Question 5.
In an ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.
LHS = a cos A+ 6 cos B + c cos C
Using sine formula, we get k sin A cos A + k sin B cos B + k sin C cos C k
= $$\frac{k}{2}$$ [2 sin A cos A + 2 sin B cos B + 2 sin C cos C]
= $$\frac{k}{2}$$ [sin 2A + sin 2B + sin 2C]
= $$\frac{k}{2}$$ [2 sin (A + B) . cos (A – B) + 2 sin C . cos C]
= $$\frac{k}{2}$$ [2 sin (A – B) . cos (A – B) + 2 sin C . cos C]
= $$\frac{k}{2}$$ [2 sin C . cos (A – B) + 2 sin C . cos C]
= k sin C [cos(A – B) + cos C]

= k sin C [cos (A – B) – cos (A + B)]
= k sin C . 2 sin A sin B
= 2k sin A . sin B sin C
= 2a sin B sin C
= RHS

Question 6.
In a ∆ ABC, ∠A = 60°. Prove that b + c = 2a cos $$\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)$$
Given ∠A = 60°
A + B + C = 180°
60° + B + C = 180°
B + C = 180° – 60° = 120°

Question 7.
In an ∆ ABC, prove the following,
(i) a sin $$\left(\frac{\mathbf{A}}{2}+\mathbf{B}\right)$$ = (b + c) . sin $$\frac{\mathbf{A}}{2}$$

(ii) a (cos B + cos C) = 2(b + c) sin2$$\frac{\mathbf{A}}{2}$$

(iii)

(iv)

(v)

Question 8.
In a triangle ∆ ABC, prove that
(a2 – b2 + c2) tan B = (a2 + b2 – c2) tan C

Question 9.
An Engineer has to develop a triangular shaped park with a perimeter 120m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.
Let ∆ A B C be the triangular-shaped park.
a, b, c be the length of the sides.
Given perimeter of the park = 120 m
2s = a + b + c = 120m —– (1)
For a fixed perimeter 2s. the area of a triangle is maximum when a = b = c.
(1) = a + a + a = 120
3a = 120
⇒ a = 40m
Length of the sides 40 m, 40 m, 40 rn.

Question 10.
A rope of length 42m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.
Let a, b, c be the lengths of the sides of the triangle.
Given the perimeter of the triangle
2s = a + b + c = 42m —-—(1)
For a fixed perimeter 2 s, the area of a triangle is maximum
when a = b = c.
(1) ⇒ a + a + a = 42
3a = 42 ⇒ a = $$\frac{42}{3}$$
a = 14m
∴ a = b = c = 14m

∴ The dimensions of the triangle are 14 m, 14 m, 14 m.
Maximum area = 49√3 sq.m.

Question 11.
Derive Projection formula from
(i) Law of sines,
(ii) Law of cosines.
To prove (a) a = b cos C + c cos B
(b) b = c cos A + a cos C
(c) c = a cos B + b cos A

(i) Using the Law of sines,

(a) b cos C + c cos B = 2 R sin B cos C + 2 R sin C cos B
= 2 R ( sin B cos C + cos B sin C)
= 2R sin (B + C)
= 2R sin (180° – A)
b cos C + c cos B = 2R sin A = a
a = b cos C + c cos B

(b) c cos A + a cos C = 2R sin C cos A + 2R sin A cos C
= 2R (sin C cos A + cos C sin A)
= 2R sin(C + A)
= 2R sin(180° – B)
= 2R sin B = b
∴ b = c cos A + a cos C

(c) a cos B + b cosA = 2R sin A cos B + 2R sin B cos A
= 2R (sin A cos B + cos A sin B)
= 2R sin (A + B)
= 2R sin (180° – C)
= 2R sin C = c
∴ c = a cos B + b cos A

(ii) Using Law of cosines.