You can Download Samacheer Kalvi 11th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.

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

Question 1.

Solution:

Question 2.

The angles of a triangle ABC, are in Arithmetic Progression and if b : c = \(\sqrt{3}: \sqrt{2}\), find ∠A.

Solution:

Question 3.

Solution:

⇒ a^{2} + b^{2} – c^{2} = a^{2} ⇒ b^{2} – c^{2} = a^{2} – a^{2
}⇒ b^{2} – c^{2} = 0 ⇒ b = c

∴ ∆ ABC is isosceles

Question 4.

Solution:

Question 5.

In an ∆ABC, prove that a cos A + b cos B + c cos C = 2a sin B sin C.

Solution:

We have A + B + C = 180°

2A + 2B + 2C = 360°

a cos A + b cos B + c cos C = 2R sin A cos A + 2R sin B cos B + 2R sin C cos C

= R sin 2A + R sin 2B + R sin 2C

= R (sin 2A + sin 2B + sin 2C)

= R[sin (360° – (2B + 2C)) + sin 2B + sin 2C]

= R [- sin(2B + 2C) + sin 2B + sin 2C]

= R [- (sin 2B cos 2C + cos 2B sin 2C) + sin 2B + sin 2C]

= R [- sin 2B cos 2C – cos 2B sin 2C + sin 2B + sin 2C]

= R[sin 2B(1 – cos 2C) + sin 2C(1 – cos 2B)]

= R [sin 2B 2 sin^{2}C + sin 2C . 2 sin^{2}B]

= 2R [2 sin B cos B sin 2C + 2 sin C cos C sin ^{2}B]

= 2R. 2 sin B sin C [sin C cos B + cos C sin B]

= 4R sinB sinC [sin (B + C)]

= 4R sinB sinC [sin (180° – A)]

= 4R sin B sin C sin A

= 2 (2R sin A) sin B sin C

= 2a sin B sin C

Question 6.

Solution:

Question 7.

In an ∆ABC, prove the following.

Solution:

Question 8.

In a ∆ABC, prove that (a^{2} – b^{2} + c^{2}) tan B = (a^{2} + b^{2} – c^{2})tan C

Solution:

Question 9.

An Engineer has to develop a triangular-shaped park with a perimeter of 120 m in a village. The park to be developed must be of maximum area. Find out the dimensions of the park.

Solution:

Given, the perimeter of a triangular-shaped park = 120 m

All sides of a triangular part would be 40 m.

i.e., a = 40 m,

b = 40 m,

c = 40 m.

Question 10.

A rope of length 12 m is given. Find the largest area of the triangle formed by this rope and find the dimensions of the triangle so formed.

Solution:

The largest triangle will be an equilateral triangle

Question 11.

Derive Projection formula from

(i) Law of sines,

(ii) Law of cosines.

Solution:

(i) To Prove a = b cos c + c cos B

Using sine formula

RHS = b cos C + c cos B

= 2R sin B cos C + 2R sin C cos B

= 2R [sin B cos C + cos B sin C]

= 2R sin (B + C) = 2R [sin π – A)

= 2R sin A = a = LHS

(ii) To prove a = b cos c + c cos B

Using cosine formula

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

Question 1.

Solution:

Question 2.

Solution:

Question 3.

Solution:

Question 4.

Solution:

Question 5.

Solution: