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

Solution:

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.

Solution:

Question 7.

In a ∆ 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 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 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: