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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:
⇒ a2 + b2 – c2 = a2 ⇒ b2 – c2 = a2 – a2
⇒ b2 – c2 = 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 sin2C + sin 2C . 2 sin2B]
= 2R [2 sin B cos B sin 2C + 2 sin C cos C sin 2B]
= 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 (a2 – b2 + c2) tan B = (a2 + b2 – c2)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: