Chapter 3 Trigonometric Functions Ex 3.2
Chapter 3 Trigonometric Functions Ex 3.2
Question 1.
Find the Cartesian co-ordinates of the point whose polar co-ordinates are :
Let the cartesian coordinates be (x, y)
Let the cartesian coordinates be (x, y)
Question 2.
Find the of the polar co-ordinates point whose Cartesian co-ordinates are.
![](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-19010-9.png)
![](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-19010-10.png)
![](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-19010-11.jpeg)
![](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-19010-12.png)
![](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/word-image-19010-13.png)
![Maharashtra Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.2 5](https://mhboardsolutions.xyz/wp-content/uploads/2022/02/maharashtra-board-12th-maths-solutions-chapter-3-t-91.png)
Question 4.
Solution:
By the sine rule,
Question 5.
Solution:
Question 6.
In ∆ABC, prove that a3sin(B – C) + b3sin(C – A) + c3sin(A – B) = 0
Solution:
By the sine rule,
∴ a = k sin A, b = k sin B, c = k sin C
LHS = a3sin (B – C) + b3sin (C – A) + c3sin (A – B)
= a3(sin B cos C – cos B sin C) + b3(sinCcos A – cos C sin A) + c3(sinAcosB – cos A sin B)
= 1/2k [a2(a2 + b2 – c2) – a2(a2 + c2 – b2) + b2(b2 + c2 – a2) – b2(a2 + b2 – c2) + c2(c2 + a2 – b2) – c2(b2 + c2 – a2)]
= 1/2k [a4 + a2b2 – a2c2 – a4 – a2c2 + a2b2 + b4 + b2c2 – a2b2 – a2b2 – b4 + b2c2 + c4 + a2c2 – b2c2 – b2c2 – c4 + a2c2]
= 1/2k(0) = 0 = RHS.
Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 are also in A.P
Solution:
By the sine rule,
∴ sin A = ka, sin B = kb, sin C = kc …(1)
Now, cot A, cotB, cotC are in A.P.
∴ cotC – cotB = cotB – cot A
∴ cotA + cotC = 2cotB
Question 8.
In ∆ABC, if a cos A = b cos B then prove that the triangle is right angled or an isosceles traingle.
Solution:
By the sine rule,
a = k sin A and b = k sin B
∴ a cos A = b cos B gives
k sin A cos A = k sin B cos B
∴ 2 sin A cos A = 2 sin B cos B
∴ sin 2A = sin 2B ∴ sin 2A – sin 2B = 0
∴ 2 cos (A + B)∙sin (A -B) = 0
∴ 2cos (π – C)∙sin(A – B) = 0 … [∵ A + B + C = π]
∴ -2 cos C∙sin (A – B) = 0
∴ cos C = 0 OR sin(A -B) = 0
∴ C = 90° OR A – B = 0
∴ C = 90° OR A = B
∴ the triangle is either rightangled or an isosceles triangle.
Question 9.
With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a2 + b2 + c2.
Solution:
LHS = 2 (bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
= b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2 = a2 + b2 + c2 = RHS.
Question 10.
In △ABC, if a = 18, b = 24, c = 30 then find the values of
(i) cos A
Solution:
Given : a = 18, b = 24 and c = 30
∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36
Solution:
Solution:
Solution:
(v) A(△ABC)
Solution:
(iv) sin A.
Solution:
Question 11.
Question 12.