Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.2
Solution & Step-by-Step Answer:
Here, r = and θ = Let the cartesian coordinates be (x, y) Then, x = rcosθ = cos = = 1 y = rsinθ = sin = = 1 ∴ the cartesian coordinates of the given point are (1, 1).
(ii)
Solution:
(iii)
Solution:
Here, r = and θ =
Let the cartesian coordinates be (x, y)
(iv)
Solution:
Here, r = and θ =
Let the cartesian coordinates be (x, y)
∴ the cartesian coordinates of the given point are
Solution & Step-by-Step Answer:
Here x = and y = ∴ the point lies in the first quadrant. Let the polar coordinates be (r, θ) Then, r2 = x2 + y2 = ( )2 + ( )2 = 2 + 2 = 4 ∴ r = 2 … [∵ r > 0] cos θ = and sin θ = ∴ tan θ = 1 Since the point lies in the first quadrant and 0 ≤ θ ≤ 2π, tan θ = 1 = tan ∴ θ = ∴ the polar coordinates of the given point are .
(ii)
Solution:
Here x = 0 and y =
the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)
Then, r2= x2+ y2= (0)2+
∴ r = …[∵ r > 0]
cosθ = = 0
and sin θ = = 1
Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π
cosθ = 0 = cos and sinθ = 1 = sin
∴ θ =
∴ the polar coordinates of the given point are .
(iii)
Solution:
Here x = 1 and y =
∴ the point lies in the fourth quadrant.
Let the polar coordinates be (r, θ).
Then, r2= x2+ y2= (1)2+ ( )2= 1 + 3 = 4
∴ r = 2 … [∵ r > 0]
∴ the polar coordinates of the given point are .
(iv)
Solution:
Solution & Step-by-Step Answer:
By the sine rule, = = ∴ and ∴ a : b : c = sinA : sinB : sinC Given ∠A = 45° and ∠B = 60° ∵ ∠A + ∠B + ∠C = 180° ∴ 45° + 60° + ∠C = 180° ∴ ∠C = 180° – 105° = 75°
Solution & Step-by-Step Answer:
By the sine rule,
Solution & Step-by-Step Answer:
Solution & Step-by-Step Answer:
By the sine rule, = = = k ∴ 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) = [a2(a2 + b2 – c2) – a2(a2 + c2 – b2) + b2(b2 + c2 – a2) – b2(a2 + b2 – c2) + c2(c2 + a2 – b2) – c2(b2 + c2 – a2)] = [a4 + a2b2 – a2c2 – a4 – a2c2 + a2b2 + b4 + b2c2 – a2b2 – a2b2 – b4 + b2c2 + c4 + a2c2 – b2c2 – b2c2 – c4 + a2c2] = (0) = 0 = RHS.
Solution & Step-by-Step Answer:
By the sine rule, = = = k ∴ 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
Solution & Step-by-Step Answer:
By the sine rule, = = k 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.
Solution & Step-by-Step Answer:
LHS = 2 (bc cos A + ac cos B + ab cos C) = 2bc cos A + 2ac cos B + 2ab cos C = 2bc + 2ac + 2ab …(By cosine rule] = b2 + c2 – a2 + c2 + a2 – b2 + a2 + b2 – c2 = a2 + b2 + c2 = RHS.
Solution & Step-by-Step Answer:
Given : a = 18, b = 24 and c = 30 ∴ 2s = a + b + c = 18 + 24 + 30 = 72 ∴ s = 36
(ii) sin
Solution:
(iii) cos
Solution:
(iv) tan
Solution:
(v) A(△ABC)
Solution:
(iv) sin A.
Solution:
Solution & Step-by-Step Answer:
(b + c – a) tan
Solution & Step-by-Step Answer:
LHS = sin ∙sin ∙sin
