Balbharti 12th Maharashtra State Board Maths Solutions Book Pdf Chapter 3 Trigonometric Functions Ex 3.2 Questions and Answers.

## Maharashtra State Board 12th Maths Solutions Chapter 3 Trigonometric Functions Ex 3.2

Question 1.

Find the Cartesian co-ordinates of the point whose polar co-ordinates are :

(i) \(\left(\sqrt{2}, \frac{\pi}{4}\right)\)

Solution:

Here, r = \(\sqrt {2}\) and θ = \(\frac{\pi}{4}\)

Let the cartesian coordinates be (x, y)

Then, x = rcosθ = \(\sqrt {2}\)cos\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1

y = rsinθ = \(\sqrt {2}\)sin\(\frac{\pi}{4}\) = \(\sqrt{2}\left(\frac{1}{\sqrt{2}}\right)\) = 1

∴ the cartesian coordinates of the given point are (1, 1).

(ii) \(\left(4, \frac{\pi}{2}\right)\)

Solution:

(iii) \(\left(\frac{3}{4}, \frac{3 \pi}{4}\right)\)

Solution:

Here, r = \(\frac{3}{4}\) and θ = \(\frac{3 \pi}{4}\)

Let the cartesian coordinates be (x, y)

(iv) \(\left(\frac{1}{2}, \frac{7 \pi}{3}\right)\)

Solution:

Here, r = \(\frac{1}{2}\) and θ = \(\frac{7 \pi}{4}\)

Let the cartesian coordinates be (x, y)

∴ the cartesian coordinates of the given point are \(\left(\frac{1}{4}, \frac{\sqrt{3}}{4}\right)\)

Question 2.

Find the of the polar co-ordinates point whose Cartesian co-ordinates are.

(i) \((\sqrt{2}, \sqrt{2})\)

Solution:

Here x = \(\sqrt {2}\) and y = \(\sqrt {2}\)

∴ the point lies in the first quadrant.

Let the polar coordinates be (r, θ)

Then, r^{2} = x^{2} + y^{2} = (\(\sqrt {2}\) )^{2} + (\(\sqrt {2}\) )^{2} = 2 + 2 = 4

∴ r = 2 … [∵ r > 0]

cos θ = \(\frac{x}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)

and sin θ = \(\frac{y}{r}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\)

∴ tan θ = 1

Since the point lies in the first quadrant and

0 ≤ θ ≤ 2π, tan θ = 1 = tan\(\frac{\pi}{4}\)

∴ θ = \(\frac{\pi}{4}\)

∴ the polar coordinates of the given point are \(\left(2, \frac{\pi}{4}\right)\).

(ii) \(\left(0, \frac{1}{2}\right)\)

Solution:

Here x = 0 and y = \(\frac{1}{2}\)

the point lies on the positive side of Y-axis. Let the polar coordinates be (r, θ)

Then, r^{2} = x^{2} + y^{2} = (0)^{2} + \(\left(\frac{1}{2}\right)^{2}=0+\frac{1}{4}=\frac{1}{4}\)

∴ r = \(\frac{1}{2}\) …[∵ r > 0]

cosθ = \(\frac{x}{r}=\frac{0}{(1 / 2)}\) = 0

and sin θ = \(\frac{y}{r}=\frac{(1 / 2)}{(1 / 2)}\) = 1

Since, the point lies on the positive side of Y-axis and 0 ≤ θ ≤ 2π

cosθ = 0 = cos\(\frac{\pi}{2}\) and sinθ = 1 = sin\(\frac{\pi}{2}\)

∴ θ = \(\frac{\pi}{2}\)

∴ the polar coordinates of the given point are \(\left(\frac{1}{2}, \frac{\pi}{2}\right)\).

(iii) \((1,-\sqrt{3})\)

Solution:

Here x = 1 and y = \(-\sqrt{3}\)

∴ the point lies in the fourth quadrant.

Let the polar coordinates be (r, θ).

Then, r^{2} = x^{2} + y^{2} = (1)^{2} + (\(-\sqrt {3}\) )^{2} = 1 + 3 = 4

∴ r = 2 … [∵ r > 0]

∴ the polar coordinates of the given point are \(\left(2, \frac{5 \pi}{3}\right)\).

(iv) \(\left(\frac{3}{2}, \frac{3 \sqrt{3}}{2}\right)\)

Solution:

Question 3.

In ∆ABC, if ∠A = 45º, ∠B = 60º then find the ratio of its sides.

Solution:

By the sine rule,

\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = \(\frac{c}{\sin \mathrm{C}}\)

∴ \(\frac{a}{b}=\frac{\sin A}{\sin B}\) and \(\frac{b}{c}=\frac{\sin B}{\sin C}\)

∴ a : b : c = sinA : sinB : sinC

Given ∠A = 45° and ∠B = 60°

∵ ∠A + ∠B + ∠C = 180°

∴ 45° + 60° + ∠C = 180°

∴ ∠C = 180° – 105° = 75°

Question 4.

In ∆ABC, prove that sin \(\left(\frac{\mathbf{B}-\mathbf{C}}{2}\right)=\left(\frac{\boldsymbol{b}-\boldsymbol{c}}{a}\right)\) cos \(\frac{A}{2}\).

Solution:

By the sine rule,

Question 5.

With usual notations prove that 2 \(\left\{a \sin ^{2} \frac{C}{2}+c \sin ^{2} \frac{A}{2}\right\}\) = a – b + c.

Solution:

Question 6.

In ∆ABC, prove that a^{3}sin(B – C) + b^{3}sin(C – A) + c^{3}sin(A – B) = 0

Solution:

By the sine rule,

\(\frac{a}{\sin A}\) = \(\frac{b}{\sin B}\) = \(\frac{c}{\sin C}\) = k

∴ a = k sin A, b = k sin B, c = k sin C

LHS = a^{3}sin (B – C) + b^{3}sin (C – A) + c^{3}sin (A – B)

= a^{3}(sin B cos C – cos B sin C) + b^{3}(sinCcos A – cos C sin A) + c^{3}(sinAcosB – cos A sin B)

= \(\frac{1}{2 k}\) [a^{2}(a^{2} + b^{2} – c^{2}) – a^{2}(a^{2} + c^{2} – b^{2}) + b^{2}(b^{2} + c^{2} – a^{2}) – b^{2}(a^{2} + b^{2} – c^{2}) + c^{2}(c^{2} + a^{2} – b^{2}) – c^{2}(b^{2} + c^{2} – a^{2})]

= \(\frac{1}{2 k}\) [a^{4} + a^{2}b^{2} – a^{2}c^{2} – a^{4} – a^{2}c^{2} + a^{2}b^{2} + b^{4} + b^{2}c^{2} – a^{2}b^{2} – a^{2}b^{2} – b^{4} + b^{2}c^{2} + c^{4} + a^{2}c^{2} – b^{2}c^{2} – b^{2}c^{2} – c^{4} + a^{2}c^{2}]

= \(\frac{1}{2 k}\)(0) = 0 = RHS.

Question 7.

In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a^{2}, b^{2}, c^{2} are also in A.P

Solution:

By the sine rule,

\(\frac{\sin \mathrm{A}}{a}\) = \(\frac{\sin \mathrm{B}}{b}\) =\(\frac{\sin \mathrm{C}}{c}\) = 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

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,

\(\frac{a}{\sin \mathrm{A}}\) = \(\frac{b}{\sin \mathrm{B}}\) = 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.

Question 9.

With usual notations prove that 2(bc cos A + ac cos B + ab cos C) = a^{2} + b^{2} + c^{2}.

Solution:

LHS = 2 (bc cos A + ac cos B + ab cos C)

= 2bc cos A + 2ac cos B + 2ab cos C

= 2bc \(\left(\frac{b^{2}+c^{2}-a^{2}}{2 b c}\right)\) + 2ac\(\left(\frac{c^{2}+a^{2}-b^{2}}{2 c a}\right)\) + 2ab\(\left(\frac{a^{2}+b^{2}-c^{2}}{2 a b}\right)\) …(By cosine rule]

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

(ii) sin\(\frac{A}{2}\)

Solution:

(iii) cos\(\frac{A}{2}\)

Solution:

(iv) tan\(\frac{A}{2}\)

Solution:

(v) A(△ABC)

Solution:

(iv) sin A.

Solution:

Question 11.

In △ABC prove that (b + c – a) tan \(\frac{A}{2}\) = (c + a – b) tan\(\frac{B}{2}\) = (a + b – c) tan\(\frac{C}{2}\).

Solution:

(b + c – a) tan \(\frac{A}{2}\)

Question 12.

In △ABC prove that sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\)∙sin \(\frac{A}{2}\) = \(\frac{[A(\triangle A B C)]^{2}}{a b c s}\)

Solution:

LHS = sin \(\frac{A}{2}\)∙sin \(\frac{B}{2}\)∙sin \(\frac{C}{2}\)