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

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, r2 = x2 + y2 = ($$\sqrt {2}$$ )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, r2 = x2 + y2 = (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, r2 = x2 + y2 = (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 a3sin(B – C) + b3sin(C – A) + c3sin(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 = 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)

= $$\frac{1}{2 k}$$ [a2(a2 + b2 – c2) – a2(a2 + c2 – b2) + b2(b2 + c2 – a2) – b2(a2 + b2 – c2) + c2(c2 + a2 – b2) – c2(b2 + c2 – a2)]
= $$\frac{1}{2 k}$$ [a4 + a2b2 – a2c2 – a4 – a2c2 + a2b2 + b4 + b2c2 – a2b2 – a2b2 – b4 + b2c2 + c4 + a2c2 – b2c2 – b2c2 – c4 + a2c2]
= $$\frac{1}{2 k}$$(0) = 0 = RHS.

Question 7.
In ∆ABC, if cot A, cot B, cot C are in A.P. then show that a2, b2, c2 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) = a2 + b2 + c2.
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]
= 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

(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 $$\frac{C}{2}$$