11. ⇒ (MHT CET 2023 9th May Morning Shift )

Two sides of a triangle are $\sqrt{3} + 1$ and $\sqrt{3} - 1$ and the included angle is $60^{\circ}$ , then the difference of the remaining angles is

A. $30^{\circ}$

B. $45^{\circ}$

C. $60^{\circ}$

D. $90^{\circ}$

Correct answer option is (D)

Let $a = \sqrt{3} + 1, b = \sqrt{3} - 1, C = 60^{\circ}$

Using cosine Rule,

$\begin{aligned} c^{2} = a^{2} + b^{2} - 2 a b \cos C \\ c^{2} = (\sqrt{3} + 1)^{2} + (\sqrt{3} - 1)^{2} - 2 (\sqrt{3} + 1) (\sqrt{3} - 1) \frac{1}{2} \\ c^{2} = 6 \end{aligned}$

$\begin{aligned} ∴ c^{2} & = 6 \\ c & = \sqrt{6} \end{aligned}$

Using sine Rule,

$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

Consider

$\begin{array}{ll} \frac{b}{\sin B} = \frac{c}{\sin C} \\ ∴ & \frac{\sqrt{3} - 1}{\sin B} = \frac{\sqrt{6}}{\frac{\sqrt{3}}{2}} \\ ∴ & \sin B = \frac{\sqrt{3} - 1}{2 \sqrt{2}} \\ ∴ & ∠ B = 15^{\circ} \\ ∴ & ∠ A = 105^{\circ} \end{array}$

.... [Remaining angle of a Triangle]

$∴ Difference = 90^{\circ}$

12. ⇒ (MHT CET 2021 21th September Morning Shift )

If the polar co-ordinates of a point are $(\sqrt{2}, \frac{π}{4})$ , then its Cartesian co-ordinates are

A. $(\sqrt{2}, 2)$

B. $(1, - 1)$

C. $(2, \sqrt{2})$

D. $(1, 1)$

Correct answer option is (D)

Polar coordinates of $z = a + i b$ are $(\sqrt{2}, \frac{π}{4})$

$\begin{aligned} ∴ \sqrt{2} = \sqrt{a^{2} + b^{2}} \Rightarrow a^{2} + b^{2} = 2 and \tan (\frac{π}{4}) = \frac{b}{a} \Rightarrow \frac{b}{a} = 1 \Rightarrow a = b \\ ∴ a^{2} + b^{2} = 2 \Rightarrow 2 a^{2} = 2 \Rightarrow a^{2} = 1 \Rightarrow a = \pm 1 \end{aligned}$

Since point lies in $1^{st}$ quadrant, $a = 1 \Rightarrow b = 1$

$∴$ Cartesian coordinates are $(1, 1)$

13. ⇒ (MHT CET 2021 21th September Evening Shift )

With usual notations, perimeter of a triangle $A B C$ is 6 times the arithmetic mean of sine of its angles. If $a = 1$ , then measure of angle $A =$

A. $\frac{π^{c}}{3}$

B. $\frac{π^{c}}{2}$

C. $\frac{π^{c}}{4}$

D. $\frac{π^{c}}{6}$

Correct answer option is (D)

$\begin{aligned} Let \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \\ ∴ \sin A = \frac{a}{k}, \sin B = \frac{b}{k}, \sin C = \frac{c}{k} \end{aligned}$

With usual notations, from the given data, we write

$\begin{aligned} a + b + c = 6 [\frac{(\frac{a}{k} + \frac{b}{k} + \frac{c}{k})}{3}] \\ ∴ (a + b + c) = \frac{2 (a + b + c)}{k} \Rightarrow k = 2 \\ ∴ \sin A = \frac{a}{k} = \frac{1}{2} \Rightarrow A = \frac{π^{c}}{6} \end{aligned}$

14. ⇒ (MHT CET 2021 20th September Morning Shift )

With usual notations if the angles of a triangle are in the ratio 1 : 2 : 3, then their corresponding sides are in the ratio.

A. 1 : 2 : 3

B. 1 : $\sqrt{3}$ : 3

C. $\sqrt{2}$ : $\sqrt{3}$ : 3

D. 1 : $\sqrt{3}$ : 2

Correct answer option is (D)

Let the angles be $x, 2 x, 3 x$

$∴ x + 2 x + 3 x = 180^{\circ} \Rightarrow 6 x = 180^{\circ} \Rightarrow x = 30^{\circ}$

Thus angles of the triangle are $30^{\circ}, 60^{\circ}, 90^{\circ}$ .

Now, $\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}$

$\begin{aligned} ∴ \frac{a}{\sin 30^{\circ}} = \frac{b}{\sin 60^{\circ}} = \frac{c}{\sin 90^{\circ}} \Rightarrow \frac{a}{(\frac{1}{2})} = \frac{b}{(\frac{\sqrt{3}}{2})} = \frac{c}{(1)} \\ ∴ 2 a = \frac{2 b}{\sqrt{3}} = c \Rightarrow a = \frac{c}{2} and b = \frac{\sqrt{3} c}{2} \\ ∴ a : b : c = \frac{c}{2} : \frac{\sqrt{3} c}{2} : c = 1 : \sqrt{3} : 2 \end{aligned}$

⇒Trigonometric Functions