Maths Solve the Following Question.(5 Marks)

$\begin{aligned} & \therefore \cos ^{-1} x=\pi, \cos ^{-1} y=\pi \text { and } \cos ^{-1} z=\pi \\ & \therefore x=y=z=\cos \pi=-1\end{aligned}$

$\begin{aligned} & \therefore x^2+y^2+z^2+2 x y z \\ & =(-1)^2+(-1)^2+(-1)^2+2(-1)(-1)(-1)\end{aligned}$

$\begin{aligned} & =1+1+1-2 \\ & =3-2=1 .\end{aligned}$

Dividing both sides by $\sqrt{(1)^2+(-1)^2}=\sqrt{2}$, we get

$\frac{1}{\sqrt{2}} \cos \theta-\frac{1}{\sqrt{2}} \sin \theta=-\frac{1}{\sqrt{2}}$

$\therefore \cos \frac{\pi}{4} \cos \theta-\sin \frac{\pi}{4} \sin \theta=-\cos \frac{\pi}{4}$

$\therefore \cos \left(\theta-\frac{\pi}{4}\right)=\cos \left(\pi-\frac{\pi}{4}\right)$

$\ldots[\because \cos (\pi-\theta)=-\cos \theta]$

$\therefore \cos \left(\theta-\frac{\pi}{4}\right)=\cos \frac{3 \pi}{4}$

$\ldots(1)$

The general solution of $\cos \theta=\cos \alpha$ is

$\theta=2 n \pi \pm \alpha, n \in Z$

$\therefore$ the general solution of $(1)$ is given by

$\theta-\frac{\pi}{4}=2 n \pi \pm \frac{3 \pi}{4}, n \in Z$

Taking positive sign, we get

$\theta-\frac{\pi}{4}=2 n \pi+\frac{3 \pi}{4}, n \in Z$.

$\therefore \theta=2 n \pi+\pi=(2 n+1) \pi, n \in Z$

Taking negative sign, we get

$\begin{aligned} & \theta-\frac{\pi}{4}=2 n \pi-\frac{3 \pi}{4}, n \in Z . \\ & \therefore \theta=2 n \pi-\frac{\pi}{2}, n \in Z\end{aligned}$

$\therefore$ the required general solution is

$\theta=(2 n+1) \pi, n \in Z$ or $\theta=2 n \pi-\frac{\pi}{2}, n \in Z$.

$\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)=\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

i.e. to show that,

$\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)=\frac{9 \pi}{8}$

Let $\sin ^{-1}\left(\frac{1}{3}\right)=x$

$\therefore \sin x=\frac{1}{3}$, where $0<x<\frac{\pi}{3}$

$\therefore \cos x>0$

Now, $\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{1}{9}}=\sqrt{\frac{8}{9}}=\frac{2 \sqrt{2}}{3}$

$\therefore x=\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

$\therefore \sin ^{-1}\left(\frac{1}{3}\right)=\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

$\ldots$ (1)

$\therefore \mathrm{LHS}=\frac{9}{4} \sin ^{-1}\left(\frac{1}{3}\right)+\frac{9}{4} \sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)$

$=\frac{9}{4}\left[\sin ^{-1}\left(\frac{1}{3}\right)+\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\right]$

$=\frac{9}{4}\left[\cos ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)+\sin ^{-1}\left(\frac{2 \sqrt{2}}{3}\right)\right] \ldots[$ By (1)]

$=\frac{9}{4}\left(\frac{\pi}{2}\right) \quad \ldots\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$=\frac{9 \pi}{8}=$ RHS.

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

$\therefore a=f k \sin A, b=k \sin B, c=k \sin C$

Now, $\left(c^2-a^2+b^2\right) \tan \mathrm{A}=\left(c^2-a^2+b^2\right) \cdot \frac{\sin \mathrm{A}}{\cos \mathrm{A}}$

$=\left(c^2+b^2-a^2\right) \times \frac{k a}{\left(\frac{c^2+b^2-a^2}{2 b c}\right)}$

$=\left(c^2+b^2-a^2\right) \times \frac{2 k a b c}{c^2+b^2-a^2}$

$=2 k a b c$

$\ldots(1)$

$\left(a^2-b^2+c^2\right) \tan \mathrm{B}=\left(a^2-b^2+c^2\right) \cdot \frac{\sin \mathrm{B}}{\cos \mathrm{B}}$

$=\left(a^2+c^2-b^2\right) \times \frac{k b}{\left(\frac{a^2+c^2-b^2}{2 a c}\right)}$

$=\left(a^2+c^2-b^2\right) \times \frac{2 k a b c}{a^2+c^2-b^2}$

$=2 \mathrm{kabc}$

$\ldots(2)$

$\left(b^2-c^2+a^2\right) \tan C=\left(b^2-c^2+a^2\right) \cdot \frac{\sin C}{\cos C}$

$=\left(a^2+b^2-c^2\right) \times \frac{k c}{\left(\frac{a^2+b^2-c^2}{2 a b}\right)}$

$\begin{aligned} & =\left(a^2+b^2-c^2\right) \times \frac{2 k a b c}{a^2+b^2-c^2} \\ & =2 k a b c\end{aligned}$

... (3)From (1), (2) and (3), we get

$\begin{aligned} & \left(c^2-a^2+b^2\right) \tan A=\left(a^2-b^2+c^2\right) \tan B \\ & =\left(b^2-c^2+a^2\right) \tan C .\end{aligned}$

Now, $\cot \frac{A}{2}+\cot \frac{C}{2}=\frac{\cos \frac{A}{2}}{\sin \frac{A}{2}}+\frac{\cos \frac{C}{2}}{\sin \frac{C}{2}}$

$=\frac{\cos \frac{A}{2} \cdot \sin \frac{C}{2}+\sin \frac{A}{2} \cdot \cos \frac{C}{2}}{\sin \frac{A}{2} \cdot \sin \frac{C}{2}}$

$=\frac{\sin \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{C}}{2}\right)}{\sin \frac{\mathrm{A}}{2} \cdot \sin \frac{\mathrm{C}}{2}}$

$=\frac{\sin \left(\frac{\pi}{2}-\frac{B}{2}\right)}{\sqrt{\frac{(s-b)(s-c)}{b c}} \cdot \sqrt{\frac{(s-a)(s-b)}{a b}}}$

$\cdots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$

$=\frac{\cos \frac{B}{2}}{\frac{(s-b)}{b} \cdot \sqrt{\frac{(s-c)(s-a)}{c a}}}$

$=\frac{b \cos \frac{\mathrm{B}}{2}}{(s-b) \cdot \sin \frac{\mathrm{B}}{2}}$

$=\frac{b}{s-b} \cdot \cot \frac{B}{2}$

$=\frac{b}{\left(\frac{a+b+c}{2}-b\right)} \cdot \cot \frac{\mathrm{B}}{2}$

$\cdots[\because 2 s=a+b+c]$

$=\frac{2 b}{(2 b-b)} \cdot \cot \frac{\mathrm{B}}{2} \quad \ldots[$ By $(1)]$

$=\frac{2 b}{b} \cdot \cot \frac{B}{2}$

$=\left(\frac{2 b}{a+c-b}\right) \cdot \cot \frac{\mathrm{B}}{2}$

$\therefore \cot \frac{A}{2}+\cot \frac{C}{2}=2 \cot \frac{B}{2}$

Hence, $\cot \frac{A}{2}, \cot \frac{B}{2}, \cot \frac{C}{2}$ are in A.P.

$\therefore \sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\sqrt{\tan ^2\left(\frac{\theta}{2}\right)}=\tan \left(\frac{\theta}{2}\right)$

$\therefore$ LHS $=\tan ^{-1}\left[\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}\right]$

$=\tan ^{-1}\left[\tan \left(\frac{\theta}{2}\right)\right]$

$\begin{aligned} & =\frac{\theta}{2} \cdots\left[\because \tan ^{-1}(\tan \theta)=\theta\right] \\ & =\mathrm{RHS} .\end{aligned}$

$=\tan ^{-1}\left(\frac{1+\tan \theta}{1-\tan \theta}\right)$

$=\tan ^{-1}\left(\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \tan \theta}\right)$

$=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+\theta\right)\right]$

$=\frac{\pi}{4}+\theta \quad \ldots\left[\because \tan ^{-1}(\tan \theta)=\theta\right]$

$=$ RHS.

$\cdots\left[\because 2 \tan ^{-1} x=\tan ^{-1}\left(\frac{2 x}{1-x^2}\right)\right]$

$=\tan ^{-1}\left(\frac{\left(\frac{2}{3}\right)}{1-\frac{1}{9}}\right)=\tan ^{-1}\left(\frac{2}{3} \times \frac{9}{8}\right)$

$=\tan ^{-1}\left(\frac{3}{4}\right)$

$=$ RHS.

Alternative Method:

LHS $=2 \tan ^{-1}\left(\frac{1}{3}\right)=\tan ^{-1}\left(\frac{1}{3}\right)+\tan ^{-1}\left(\frac{1}{3}\right)$

$=\tan ^{-1}\left(\frac{\frac{1}{3}+\frac{1}{3}}{1-\frac{1}{3} \times \frac{1}{3}}\right)$

$=\tan ^{-1}\left(\frac{3+3}{9-1}\right)=\tan ^{-1}\left(\frac{6}{8}\right)$

$\begin{aligned} & =\tan ^{-1}\left(\frac{3}{4}\right) \\ & =\mathrm{RHS}\end{aligned}$

$=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2} \times \frac{1}{3}}\right)$

$=\tan ^{-1}\left(\frac{3+2}{6-1}\right)=\tan ^{-1}(1)$

$=\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4}$

$\therefore \cos x=\left(\frac{3}{5}\right)$, where $0<x<\frac{\pi}{2} \therefore \sin x>0$

Now, $\sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

$\therefore x=\sin ^{-1}\left(\frac{4}{5}\right)$

$\therefore \cos ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1}\left(\frac{4}{5}\right)$$\ldots$ (1)

$\mathrm{LHS}=\cos ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{4}{5}\right)$

$=\sin ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{4}{5}\right) \quad \ldots[\mathrm{By}(1)]$

$=\frac{\pi}{2} \quad \therefore\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]$

$=\mathrm{RHS}$.

Then $\sin x=\frac{3}{5}$, where $0<x<\frac{\pi}{2}$

$\cos y=\frac{12}{13}$, where $0<y<\frac{\pi}{2}$

and $\sin z=\frac{56}{65}$, where $0<z<\frac{\pi}{2}$

$\therefore \cos x>0, \sin y>0$

Now, $\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

and $\sin y=\sqrt{1-\cos ^2 y}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$

We have to prove that, $x+y=z$

Now, $\sin (x+y)=\sin x \cos y+\cos x \sin y$

$=\left(\frac{3}{5}\right)\left(\frac{12}{13}\right)+\left(\frac{4}{5}\right)\left(\frac{5}{13}\right)$

$=\frac{36}{65}+\frac{20}{65}=\frac{56}{65}$

$\therefore \sin (x+y)=\sin z \quad \therefore x+y=z$

Hence, $\sin ^{-1}\left(\frac{3}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\sin ^{-1}\left(\frac{56}{65}\right)$.

$\therefore \sin \alpha=-\frac{1}{2}=-\sin \frac{\pi}{6}$

$\therefore \sin \alpha=\sin \left(-\frac{\pi}{6}\right) \quad \ldots[\because \sin (-\theta)=-\sin \theta]$

$\therefore \alpha=-\frac{\pi}{6} \quad \quad \cdots\left[\because-\frac{\pi}{2} \leqslant-\frac{\pi}{6} \leqslant \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$$\ldots(1)$

Let $\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\beta$, where $0 \leqslant \beta \leqslant \pi$

$\therefore \cos \beta=-\frac{\sqrt{3}}{2}=-\cos \frac{\pi}{6}$

$\cos \beta=\cos \left(\pi-\frac{\pi}{6}\right) \quad \ldots[\because \cos (\pi-\theta)=-\cos \theta]$

$\therefore \cos \beta=\cos \frac{5 \pi}{6}$

$\therefore \beta=\frac{5 \pi}{6} \quad \cdots\left[\because 0 \leqslant \frac{5 \pi}{6} \leqslant \pi\right]$

$\therefore \cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=\frac{5 \pi}{6}$$\ldots(2)$

Let $\cos ^{-1}\left(-\frac{1}{2}\right)=\gamma$, where $0 \leqslant \gamma \leqslant \pi$

$\therefore \cos \gamma=-\frac{1}{2}=-\cos \frac{\pi}{3}$

$\therefore \cos \gamma=\cos \left(\pi-\frac{\pi}{3}\right) \quad \ldots[\because \cos (\pi-\theta)=-\cos \theta]$

$\therefore \cos \gamma=\cos \frac{2 \pi}{3}$

$\therefore \gamma=\frac{2 \pi}{3} \quad \ldots\left[\because 0 \leqslant \frac{2 \pi}{3} \leqslant \pi\right]$

$\therefore \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}$

$\ldots(3)$

$\mathrm{LHS}=\sin ^{-1}\left(-\frac{1}{2}\right)+\cos ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$

$=-\frac{\pi}{6}+\frac{5 \pi}{6} \quad \ldots .[\mathrm{By}(1)$ and (2)]

$\begin{aligned} & =\frac{4 \pi}{6}=\frac{2 \pi}{3} \\ & =\cos ^{-1}\left(-\frac{1}{2}\right) \\ & =\text { RHS. }\end{aligned}$

... [By (3)]

$\therefore \sin \alpha=\frac{1}{\sqrt{2}}=\sin \frac{\pi}{4}$

$\therefore x=\frac{\pi}{4} \quad \cdots\left[\because-\frac{\pi}{2} \leqslant \frac{\pi}{4} \leqslant \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4}$$\ldots$ (1)

Let $\sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\beta$, where $-\frac{\pi}{2} \leqslant \beta \leqslant \frac{\pi}{2}$

$\therefore \sin \beta=\frac{\sqrt{3}}{2}=\sin \frac{\pi}{3}$

$\therefore \beta=\frac{\pi}{3} \quad \cdots\left[\because-\frac{\pi}{2} \leqslant \frac{\pi}{3} \leqslant \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}$

$\ldots(2)$

$\mathrm{LHS}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)-3 \sin ^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$=\frac{\pi}{4}-3\left(\frac{\pi}{3}\right) \quad \ldots$ [By (1) and (2)]

$=\frac{\pi}{4}-\pi=-\frac{3 \pi}{4}=$ RHS.

$\therefore \operatorname{cosec} \alpha=-\sqrt{2}=-\operatorname{cosec} \frac{\pi}{4}$

$\therefore \operatorname{cosec} \alpha=\operatorname{cosec}\left(-\frac{\pi}{4}\right)$$\ldots[\because \operatorname{cosec}(-\theta)=-\operatorname{cosec} \theta]$

$\therefore \alpha=-\frac{\pi}{4} \quad \quad \ldots\left[\frac{-\pi}{2} \leqslant \frac{-\pi}{4} \leqslant \frac{\pi}{2}\right]$

$\therefore \operatorname{cosec}^{-1}(-\sqrt{2})=-\frac{\pi}{4}$$\ldots(1)$

Let $\cot ^{-1}(\sqrt{3})=\beta$, where $0<\beta<\pi$

$\therefore \cot \beta=\sqrt{3}=\cot \frac{\pi}{6}$

$\therefore \beta=\frac{\pi}{6} \quad \cdots\left[\because 0<\frac{\pi}{6}<\pi\right]$

$\therefore \cot ^{-1}(\sqrt{3})=\frac{\pi}{6}$

$\ldots(2)$

$\therefore \operatorname{cosec}^{-1}(-\sqrt{2})+\cot ^{-1}(\sqrt{3})$

$=-\frac{\pi}{4}+\frac{\pi}{6} \quad \ldots .[$ By (1) and (2)]

$=\frac{-3 \pi+2 \pi}{12}=-\frac{\pi}{12}$

$\therefore \tan \alpha=\sqrt{3}=\tan \frac{\pi}{3}$

$\therefore \alpha=\frac{\pi}{3} \quad \cdots\left[\because \frac{-\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}\right]$

$\therefore \tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$$\ldots(1)$

Let $\sec ^{-1}(-2)=\beta$, where $0 \leqslant \beta \leqslant \pi, \beta \neq \frac{\pi}{2}$

$\therefore \sec \beta=-2=-\sec \frac{\pi}{3}$

$\therefore \sec \beta=\sec \left(\pi-\frac{\pi}{3}\right) \ldots[\because \sec (\pi-\theta)=-\sec \theta]$

$\therefore \sec \beta=\sec \frac{2 \pi}{3}$

$\therefore \beta=\frac{2 \pi}{3} \quad \ldots\left[\left[\because 0 \leqslant \frac{2 \pi}{3} \leqslant \pi\right]\right.$

$\therefore \sec ^{-1}(-2)=\frac{2 \pi}{3}$

$\ldots(2)$

$\begin{aligned} & \therefore \tan ^{-1} \sqrt{3}-\sec ^{-1}(-2) \\ & =\frac{\pi}{3}-\frac{2 \pi}{3} \ldots[B y(1) \text { and }(2)] \\ & =-\frac{\pi}{3} .\end{aligned}$

$\therefore \cos \alpha=\frac{1}{2}=\cos \frac{\pi}{3}$

$\therefore \alpha=\frac{\pi}{3} \quad \cdots\left[\because 0<\frac{\pi}{3}<\pi\right]$

$\therefore \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$$\ldots(1)$

Let $\sin ^{-1}\left(\frac{1}{2}\right)=\beta$, where $\frac{-\pi}{2} \leq \beta \leq \frac{\pi}{2}$

$\therefore \sin \beta=\frac{1}{2}=\sin \frac{\pi}{6}$

$\therefore \beta=\frac{\pi}{6} \quad \cdots\left[\because \frac{-\pi}{2} \leq \frac{\pi}{6} \leq \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

...(2)

$\cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$ and $\sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$

$\therefore \cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$

$\begin{aligned} & =\frac{\pi}{3}+2\left(\frac{\pi}{6}\right) \\ & =\frac{\pi}{3}+\frac{\pi}{3} \\ & =\frac{2 \pi}{3}\end{aligned}$

$\therefore \tan \alpha=1=\tan \frac{\pi}{4}$

$\therefore \quad \alpha=\frac{\pi}{4}$$\left[\because \frac{-\pi}{2}<\frac{\pi}{4}<\frac{\pi}{2}\right]$

$\therefore \tan ^{-1}(1)=\frac{\pi}{4}$... (1)

Let $\cos ^{-1}\left(\frac{1}{2}\right)=\beta$, where $0 \leqslant \beta \leqslant \pi$

$\therefore \cos \beta=\frac{1}{2}=\cos \frac{\pi}{3}$

$\therefore \beta=\frac{\pi}{3} \quad \ldots\left[\because 0<\frac{\pi}{3}<\pi\right]$

$\therefore \cos ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{3}$$\ldots(2)$

Let $\sin ^{-1}\left(\frac{1}{2}\right)=\gamma$, where $\frac{-\pi}{2} \leqslant \gamma \leqslant \frac{\pi}{2}$

$\therefore \sin y=\frac{1}{2}=\sin \frac{\pi}{6}$

$\therefore \gamma=\frac{\pi}{6} \quad \ldots\left[\because \frac{-\pi}{2} \leqslant \frac{\pi}{6} \leqslant \frac{\pi}{2}\right]$

$\therefore \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}$$\ldots$ (3)

$\therefore \tan ^{-1}(1)+\cos ^{-1}\left(\frac{1}{2}\right)+\sin ^{-1}\left(\frac{1}{2}\right)$

$=\frac{\pi}{4}+\frac{\pi}{3}+\frac{\pi}{6} \quad \ldots[$ By (1), (2) and (3)]

$=\frac{3 \pi+4 \pi+2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}$

$\begin{aligned} \cos \mathrm{A} & =\frac{b^2+c^2-a^2}{2 b c}=\frac{(24)^2+(30)^2-(18)^2}{2(24)(30)} \\ & =\frac{576+900-324}{1440}=\frac{1152}{1440}=\frac{4}{5} .\end{aligned}$

2. $\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}}=\sqrt{\frac{(36-24)(36-30)}{(24)(30)}}$

$=\sqrt{\frac{12 \times 6}{24 \times 30}}=\sqrt{\frac{1}{10}}=\frac{1}{\sqrt{10}}$

3. $\cos \frac{\mathrm{A}}{2}=\sqrt{\frac{s(s-a)}{b c}}=\sqrt{\frac{36(36-18)}{(24)(30)}}$

$=\sqrt{\frac{36 \times 18}{24 \times 30}}=\sqrt{\frac{9}{10}}=\frac{3}{\sqrt{10}}$

4. $\tan \frac{A}{2}=\frac{\sin \frac{A}{2}}{\cos \frac{A}{2}}=\frac{1 / \sqrt{10}}{3 / \sqrt{10}}=\frac{1}{3}$

5. $\begin{aligned} A(\Delta \mathrm{ABC}) & =\sqrt{s(s-a)(s-b)(s-c)} \\ & =\sqrt{36(36-18)(36-24)(36-30)}\end{aligned}$

$\begin{aligned} & =\sqrt{36 \times 18 \times 12 \times 6} \\ & =\sqrt{36 \times 18 \times 4 \times 18} \\ & =6 \times 18 \times 2=216 \text { sq units. }\end{aligned}$

6. $\begin{aligned} & A(\triangle A B C)=\frac{1}{2} b c \sin \mathrm{A} \\ & \therefore 216=\frac{1}{2}(24)(30) \sin \mathrm{A}\end{aligned}$

$\therefore \sin A=\frac{216}{12 \times 30}=\frac{216}{360}=\frac{3}{5}$.