MHT CET 2025 20th April Evening Shift (Mathematics) Question Paper with Solutions

Given, $3\sin 2𝜃=2\sin 3𝜃,0<𝜃<𝜋$

$\begin{array}{cc}& \Rightarrow 3(2\sin 𝜃\cos 𝜃)=2(3\sin 𝜃-4{\sin }^3𝜃)\\ & \Rightarrow 6\sin 𝜃\cos 𝜃=6\sin 𝜃-8{\sin }^3𝜃\end{array}$

Dividing both sides by $\sin 𝜃$, we get

$\begin{array}{cc}& 6\cos 𝜃=6-8{\sin }^2𝜃\\ & \Rightarrow 6\cos 𝜃=6-8(1-{\cos }^2𝜃)\\ & \Rightarrow 8{\cos }^2𝜃-6\cos 𝜃-2=0\\ & \Rightarrow 4{\cos }^2𝜃-3\cos 𝜃-1=0\\ & \Rightarrow (\cos 𝜃-1)(4\cos 𝜃+1)=0\\ & \Rightarrow \cos 𝜃=1\text{ or }\cos 𝜃=\frac{-1}{4}\end{array}$

$\begin{array}{cc}\mathrm{But}0& <𝜃<𝜋\\ \therefore \cos 𝜃& \ne 1\\ \therefore \cos 𝜃& =\frac{-1}{4}\\ & \\ & {\sin }^2𝜃=1-{\cos }^2𝜃\\ & =1-{(-\frac{1}{4})}^2\\ \therefore & {\sin }^2𝜃=\frac{15}{16}\\ & \Rightarrow \sin 𝜃=\frac{\sqrt{\surd 15}}{4}\dots [0<𝜃<𝜋]\end{array}$

Given, ${𝑥}^2+{𝑦}^2+k𝑥+(1-k)𝑦+5=0$

Comparing with

$𝑔=\frac{𝑘}{2},𝑓=\frac{1-𝑘}{2},𝑐=5$

According to the given condition, Radius $\le 5$

$\begin{array}{cc}& \Rightarrow \sqrt{\surd {\left(\frac{𝑘}{2}\right)}^2+{\left(\frac{1-𝑘}{2}\right)}^2-5}\le 5\\ & \Rightarrow \sqrt{\surd \frac{2{𝑘}^2-2𝑘-19}{4}}\le 5\end{array}$

Squaring on both sides, we get

$\begin{array}{cc}& \frac{2{𝑘}^2-2𝑘-19}{4}\le 25\\ & \Rightarrow 2{𝑘}^2-2𝑘-119\le 0\end{array}$

Consider,

$2{𝑘}^2-2𝑘-119=0$

$\begin{array}{cc}& \Rightarrow 𝑘=\frac{1\pm \sqrt{\surd 239}}{2}=\frac{1\pm 15.46}{2}\\ & \Rightarrow 𝑘=8.23\text{ or }𝑘=-7.23\\ & \Rightarrow -7.23\le 𝑘\le 8.23\end{array}$

$\therefore$ The integral values of k are

$-7,-6,-5,\dots .,6,7,8$

$\therefore$ The number of integral values of $k=16$

Let $\overline{\mathrm{AB}}=c=10,\overline{\mathrm{BC}}=a=8,\overline{\mathrm{CA}}=b=7$

By using cosine rule,

$\begin{array}{cc}\cos A& =\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}\\ & =\frac{7^2+{10}^2-8^2}{2\times 7\times 10}\\ & =\frac{85}{140}\\ & =\frac{17}{28}\\ \overline{\mathrm{AB}}\cdot \overline{\mathrm{AC}}& =\left|\overline{\mathrm{AB}}\right|\left|\overline{\mathrm{AC}}\right|\cos A\\ & =10\times 7\times \frac{17}{28}\\ & =\frac{85}{2}\end{array}$

Projection of $\overline{\mathrm{AB}}$ on $\overline{\mathrm{AC}}=\frac{\overline{\mathrm{AB}}\cdot \overline{\mathrm{AC}}}{\left|\overline{\mathrm{AC}}\right|}$

$=\frac{\frac{85}{2}}{7}=\frac{85}{14}\text{ units }$

Let $\bar{r}=𝑥\hat{i}+𝑦\hat{j}+z\hat{k}$, then

$\begin{array}{cc}& \bar{r}\times \bar{a}=\bar{b}\times \bar{a}\Rightarrow (\bar{r}-\bar{b})\times \bar{a}=0\\ \therefore & \mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 𝑥-2& 𝑦& 𝑧+1\\ 1& 1& 0\end{array}\mid =0\\ & \Rightarrow (-z-1)\hat{i}-(-z-1)\hat{j}+(𝑥-𝑦-2)\hat{k}=0\\ & \Rightarrow z=-1,𝑥-𝑦=2...\left(𝑖\right)\\ & \text{ Now, }\bar{r}\times \bar{b}=\bar{a}\times \bar{b}\Rightarrow (\bar{r}-\bar{a})\times \bar{b}=0\\ \therefore & \mid \begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 𝑥-1& 𝑦-1& z\\ 2& 0& -1\end{array}\mid =0\\ & \Rightarrow (1-𝑦)\hat{i}-(1-𝑥-2z)\hat{j}+(2-2𝑦)\hat{k}=0\\ & \Rightarrow 𝑦=1,𝑥+2z=1...\left(𝑖𝑖\right)\end{array}$

Solving (i) and (ii), we get

$𝑥=3,𝑦=1,z=-1$

At $A(5,5),z=550\left(5\right)+300\left(5\right)=4250$

At $B(5,10),z=550\left(5\right)+300\left(10\right)=5750$

At $C(11,4),z=550\left(11\right)+300\left(4\right)=7250$

At $D(6,4),z=550\left(6\right)+300\left(4\right)=4500$

$\therefore$ The maximum amount of scholarship is 7250 at the point $C(11,4)$.

$\begin{array}{cc}& \text{ Let }𝐼=\int \log (2+𝑥{)}^{2+𝑥}𝑑𝑥\\ & \text{ Put }2+𝑥=𝑡\Rightarrow 𝑑𝑥=𝑑𝑡\\ \therefore & 𝐼=\int 𝑡\log {𝑡}^{\prime }𝑑𝑡\\ & =\int \log 𝑡𝑑𝑡\\ & =\log 𝑡\left(\frac{{𝑡}^2}{2}\right)-\int (\frac{1}{𝑡}\times \frac{{𝑡}^2}{2})𝑑𝑡\\ & =\frac{{𝑡}^2\log 𝑡}{2}-\frac{1}{2}\int 𝑡𝑑𝑡\\ & =\frac{{𝑡}^2\log 𝑡}{2}-\frac{1}{2}\left(\frac{{𝑡}^2}{2}\right)+𝑐\\ & =\frac{{𝑡}^2}{2}(\log 𝑡-\frac{1}{2})+𝑐\\ & =\frac{{𝑡}^2}{2}(\log 𝑡-\log \sqrt{\surd 𝑒})+𝑐\\ & =\frac{{𝑡}^2}{2}\log \frac{𝑡}{\sqrt{\surd 𝑒}}+𝑐\\ & =\frac{(2+𝑥{)}^2}{2}\log \left(\frac{2+𝑥}{\sqrt{\surd 𝑒}}\right)+𝑐\end{array}$

Projection of $\overline{𝑎}$ on $\overline{𝑏}+\overline{𝑐}=\frac{\overline{𝑎}\cdot (\overline{𝑏}+\overline{𝑐})}{|\overline{𝑏}+\overline{𝑐}|}$

Projection of $\bar{b}+\bar{c}$ on $\bar{a}=\frac{(\bar{b}+\bar{c})\cdot \bar{a}}{\left|\bar{a}\right|}$

According to the given condition,

$\begin{array}{cc}& \frac{\overline{𝑎}\cdot (\overline{𝑏}+\overline{𝑐})}{|\overline{𝑏}+\overline{𝑐}|}=\frac{2(\overline{𝑏}+\overline{𝑐})\cdot \overline{𝑎}}{\left|\overline{𝑎}\right|}\\ & \Rightarrow \frac{1}{|\overline{𝑏}+\overline{𝑐}|}=\frac{2}{\left|\overline{𝑎}\right|}\\ & \Rightarrow \left|\overline{𝑎}\right|=2|\overline{𝑏}+\overline{𝑐}|...\left(𝑖\right)\\ & \text{ Now, }|\overline{𝑏}+\overline{𝑐}|=\sqrt{\surd |\overline{𝑏}{|}^2+|\overline{𝑐}{|}^2+2\left|\overline{𝑏}\right|\left|\overline{𝑐}\right|\cos 𝜃}\\ & =\sqrt{\surd (2\sqrt{\surd 2}{)}^2+(4{)}^2+2\times 2\sqrt{\surd 2}\times 4\times \cos \frac{𝜋}{4}}\\ & =\sqrt{\surd 8+16+16\sqrt{\surd 2}\times \frac{1}{\sqrt{\surd 2}}}\\ \therefore & |\overline{𝑏}+\overline{𝑐}|=\sqrt{\surd 40}=2\sqrt{\surd 10}\\ & \left|\overline{𝑎}\right|=2\left(2\sqrt{\surd 10}\right)\dots \left[𝐹𝑟𝑜𝑚\right(𝑖\left)\right]\\ & =4\sqrt{\surd 10}\end{array}$

Equation of the plane passing through the intersection of given planes is

$(2𝑥-𝑦+𝑧-3)+𝜆(4𝑥-3𝑦+5𝑧+9)=0$

$\begin{array}{c}\Rightarrow (2+4𝜆)𝑥+(-1-3𝜆)𝑦+(1+5𝜆)𝑧-3+9𝜆=0...\left(𝑖\right)\end{array}$

Direction ratios of the line $\frac{𝑥+1}{2}=\frac{𝑦+3}{4}=\frac{𝑧-3}{5}$ are $2,4,5$.

Since the plane is parallel to the given line,

$\begin{array}{cc}& (2+4𝜆)\left(2\right)+(-1-3𝜆)\left(4\right)+(1+5𝜆)\left(5\right)=0\\ & \Rightarrow 4+8𝜆-4-12𝜆+5+25𝜆=0\\ & \Rightarrow 5+21𝜆=0\\ & \Rightarrow 𝜆=\frac{-5}{21}\end{array}$

Substituting $𝜆=\frac{-5}{21}$ in equation (i), we get

$\begin{array}{cc}& \frac{22}{21}𝑥-\frac{6}{21}𝑦-\frac{4}{21}𝑧-\frac{108}{21}=0\\ & \Rightarrow 11𝑥-3𝑦-2𝑧-54=0\end{array}$

Comparing with $𝛼𝑥+𝛽𝑦+𝛾𝑧+d=0$, we get

$\begin{array}{cc}& 𝛼=11,𝛽=-3,𝛾=-2,𝑑=-54\\ & \Rightarrow 𝛼+𝛽+𝛾+𝑑=11-3-2-54=-48\end{array}$

$\begin{array}{cc}& \begin{array}{cc}& \overline{\mathrm{AB}}=\hat{i}+(p-2)\hat{k}\\ & \overline{\mathrm{AC}}=-\hat{j}+\hat{k}\\ & \overline{\mathrm{AD}}=\hat{i}+\hat{j}-2\hat{k}\end{array}\\ & \text{ Since A, B, C, D are coplanar, }\\ & \left[\begin{array}{ccc}\overline{\mathrm{AB}}& \overline{\mathrm{AC}}& \overline{\mathrm{AD}}\end{array}\right]=0\end{array}$

$\begin{array}{cc}& \Rightarrow \mid \begin{array}{ccc}1& 0& p-2\\ 0& -1& 1\\ 1& 1& -2\end{array}\mid =0\\ & \Rightarrow 1\left(1\right)-0+(p-2)\left(1\right)=0\\ & \Rightarrow 1+p-2=0\\ & \Rightarrow p=1\end{array}$

$\begin{array}{cc}& \text{ Let }I=\int \frac{e^{{\tan }^{-1}2𝑥}}{1+4{𝑥}^2}d𝑥\\ & \text{ Put }{\tan }^{-1}2𝑥=t\\ & \Rightarrow \frac{2}{1+4{𝑥}^2}d𝑥=\mathrm{dt}\\ \therefore & I=\frac{1}{2}\int e^t\mathrm{dt}\\ & =\frac{e^t}{2}+c=\frac{e^{{\tan }^{-1}2𝑥}}{2}+c\end{array}$

Let $𝑀$ be the midpoint of $𝐵𝐶$.

Then, AM is the median through A .

$\begin{array}{cc}\therefore M=(\frac{{𝑥}_1+{𝑥}_2}{2},\frac{{𝑦}_1+{𝑦}_2}{2},& =(\frac{2+(-4)}{2},\frac{1+5}{2})\\ & =(\frac{-2}{2},\frac{6}{2})\\ & =(-1,3)\end{array}$

Since the line $𝑥+4𝑦=p$ passes through the midpoint $M(-1,3)$, we get

$(-1)+4\left(3\right)=𝑝\Rightarrow 𝑝=11$

Also, the line $𝑥+4𝑦=p$ passes through the point $𝐴(3,𝑘)$.

$\begin{array}{cc}\therefore & 3+4𝑘=11\\ & \Rightarrow 4𝑘=8\\ & \Rightarrow 𝑘=2\end{array}$

$\begin{array}{cc}& \bar{a}\text{ and }\bar{b}\text{ are unit vectors. }\\ & \begin{array}{cc}\therefore & \left|\overrightarrow{𝑎}\right|=1,\left|\overrightarrow{𝑏}\right|=1\\ & |\overrightarrow{𝑎}+\overrightarrow{𝑏}{|}^2=|\overrightarrow{𝑎}{|}^2+2\overrightarrow{𝑎}\cdot \overline{𝑏}+|\overrightarrow{𝑏}{|}^2\\ & =|\overrightarrow{𝑎}{|}^2+2|\overrightarrow{𝑎}\left|\right|\overrightarrow{𝑏}|\cos 𝜃+|\overrightarrow{𝑏}{|}^2\\ & =1+2\cos 𝜃+1\\ & =2(1+\cos 𝜃)\\ & \Rightarrow |\overrightarrow{𝑎}+\overrightarrow{𝑏}|=\sqrt{\surd 2(1+\cos 𝜃)}\\ & \text{ Similarly, }|\overrightarrow{𝑎}-\overrightarrow{𝑏}|=\sqrt{\surd 2(1-\cos 𝜃)}\\ \therefore & |\overline{𝑎}-\overline{𝑏}|=\frac{\sqrt{\surd 2(1-\cos 𝜃)}}{\sqrt{\surd 2(1+\cos 𝜃)}}=\sqrt{\surd \frac{1-\cos 𝜃}{1+\cos 𝜃}}\\ & =\sqrt{\surd \frac{2{\sin }^2\frac{𝜃}{2}}{2{\cos }^2\frac{𝜃}{2}}}=\tan \frac{𝜃}{2}\end{array}\end{array}$

$\begin{array}{cc}& \text{ Given, }\\ & \text{ Cost price of }𝑥\text{ items }={𝑥}^2+78𝑥+2500\\ & \text{ Price per unit }\left(p\right)=600-8𝑥\\ & \text{ Selling price of }𝑥\text{ items }=(600-8𝑥)𝑥\\ & \text{ Now, }\\ & \text{ Profit }P\left(𝑥\right)=\text{ Selling Price }-\text{ Cost Price }\\ \therefore & P\left(𝑥\right)=(600-8𝑥)𝑥-({𝑥}^2+78𝑥+2500)\\ & =600𝑥-8{𝑥}^2-{𝑥}^2-78𝑥-2500\end{array}$

$\begin{array}{cc}\therefore & P\left(𝑥\right)=-9{𝑥}^2+522𝑥-2500\\ \therefore & P^{\prime }\left(𝑥\right)=-18𝑥+522\\ & \text{ For maxima, }{P}^{\prime }\left(𝑥\right)=0\\ & \Rightarrow -18𝑥+522=0\\ & \Rightarrow 𝑥=\frac{522}{18}\\ & \Rightarrow 𝑥=29\\ & P^{\prime \prime }\left(𝑥\right)=-18<0\\ & \Rightarrow P\text{ is maximum at }𝑥=29\\ \therefore & \text{ Maximum profit }\\ & =-9(29{)}^2+522(29)-2500\\ & =5069\end{array}$

Given equation is $k𝑥𝑦+10𝑥+8𝑦+16=0$

Comparing with

$\begin{array}{cc}& 𝑎{𝑥}^2+2ℎ𝑥𝑦+𝑏{𝑦}^2+2𝑔𝑥+2𝑓𝑦+𝑐=0,\text{ we get }\\ & 𝑎=0,ℎ=\frac{𝑘}{2},𝑏=0,𝑔=5,𝑓=4,𝑐=16\end{array}$

Since the given equation represents a pair of lines,

$\begin{array}{cc}& \mathrm{abc}+2\mathrm{fgh}-{\mathrm{af}}^2-{\mathrm{bg}}^2-{\mathrm{ch}}^2=0\\ & \Rightarrow 0+2\left(4\right)\left(5\right)\left(\frac{k}{2}\right)-0-0-16{\left(\frac{k}{2}\right)}^2=0\\ & \Rightarrow 20k-4k^2=0\\ & \Rightarrow k(20-4k)=0\\ & \Rightarrow k=0\text{ or }k=5\end{array}$

If $k=0$, then the equation becomes $10𝑥+8𝑦+16=0$, which does not represent a pair of lines.

$\therefore k=5$

$\begin{array}{cc}& \frac{d𝑦}{d𝑥}+\frac{𝑥}{𝑦}=\frac{a}{𝑦}\\ & \Rightarrow 𝑦\frac{d𝑦}{d𝑥}=a-𝑥\\ & \Rightarrow 𝑦d𝑦=(a-𝑥)d𝑥\end{array}$

Integrating on both sides, we get

$\begin{array}{cc}& \int 𝑦d𝑦=\int (a-𝑥)d𝑥\\ & \Rightarrow \frac{{𝑦}^2}{2}=a𝑥-\frac{{𝑥}^2}{2}+c\\ & \Rightarrow {𝑥}^2+{𝑦}^2-2a𝑥-2c=0\end{array}$

The above equation represents a circle with radius $\left(𝑟\right)=\sqrt{\surd {𝑎}^2+2𝑐}$

$2 \mathrm{f}(x)+3 \mathrm{f}\left(\frac{1}{x}\right)=x^2+1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Replacing $x$ by $\frac{1}{x}$, we get

$2 f\left(\frac{1}{x}\right)+3 f(x)=\left(\frac{1}{x}\right)^2+1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Multiplying equation (i) by 2 and equation (ii) by 3 , and then subtracting, we get

$$\begin{aligned} & 5 \mathrm{f}(x)=\frac{3}{x^2}+3-2 x^2-2 \\ & \Rightarrow \mathrm{f}(x)=\frac{-2 x^2+\frac{3}{x^2}+1}{5} \\ & y=5 x^2 \mathrm{f}(x) \\ & \Rightarrow y=5 x^2\left(\frac{-2 x^2+\frac{3}{x^2}+1}{5}\right) \\ & \Rightarrow y=-2 x^4+x^2+3 \end{aligned}$$

$\therefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x}=-8 x^3+2 x$

$y$ is strictly increasing

$$\begin{aligned} \Rightarrow & \frac{\mathrm{d} y}{\mathrm{~d} x}>0 \\ \Rightarrow & -8 x^3+2 x>0 \\ \Rightarrow & x\left(1-4 x^2\right)>0 \\ \Rightarrow & x>0 \text { and } 1-4 x^2>0 \text { OR } \\ & x<0 \text { and } 1-4 x^2<0 \end{aligned}$$

Case I: $x>0$ and $1-4 x^2>0$

$$\begin{aligned} & \Rightarrow x>0 \text { and }|x|<\frac{1}{2} \\ & \Rightarrow 0<x<\frac{1}{2} \end{aligned}$$

Case II: $x<0$ and $1-4 x^2<0$

$$\begin{aligned} & \Rightarrow x<0 \text { and }|x|>\frac{1}{2} \\ & \Rightarrow x<-\frac{1}{2} \end{aligned}$$

$\therefore \quad y$ is strictly increasing when

$x \in\left(-\infty,-\frac{1}{2}\right) \cup\left(0, \frac{1}{2}\right) .$

$\begin{array}{cc}& \sin 7𝜃+\sin 𝜃+\sin 4𝜃=0\\ & \Rightarrow 2\sin 4𝜃\cos 3𝜃+\sin 4𝜃=0\\ & \Rightarrow \sin 4𝜃(2\cos 3𝜃+1)=0\\ & \Rightarrow \sin 4𝜃=0\text{ or }\cos 3𝜃=\frac{-1}{2}\\ & \Rightarrow \sin 4𝜃=0\text{ or }\cos 3𝜃=\cos \left(\frac{2𝜋}{3}\right)\\ & \Rightarrow 4𝜃=𝑛𝜋\text{ or }3𝜃=2𝑛𝜋\pm \frac{2𝜋}{3}\end{array}$

$\begin{array}{ccc}& \Rightarrow 𝜃=\frac{n𝜋}{4}\text{ or }& 𝜃=\frac{2n𝜋}{3}\pm \frac{2𝜋}{9}\\ \therefore & 𝜃=\frac{2𝜋}{9},\frac{𝜋}{4},\frac{4𝜋}{9},\frac{𝜋}{2},\frac{3𝜋}{4},\frac{8𝜋}{9}& \dots [\because 𝜃\in (0,𝜋\left)\right]\end{array}$

$\begin{array}{cc}\text{ Total area of ellipse }& =𝜋\mathrm{ab}\\ & =𝜋\left(5\right)\left(3\right)\\ & =15𝜋\text{ sq. units }\\ \text{ Area of a quarter ellipse }& =\frac{1}{4}\left(15𝜋\right)=\frac{15𝜋}{4}\end{array}$

Now,

$\begin{array}{cc}𝐴𝑟𝑒𝑎𝑜𝑓\Delta \mathrm{AOB}& =\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB}\\ & =\frac{1}{2}\times 5\times 3=\frac{15}{2}\end{array}$

$\therefore$ Required area

$=$ Area of a quarter ellipse - Area of $△\mathrm{AOB}$

$=\frac{15𝜋}{4}-\frac{15}{2}$

$=\frac{15}{4}(𝜋-2)$ sq. units

Given, ${𝑥}^3-11{𝑥}^2+38𝑥-40=0$

$\begin{array}{cc}& \Rightarrow (𝑥-4)({𝑥}^2-7𝑥+10)=0\\ & \Rightarrow (𝑥-4)(𝑥-5)(𝑥-2)=0\end{array}$

$a,b,c$ are the roots of the given equation.

$\Rightarrow 𝑎=4,𝑏=5,𝑐=2$

By using cosine rule, we get

$\begin{array}{cc}& \frac{\cos 𝐴}{𝑎}+\frac{\cos 𝐵}{𝑏}+\frac{\cos 𝐶}{𝑐}\\ & =\frac{1}{𝑎}\left(\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}\right)+\frac{1}{𝑏}\left(\frac{{𝑐}^2+{𝑎}^2-{𝑏}^2}{2𝑐𝑎}\right)+\frac{1}{𝑐}\left(\frac{{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏}\right)\\ & =\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2+{𝑐}^2+{𝑎}^2-{𝑏}^2+{𝑎}^2+{𝑏}^2-{𝑐}^2}{2𝑎𝑏𝑐}\\ & =\frac{{𝑎}^2+{𝑏}^2+{𝑐}^2}{2𝑎𝑏𝑐}\\ & =\frac{4^2+5^2+2^2}{2\times 4\times 5\times 2}\\ & =\frac{45}{80}=\frac{9}{16}\end{array}$

$\begin{array}{cc}𝑠& =\frac{𝑎+𝑏+𝑐}{2}=\frac{13+14+15}{2}=\frac{42}{2}=21\\ \Delta & =\sqrt{\surd 𝑠(𝑠-𝑎)(𝑠-𝑏)(𝑠-𝑐)}\\ & =\sqrt{\surd 21(21-13)(21-14)(21-15)}\\ & =\sqrt{\surd 21\times 8\times 7\times 6}\\ & =\sqrt{\surd 7056}\\ & =84\end{array}$

$\begin{array}{cc}& \therefore \sin 𝐴=\frac{2\Delta }{𝑏𝑐}=\frac{2\times 84}{14\times 15}=\frac{4}{5}\\ & \text{ Alternate method: }\\ & \begin{array}{cc}\cos 𝐴& =\frac{{𝑏}^2+{𝑐}^2-{𝑎}^2}{2𝑏𝑐}\\ & =\frac{(14{)}^2+(15{)}^2-(13{)}^2}{2\left(14\right)\left(15\right)}\\ & =\frac{196+225-169}{420}\\ & =\frac{252}{420}=\frac{21}{35}\\ {\sin }^2𝐴& =1-{\cos }^2𝐴=1-{\left(\frac{21}{35}\right)}^2\\ & =1-\frac{441}{1225}\\ & =\frac{784}{1225}\\ \therefore \sin 𝐴& =\frac{28}{35}=\frac{4}{5}\end{array}\end{array}$