# Answer for 2019 Sample Question : Section (C)

ဒီေနရာမွာ တင္ေပးခဲ့ေသာ Section (C) ေမးခြန္းရဲ့ အေျဖျဖစ္ပါတယ္။ Section (A) ရဲ့ အေျဖကိုေတာ့ ဒီေနရာမွာ တင္ေပးခဲ့ၿပီး  Section (B) အေျဖကို ဒီေနရာမွာ တင္ေပးထားပါတယ္။

Section (C)
Solution

11.    (a) Two unequal circles are tangent externally at $\displaystyle O$. $\displaystyle AB$ the chord of the first circle is tangent to the second circle at $\displaystyle C$, and $\displaystyle AO$ meets this circle at $\displaystyle E$. Prove that $\displaystyle ∠BOC=∠COE$.
(5 marks)

Show/Hide Solution
Draw a common tangent at $\displaystyle O$ to meet $\displaystyle BC$ at $\displaystyle D.$

$\displaystyle \begin{array}{l}\ \ \ \text{In}\odot P,\\\\\ \ \ \alpha =\angle A\ \ \ \text{( }\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{= }\angle \text{in alternate segment)}\\\\\ \ \ OD=CD\ \ \text{(tangents from same exterior point)}\\\\\therefore \beta =\gamma \\\\\therefore \alpha +\beta =\angle A+\gamma \\\\\ \ \ \text{But }\alpha +\beta =\angle BOC\\\\\ \ \ \text{In}\ \vartriangle AOC,\\\text{ }\\\ \ \ \angle A+\gamma =\angle COE\\\\\therefore \angle BOC=\angle COE\end{array}$

11.    (b) In $\displaystyle ΔABC$, $\displaystyle D$ is a point of $\displaystyle AC$ such that $\displaystyle AD=2CD$. $\displaystyle E$ is on $\displaystyle BC$ such that $\displaystyle DE \parallel AB$. Compare the areas of $\displaystyle ΔCDE$ and $\displaystyle ΔABC$. If $\displaystyle α (ABED)=40$, what is $\displaystyle α (ΔABC)$?
(5 marks)
Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ AD=2CD\ \ \ (\text{given)}\\\\\ \ \ \text{Since}\ DE\parallel AB,\\\\\ \ \ \vartriangle CAB\sim \vartriangle CDE\end{array}$

$\displaystyle \therefore \frac{{\alpha (\Delta CAB)}}{{\alpha (\Delta CDE)}}=\frac{{A{{C}^{2}}}}{{C{{D}^{2}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(AD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{{{{(2CD+CD)}}^{2}}}}{{C{{D}^{2}}}}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =9\\\\\ \ \ \text{Let }\alpha (\Delta CDE)=x\ \text{and}\ \alpha (\Delta CAB)=9x.\\\\\therefore \alpha (ABED)=9x-x=8x\\\\\therefore 8x=40\Rightarrow x=5\\\\\therefore \alpha (\Delta CAB)=9\times 5=45\ \text{sq-units}\text{.}\end{array}$

12.    (a) The position vectors, relative to an origin $\displaystyle O$, of three nonlinear points $\displaystyle A, B$ and $\displaystyle C$ are $-2\hat{\text{i}}+3\hat{\text{j}}$, $3\hat{\text{i}}+2\hat{\text{j}}$ and $-\hat{\text{i}}-5\hat{\text{j}}$. Show that $\displaystyle ΔABC$ is isosceles.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \overrightarrow{{OA}}=-2\widehat{\text{i}}+3\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OB}}=3\widehat{\text{i}}+2\widehat{\text{j}}\\\\\ \ \ \ \overrightarrow{{OC}}=-\widehat{\text{i}}-5\widehat{\text{j}}\\\\\therefore \ \ \overrightarrow{{AB}}=\overrightarrow{{OB}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =3\widehat{\text{i}}+2\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =5\widehat{\text{i}}-\widehat{\text{j}}\\\\\therefore \ \ AB=\sqrt{{{{5}^{2}}+{{{(-1)}}^{2}}}}=\sqrt{{26}}\\\\\therefore \ \ \overrightarrow{{BC}}=\overrightarrow{{OC}}-\overrightarrow{{OB}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}-3\widehat{\text{i}}-2\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =-4\widehat{\text{i}}-7\widehat{\text{j}}\\\\\therefore \ \ BC=\sqrt{{{{{(-4)}}^{2}}+{{{(-7)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ \overrightarrow{{AC}}=\overrightarrow{{OC}}-\overrightarrow{{OA}}\\\\\ \ \ \ \ \ \ \ \ \ =-\widehat{\text{i}}-5\widehat{\text{j}}+2\widehat{\text{i}}-3\widehat{\text{j}}\\\\\ \ \ \ \ \ \ \ \ \ =\widehat{\text{i}}-8\widehat{\text{j}}\\\\\therefore \ \ AC=\sqrt{{{{1}^{2}}+{{{(-8)}}^{2}}}}=\sqrt{{65}}\\\\\therefore \ \ AC=BC\\\\\therefore \ \ \ \Delta ABC\ \text{is an isosceles triangle}\text{.}\end{array}$

12.    (b) The tangent at the point $\displaystyle C$ on the circle meets the diameter $\displaystyle AB$ produced at $\displaystyle T$, If $\displaystyle ∠BCT=27°$, calculate $\displaystyle ∠CTA$. If $\displaystyle CT=t$ and $\displaystyle BT=x$, prove that the radius of the circle is $\large {\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}}$.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \angle BAC=\angle BCT=\text{ }27{}^\circ \ \ \ (\angle \text{ between tangent and chord}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{=}\ \angle \ \text{in alternate segment)}\\\\\ \ \ \ \angle ACB=90{}^\circ \ \ \ \ \ \ \ (\angle \text{ in semicircle)}\\\\\ \ \ \ \text{In}\ \vartriangle \text{ACT,}\\\\\ \ \ \ \angle BAC+\angle ACT+\angle CTA=180{}^\circ \\\\\therefore \ \ 27{}^\circ +27{}^\circ +90{}^\circ +\angle CTA=180{}^\circ \\\\\therefore \ \ \angle CTA=36{}^\circ \\\\\ \ \ \ \text{Let}\ AB=2r,\text{where }r\ \text{is the radius of the circle}\text{.}\\\\\ \ \ \ \text{Since}\ AT\cdot BT=C{{T}^{2}},\\\\\ \ \ \ (2r+x)\cdot x={{t}^{2}}\\\\\therefore \ \ 2xr+{{x}^{2}}={{t}^{2}}\end{array}$

$\displaystyle \therefore \ \ r=\frac{{{{t}^{2}}-{{x}^{2}}}}{{2x}}$

13.    (a) Find the angles of a triangle, given that $\displaystyle \angle A$ is obtuse and $\sec (B+C)=\operatorname{cosec}(B-C)=2$.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \text{In}\ \vartriangle ABC,\angle A\ \text{is obtuse and}\\\\\ \ \ \ \sec (B+C)=\operatorname{cosec}(B-C)=2\end{array}$

$\displaystyle \therefore \ \ \cos (B+C)=\sin (B-C)=\frac{1}{2}$

$\displaystyle \begin{array}{l}\ \ \ \ \text{Since}\ \angle A\ \text{is obtuse,}\\\\\ \ \ \angle B+\angle C=60{}^\circ \\\\\ \ \ \angle B-\angle C=30{}^\circ \\\\\therefore \ \ 2\angle B=90{}^\circ \Rightarrow \angle B=45{}^\circ \\\\\therefore \ \ 45{}^\circ +\angle C=60{}^\circ =\angle C=15{}^\circ \\\\\ \ \ \ \text{Since}\ \angle A+\angle B+\angle C=180{}^\circ ,\\\\\ \ \ \ \angle A+45{}^\circ +15{}^\circ =180{}^\circ \Rightarrow \angle A=120{}^\circ \end{array}$

13.    (b) Differentiate $\displaystyle \frac{1}{\sqrt{x}}$ with respect to $\displaystyle x$ from the first principles.
(5 marks)

Show/Hide Solution
$\displaystyle \ \ \ \ \text{Let}\ f(x)=\frac{1}{{\sqrt{x}}}.$

$\displaystyle \therefore \ \ f(x+\delta x)=\frac{1}{{\sqrt{{x+\delta x}}}}$

$\displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{1}{{\sqrt{{x+\delta x}}}}-\frac{1}{{\sqrt{x}}}$

$\displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}$

$\displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{\sqrt{x}-\sqrt{{x+\delta x}}}}{{\sqrt{x}\sqrt{{x+\delta x}}}}\times \frac{{\sqrt{x}+\sqrt{{x+\delta x}}}}{{\sqrt{x}+\sqrt{{x+\delta x}}}}$

$\displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{x-x-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$\displaystyle \therefore \ \ f(x+\delta x)-f(x)=\frac{{-\delta x}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$\displaystyle \therefore \ \ \frac{{f(x+\delta x)-f(x)}}{{\delta x}}=\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$\displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{f(x+\delta x)-f(x)}}{{\delta x}}$

$\displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+\delta x}}\left[ {\sqrt{x}+\sqrt{{x+\delta x}}} \right]}}$

$\displaystyle \therefore {f}'(x)=\underset{{x\to 0}}{\mathop{{\lim }}}\,\frac{{-1}}{{\sqrt{x}\sqrt{{x+0}}\left[ {\sqrt{x}+\sqrt{{x+0}}} \right]}}=-\frac{1}{{2x\sqrt{x}}}$

14.    (a) A cruise ship travels at a bearing of $\displaystyle 45°$ at $\displaystyle 15$ mph for $\displaystyle 3$ hours, and changes course to a bearing of $\displaystyle 120°$. It then travels $\displaystyle 10$ mph for $\displaystyle 2$ hours. Find the distance of the ship from its original position and also its bearing from the original position.
(5 marks)

Show/Hide Solution

$\displaystyle \begin{array}{l}AB=3\ \text{hours}\ \times 15\ \text{mph}=45\ \text{mi}\\\\BC=2\ \text{hours}\ \times 10\ \text{mph}=20\ \text{mi}\\\\{{\beta }_{2}}=45{}^\circ ,{{\beta }_{1}}=180{}^\circ -120{}^\circ =60{}^\circ \\\\\beta =45{}^\circ +60{}^\circ =105{}^\circ \\\\\text{By the law of cosines,}\\\\A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}-2(AB)(BC)\cos \beta \\\\\ \ \ \ \ \ \ \,\ =\text{ }{{45}^{2}}+{{20}^{2}}-2\left( {45} \right)\left( {20} \right)\cos 105{}^\circ \\\\\ \ \ \ \ \ \ \ \ =\text{ }2025+400+1800\cos 75{}^\circ ~\\\\\ \ \ \ \ \ \ \ \ =\text{ }2425+465.9\\\\\ \ \ \ \ \ \ \ \ =\text{ }2890.9\\\\AC~~~~=\text{ }53.77\text{ mi}\end{array}$

 No Log 1800 cos 75° 3.2553 $\displaystyle \overline{1}.4130$ 465.9 2.6683

$\displaystyle \ \ \ \ \text{By the law of sines,}$

$\displaystyle \ \ \ \ \frac{{\sin \alpha }}{{BC}}=\frac{{\sin \beta }}{{AC}}$

$\displaystyle \ \ \ \ \sin \alpha ~~=~\frac{{BC}}{{AC}}\times \sin \beta$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 105{}^\circ$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{20}}{{53.77}}\times \sin 75{}^\circ \ (\sin 105{}^\circ =\sin 75{}^\circ )$

$\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ =\sin 21{}^\circ 3'$

$\displaystyle \therefore \ \ \alpha =21{}^\circ 3'$

$\displaystyle \therefore \ \ \theta =45{}^\circ +21{}^\circ 3'=66{}^\circ 3'$

 No Log 20 sin 75° 1.3010 $\displaystyle \overline{1}.9849$ 53.77 1.02859 1.7305 sin21°3′ $\displaystyle \overline{1}.5554$

Hence, the ship is $\displaystyle 53.77$ mi from the original position.

It is in the direction $\displaystyle N 66° 3' E.$

14.    (b) Find the two positive numbers $\displaystyle x$ and $\displaystyle y$ such that their sum is $\displaystyle 60$ and $\displaystyle xy^3$ is maximum.
(5 marks)

Show/Hide Solution
$\displaystyle \begin{array}{l}\ \ \ \ \ \ x+y=60\\\\\therefore \ \ \ \ y=60-x\\\\\ \ \ \ \ \ \text{Let}\ z=x{{y}^{3}}\\\\\therefore \ \ \ \ z=x{{\left( {60-x} \right)}^{3}}\end{array}$

$\displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}={{\left( {60-x} \right)}^{3}}-3x{{\left( {60-x} \right)}^{2}}$

$\displaystyle \begin{array}{l}\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-x-3x} \right)\\\\\ \ \ \ \ \ \ \ \ \ ={{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)\end{array}$

$\displaystyle \ \ \ \ \ \frac{{dz}}{{dx}}=0,\text{when}$

$\displaystyle \begin{array}{l}\ \ \ \ \ {{\left( {60-x} \right)}^{2}}\left( {60-4x} \right)=0\\\\\ \ \ \ \ x=60\ \ \text{or}\,\ x=15\\\\\ \ \ \ \ \text{When}\ x=60,\ y=0\ \text{which is impossible as }y>0.\\\\\ \ \ \ \ \text{When}\ x=15,\ y=45\ \text{which is possible}\text{.}\end{array}$

$\displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-x} \right)}^{2}}-2\left( {60-x} \right)\left( {60-4x} \right)$

$\displaystyle \ \ \ \ \ \text{When}\ x=15,$

$\displaystyle \ \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}-2\left( {60-15} \right)\left( {60-60} \right)$

$\displaystyle \therefore \ \ \ \ \frac{{{{d}^{2}}z}}{{d{{x}^{2}}}}=-4{{\left( {60-15} \right)}^{2}}<0$

$\displaystyle \therefore \ \ \ \text{z is maximum}\ \text{when}\ x=15\ \text{and}\ y=45.$