\displaystyle \textbf{Question 1: }\text{In }\triangle ABC,\text{ if }a=5,\ b=6\text{ and }C=60^\circ,\text{ show that the area is}
\displaystyle \frac{15\sqrt{3}}{2}\text{ sq. units.}
\displaystyle \text{Answer:}
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}ab\sin C
\displaystyle =\frac{1}{2}\times5\times6\times\sin60^\circ
\displaystyle =15\times\frac{\sqrt{3}}{2}
\displaystyle =\frac{15\sqrt{3}}{2}\text{ sq. units.}
\displaystyle \therefore \text{The area of }\triangle ABC\text{ is }\frac{15\sqrt{3}}{2}\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{In }\triangle ABC,\text{ if }a=\sqrt{2},\ b=\sqrt{3}\text{ and }c=\sqrt{5},\text{ show that the area is}
\displaystyle \frac{\sqrt{6}}{2}\text{ sq. units.}
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule,}
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}
\displaystyle =\frac{2+3-5}{2\sqrt{2}\sqrt{3}}=0
\displaystyle \therefore \sin C=\sqrt{1-\cos^2C}=\sqrt{1-0}=1
\displaystyle \text{Area of }\triangle ABC=\frac{1}{2}ab\sin C
\displaystyle =\frac{1}{2}\times\sqrt{2}\times\sqrt{3}\times1
\displaystyle =\frac{\sqrt{6}}{2}\text{ sq. units.}
\displaystyle \therefore \text{The area of }\triangle ABC\text{ is }\frac{\sqrt{6}}{2}\text{ sq. units.}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The sides of a triangle are }a=4,\ b=6\text{ and }c=8.\text{ Show that:}
\displaystyle 8\cos A+16\cos B+4\cos C=17.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=4,\ b=6\text{ and }c=8
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}
\displaystyle =\frac{36+64-16}{2\times6\times8}
\displaystyle =\frac{84}{96}=\frac{7}{8}
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}
\displaystyle =\frac{16+64-36}{2\times4\times8}
\displaystyle =\frac{44}{64}=\frac{11}{16}
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}
\displaystyle =\frac{16+36-64}{2\times4\times6}
\displaystyle =\frac{-12}{48}=-\frac{1}{4}
\displaystyle \text{LHS}=8\cos A+16\cos B+4\cos C
\displaystyle =8\left(\frac{7}{8}\right)+16\left(\frac{11}{16}\right)+4\left(-\frac{1}{4}\right)
\displaystyle =7+11-1
\displaystyle =17=\text{RHS}
\displaystyle \therefore 8\cos A+16\cos B+4\cos C=17.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{In }\triangle ABC,\text{ if }a=18,\ b=24,\ c=30,\text{ find }\cos A,\ \cos B\text{ and }\cos C.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }a=18,\ b=24\text{ and }c=30
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}
\displaystyle =\frac{576+900-324}{2\times24\times30}
\displaystyle =\frac{1152}{1440}=\frac{4}{5}
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}
\displaystyle =\frac{324+900-576}{2\times18\times30}
\displaystyle =\frac{648}{1080}=\frac{3}{5}
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}
\displaystyle =\frac{324+576-900}{2\times18\times24}=0
\displaystyle \therefore \cos A=\frac{4}{5},\ \cos B=\frac{3}{5}\text{ and }\cos C=0.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Prove that }b(c\cos A-a\cos C)=c^2-a^2.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B,\ c=k\sin C.
\displaystyle \text{RHS}=c^2-a^2
\displaystyle =k^2\sin^2C-k^2\sin^2A
\displaystyle =k^2(\sin^2C-\sin^2A)
\displaystyle =k^2\sin(C+A)\sin(C-A)
\displaystyle =k^2\sin(\pi-B)\sin(C-A)
\displaystyle =k^2\sin B\sin(C-A)
\displaystyle =bk\sin(C-A)
\displaystyle =bk(\sin C\cos A-\cos C\sin A)
\displaystyle =b(k\sin C\cos A-k\sin A\cos C)
\displaystyle =b(c\cos A-a\cos C)
\displaystyle =\text{LHS}
\displaystyle \therefore b(c\cos A-a\cos C)=c^2-a^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Prove that }c(a\cos B-b\cos A)=a^2-b^2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=c(a\cos B-b\cos A)
\displaystyle =ac\cos B-bc\cos A
\displaystyle =ac\left(\frac{a^2+c^2-b^2}{2ac}\right)-bc\left(\frac{b^2+c^2-a^2}{2bc}\right)
\displaystyle =\frac{1}{2}\left(a^2+c^2-b^2-b^2-c^2+a^2\right)
\displaystyle =\frac{1}{2}(2a^2-2b^2)
\displaystyle =a^2-b^2=\text{RHS}
\displaystyle \therefore c(a\cos B-b\cos A)=a^2-b^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Prove that }2(bc\cos A+ca\cos B+ab\cos C)=a^2+b^2+c^2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=2(bc\cos A+ca\cos B+ab\cos C)
\displaystyle =2bc\left(\frac{b^2+c^2-a^2}{2bc}\right)+2ac\left(\frac{a^2+c^2-b^2}{2ac}\right)+2ab\left(\frac{a^2+b^2-c^2}{2ab}\right)
\displaystyle =(b^2+c^2-a^2)+(a^2+c^2-b^2)+(a^2+b^2-c^2)
\displaystyle =a^2+b^2+c^2=\text{RHS}
\displaystyle \therefore 2(bc\cos A+ca\cos B+ab\cos C)=a^2+b^2+c^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{Prove that}
\displaystyle (c^2-a^2+b^2)\tan A=(a^2-b^2+c^2)\tan B   =(b^2-c^2+a^2)\tan C.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, }\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k.
\displaystyle \therefore \sin A=ak,\quad\sin B=bk,\quad\sin C=ck.
\displaystyle \text{Also, by the cosine rule,}
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}=\frac{c^2-a^2+b^2}{2bc}.
\displaystyle (c^2-a^2+b^2)\tan A
\displaystyle =(c^2-a^2+b^2)\frac{\sin A}{\cos A}
\displaystyle =(c^2-a^2+b^2)\frac{ak\times2bc}{c^2-a^2+b^2}
\displaystyle =2abck.
\displaystyle \cos B=\frac{a^2+c^2-b^2}{2ac}=\frac{a^2-b^2+c^2}{2ac}.
\displaystyle (a^2-b^2+c^2)\tan B
\displaystyle =(a^2-b^2+c^2)\frac{\sin B}{\cos B}
\displaystyle =(a^2-b^2+c^2)\frac{bk\times2ac}{a^2-b^2+c^2}
\displaystyle =2abck.
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}=\frac{b^2-c^2+a^2}{2ab}.
\displaystyle (b^2-c^2+a^2)\tan C
\displaystyle =(b^2-c^2+a^2)\frac{\sin C}{\cos C}
\displaystyle =(b^2-c^2+a^2)\frac{ck\times2ab}{b^2-c^2+a^2}
\displaystyle =2abck.
\displaystyle \therefore (c^2-a^2+b^2)\tan A=(a^2-b^2+c^2)\tan B
\displaystyle =(b^2-c^2+a^2)\tan C.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Prove that }\frac{c-b\cos A}{b-c\cos A}=\frac{\cos B}{\cos C}.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B,\ c=k\sin C.
\displaystyle \text{LHS}=\frac{c-b\cos A}{b-c\cos A}
\displaystyle =\frac{k\sin C-k\sin B\cos A}{k\sin B-k\sin C\cos A}
\displaystyle =\frac{\sin C-\sin B\cos A}{\sin B-\sin C\cos A}.
\displaystyle \text{Since }A+B+C=\pi,
\displaystyle \sin C=\sin(A+B)\quad\text{and}\quad\sin B=\sin(A+C).
\displaystyle \therefore \text{LHS}=\frac{\sin(A+B)-\sin B\cos A}{\sin(A+C)-\sin C\cos A}
\displaystyle =\frac{\sin A\cos B+\cos A\sin B-\sin B\cos A}{\sin A\cos C+\cos A\sin C-\sin C\cos A}
\displaystyle =\frac{\sin A\cos B}{\sin A\cos C}
\displaystyle =\frac{\cos B}{\cos C}=\text{RHS}.
\displaystyle \therefore \frac{c-b\cos A}{b-c\cos A}=\frac{\cos B}{\cos C}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Prove that}
\displaystyle a(\cos B+\cos C-1)+b(\cos C+\cos A-1)   +c(\cos A+\cos B-1)=0.
\displaystyle \text{Answer:}
\displaystyle \text{In }\triangle ABC,\text{ we have}
\displaystyle a=b\cos C+c\cos B,
\displaystyle b=c\cos A+a\cos C,
\displaystyle c=a\cos B+b\cos A.
\displaystyle \text{LHS}=a(\cos B+\cos C-1)+b(\cos C+\cos A-1)
\displaystyle \qquad+c(\cos A+\cos B-1)
\displaystyle =a\cos B+a\cos C+b\cos C+b\cos A
\displaystyle \qquad+c\cos A+c\cos B-(a+b+c)
\displaystyle =(b\cos C+c\cos B)+(c\cos A+a\cos C)
\displaystyle \qquad+(a\cos B+b\cos A)-(a+b+c)
\displaystyle =a+b+c-(a+b+c)
\displaystyle =0=\text{RHS}.
\displaystyle \therefore a(\cos B+\cos C-1)+b(\cos C+\cos A-1)
\displaystyle \qquad+c(\cos A+\cos B-1)=0.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Prove that }a\cos A+b\cos B+c\cos C=2b\sin A\sin C.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }a=k\sin A,\ b=k\sin B,\ c=k\sin C.
\displaystyle \text{LHS}=a\cos A+b\cos B+c\cos C
\displaystyle =k\sin A\cos A+k\sin B\cos B+k\sin C\cos C
\displaystyle =\frac{k}{2}(\sin2A+\sin2B+\sin2C)
\displaystyle =\frac{k}{2}\left[(\sin2A+\sin2C)+\sin2B\right]
\displaystyle =\frac{k}{2}\left[2\sin(A+C)\cos(A-C)+2\sin B\cos B\right]
\displaystyle =k\left[\sin B\cos(A-C)+\sin B\cos B\right]
\displaystyle =k\sin B\left[\cos(A-C)+\cos B\right]
\displaystyle \text{Since }B=\pi-(A+C),\quad\cos B=-\cos(A+C).
\displaystyle \therefore \text{LHS}=k\sin B\left[\cos(A-C)-\cos(A+C)\right]
\displaystyle =k\sin B(2\sin A\sin C)
\displaystyle =2(k\sin B)\sin A\sin C
\displaystyle =2b\sin A\sin C
\displaystyle =\text{RHS}.
\displaystyle \therefore a\cos A+b\cos B+c\cos C=2b\sin A\sin C.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{Prove that }a^2=(b+c)^2-4bc\cos^2\frac{A}{2}.
\displaystyle \text{Answer:}
\displaystyle \text{By the cosine rule,}
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}
\displaystyle \Rightarrow 2bc\cos A=b^2+c^2-a^2
\displaystyle \Rightarrow a^2=b^2+c^2-2bc\cos A
\displaystyle =b^2+c^2-2bc\left(2\cos^2\frac{A}{2}-1\right)
\displaystyle =b^2+c^2+2bc-4bc\cos^2\frac{A}{2}
\displaystyle =(b+c)^2-4bc\cos^2\frac{A}{2}
\displaystyle \therefore a^2=(b+c)^2-4bc\cos^2\frac{A}{2}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Prove that }    \\ 4\left(bc\cos^2\frac{A}{2}+ca\cos^2\frac{B}{2}+ab\cos^2\frac{C}{2}\right)   =(a+b+c)^2.
\displaystyle \text{Answer:}
\displaystyle \text{LHS}=4\left(bc\cos^2\frac{A}{2}+ca\cos^2\frac{B}{2}+ab\cos^2\frac{C}{2}\right)
\displaystyle =2\left(2bc\cos^2\frac{A}{2}+2ca\cos^2\frac{B}{2}+2ab\cos^2\frac{C}{2}\right)
\displaystyle =2\left[bc(1+\cos A)+ca(1+\cos B)+ab(1+\cos C)\right]
\displaystyle =2bc+2bc\cos A+2ca+2ca\cos B+2ab+2ab\cos C
\displaystyle =2bc+(b^2+c^2-a^2)+2ca+(c^2+a^2-b^2)
\displaystyle \qquad+2ab+(a^2+b^2-c^2)
\displaystyle =a^2+b^2+c^2+2ab+2bc+2ca
\displaystyle =(a+b+c)^2
\displaystyle =\text{RHS}.
\displaystyle \therefore 4\left(bc\cos^2\frac{A}{2}+ca\cos^2\frac{B}{2}+ab\cos^2\frac{C}{2}\right)
\displaystyle \qquad=(a+b+c)^2.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In }\triangle ABC,\text{ prove that}
\displaystyle \sin^3A\cos(B-C)+\sin^3B\cos(C-A)   +\sin^3C\cos(A-B)=3\sin A\sin B\sin C.
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule, let }\sin A=ka,\quad\sin B=kb,\quad\sin C=kc.
\displaystyle \text{LHS}=\sin^3A\cos(B-C)+\sin^3B\cos(C-A)
\displaystyle \qquad+\sin^3C\cos(A-B)
\displaystyle =\sin^2A\sin(B+C)\cos(B-C)
\displaystyle \qquad+\sin^2B\sin(C+A)\cos(C-A)
\displaystyle \qquad+\sin^2C\sin(A+B)\cos(A-B)
\displaystyle =\frac{1}{2}\left[\sin^2A(\sin2B+\sin2C)\right.
\displaystyle \qquad+\sin^2B(\sin2C+\sin2A)
\displaystyle \left.\qquad+\sin^2C(\sin2A+\sin2B)\right]
\displaystyle =\sin^2A(\sin B\cos B+\sin C\cos C)
\displaystyle \qquad+\sin^2B(\sin C\cos C+\sin A\cos A)
\displaystyle \qquad+\sin^2C(\sin A\cos A+\sin B\cos B)
\displaystyle =k^3\left[a^2b\cos B+a^2c\cos C+b^2c\cos C\right.
\displaystyle \left.\qquad+b^2a\cos A+c^2a\cos A+c^2b\cos B\right]
\displaystyle =k^3ab(a\cos B+b\cos A)
\displaystyle \qquad+k^3ac(a\cos C+c\cos A)
\displaystyle \qquad+k^3bc(b\cos C+c\cos B)
\displaystyle \text{Using }c=a\cos B+b\cos A,\quad b=a\cos C+c\cos A,
\displaystyle \text{and }a=b\cos C+c\cos B,
\displaystyle \text{LHS}=k^3abc+k^3abc+k^3abc
\displaystyle =3(ka)(kb)(kc)
\displaystyle =3\sin A\sin B\sin C
\displaystyle =\text{RHS}.
\displaystyle \therefore \sin^3A\cos(B-C)+\sin^3B\cos(C-A)
\displaystyle \qquad+\sin^3C\cos(A-B)=3\sin A\sin B\sin C.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{In }\triangle ABC,\text{ if }\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15},
\displaystyle \text{prove that }\frac{\cos A}{2}=\frac{\cos B}{7}=\frac{\cos C}{11}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\frac{b+c}{12}=\frac{c+a}{13}=\frac{a+b}{15}=k.
\displaystyle \therefore b+c=12k,\quad c+a=13k,\quad a+b=15k.
\displaystyle \text{Adding the three equations,}
\displaystyle 2(a+b+c)=12k+13k+15k=40k
\displaystyle \Rightarrow a+b+c=20k.
\displaystyle a=20k-(b+c)=20k-12k=8k
\displaystyle b=20k-(c+a)=20k-13k=7k
\displaystyle c=20k-(a+b)=20k-15k=5k.
\displaystyle \cos A=\frac{b^2+c^2-a^2}{2bc}
\displaystyle =\frac{49k^2+25k^2-64k^2}{2\times7k\times5k}
\displaystyle =\frac{10k^2}{70k^2}=\frac{1}{7}.
\displaystyle \cos B=\frac{c^2+a^2-b^2}{2ca}
\displaystyle =\frac{25k^2+64k^2-49k^2}{2\times5k\times8k}
\displaystyle =\frac{40k^2}{80k^2}=\frac{1}{2}.
\displaystyle \cos C=\frac{a^2+b^2-c^2}{2ab}
\displaystyle =\frac{64k^2+49k^2-25k^2}{2\times8k\times7k}
\displaystyle =\frac{88k^2}{112k^2}=\frac{11}{14}.
\displaystyle \therefore \frac{\cos A}{2}=\frac{1}{14},\quad\frac{\cos B}{7}=\frac{1}{14},
\displaystyle \frac{\cos C}{11}=\frac{1}{14}.
\displaystyle \therefore \frac{\cos A}{2}=\frac{\cos B}{7}=\frac{\cos C}{11}.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{In }\triangle ABC,\text{ if }\angle B=60^\circ,\text{ prove that }   (a+b+c)(a-b+c)=3ca.
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\angle B=60^\circ
\displaystyle \therefore \cos B=\frac{1}{2}
\displaystyle \frac{a^2+c^2-b^2}{2ac}=\frac{1}{2}
\displaystyle \Rightarrow a^2+c^2-b^2=ac
\displaystyle \Rightarrow b^2=a^2+c^2-ac.
\displaystyle \text{LHS}=(a+b+c)(a-b+c)
\displaystyle =[(a+c)+b][(a+c)-b]
\displaystyle =(a+c)^2-b^2
\displaystyle =a^2+2ac+c^2-(a^2+c^2-ac)
\displaystyle =3ac
\displaystyle =\text{RHS}.
\displaystyle \therefore (a+b+c)(a-b+c)=3ca.
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{In }\triangle ABC,\text{ if }\cos^2A+\cos^2B+\cos^2C=1,
\displaystyle \text{prove that the triangle is right-angled.}
\displaystyle \text{Answer:}
\displaystyle \text{Given, }\cos^2A+\cos^2B+\cos^2C=1
\displaystyle \Rightarrow \cos^2A+\cos^2B-(1-\cos^2C)=0
\displaystyle \Rightarrow \cos^2A+\cos^2B-\sin^2C=0.
\displaystyle \text{Since }C=\pi-(A+B),\quad\sin C=\sin(A+B).
\displaystyle \therefore \cos^2A+\cos^2B-\sin^2(A+B)=0
\displaystyle \Rightarrow \cos^2A+\cos^2B
\displaystyle \qquad-(\sin A\cos B+\cos A\sin B)^2=0
\displaystyle \Rightarrow \cos^2A+\cos^2B-\sin^2A\cos^2B
\displaystyle \qquad-\cos^2A\sin^2B-2\sin A\sin B\cos A\cos B=0
\displaystyle \Rightarrow 2\cos^2A\cos^2B-2\sin A\sin B\cos A\cos B=0
\displaystyle \Rightarrow 2\cos A\cos B(\cos A\cos B-\sin A\sin B)=0
\displaystyle \Rightarrow 2\cos A\cos B\cos(A+B)=0
\displaystyle \Rightarrow -2\cos A\cos B\cos C=0
\displaystyle \Rightarrow \cos A\cos B\cos C=0.
\displaystyle \therefore \cos A=0,\quad\cos B=0\quad\text{or}\quad\cos C=0.
\displaystyle \therefore A=90^\circ,\quad B=90^\circ\quad\text{or}\quad C=90^\circ.
\displaystyle \therefore \triangle ABC\text{ is a right-angled triangle.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{In }\triangle ABC,\text{ if }\cos C=\frac{\sin A}{2\sin B},   \text{ prove that the triangle is isosceles.}
\displaystyle \text{Answer:}
\displaystyle \text{By the Sine Rule,}
\displaystyle \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}=k
\displaystyle \therefore \sin A=ak,\quad\sin B=bk,\quad\sin C=ck.
\displaystyle \text{Given, }\cos C=\frac{\sin A}{2\sin B}
\displaystyle \Rightarrow 2\sin B\cos C=\sin A
\displaystyle \Rightarrow 2bk\left(\frac{a^2+b^2-c^2}{2ab}\right)=ak
\displaystyle \Rightarrow \frac{k(a^2+b^2-c^2)}{a}=ak
\displaystyle \Rightarrow a^2+b^2-c^2=a^2
\displaystyle \Rightarrow b^2=c^2
\displaystyle \Rightarrow b=c.
\displaystyle \therefore \triangle ABC\text{ is an isosceles triangle.}
\displaystyle \text{Hence proved.}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{Two ships leave a port at the same time. One goes }24\text{ km/hr in the}
\displaystyle \text{direction of }N38^\circ E\text{ and the other travels }32\text{ km/hr in the direction of}
\displaystyle S52^\circ E.\text{ Find the distance between the ships at the end of }3\text{ hours.}
\displaystyle \text{Answer:} \displaystyle \text{Let }P\text{ and }Q\text{ be the positions of the two ships after }3\text{ hours.}
\displaystyle OP=3\times24=72\text{ km}
\displaystyle OQ=3\times32=96\text{ km}
\displaystyle \angle POQ=38^\circ+52^\circ=90^\circ
\displaystyle \text{Using the cosine rule in }\triangle OPQ,
\displaystyle PQ^2=OP^2+OQ^2-2(OP)(OQ)\cos90^\circ
\displaystyle =72^2+96^2-2\times72\times96\times0
\displaystyle =5184+9216=14400
\displaystyle \therefore PQ=\sqrt{14400}=120\text{ km}
\displaystyle \therefore \text{The distance between the two ships after }3\text{ hours is }120\text{ km.}
\displaystyle \\

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