\displaystyle \textbf{Exercise - 24(A)}

\displaystyle \textbf{Question 1: }\text{Find }x\text{:} \displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{In the given right-angled triangle,}
\displaystyle \sin60^\circ=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{20}{x}
\displaystyle \frac{\sqrt3}{2}=\frac{20}{x}
\displaystyle x=\frac{40}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{40\sqrt3}{3}
\displaystyle \therefore x=\frac{40\sqrt3}{3}.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \tan30^\circ=\frac{\text{Perpendicular}}{\text{Base}}=\frac{20}{x}
\displaystyle \frac{1}{\sqrt3}=\frac{20}{x}
\displaystyle x=20\sqrt3
\displaystyle \therefore x=20\sqrt3.

\displaystyle \text{(iii)}
\displaystyle \text{Answer:}
\displaystyle \sin45^\circ=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{20}{x}
\displaystyle \frac{1}{\sqrt2}=\frac{20}{x}
\displaystyle x=20\sqrt2
\displaystyle \therefore x=20\sqrt2.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Find angle }A\text{:} \displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \cos A=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{10}{20}=\frac{1}{2}
\displaystyle \cos A=\cos60^\circ
\displaystyle \therefore A=60^\circ.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \sin A=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\frac{10}{\sqrt2}}{10}
\displaystyle \sin A=\frac{1}{\sqrt2}=\sin45^\circ
\displaystyle \therefore A=45^\circ.

\displaystyle \text{(iii)}
\displaystyle \text{Answer:}
\displaystyle \tan A=\frac{\text{Perpendicular}}{\text{Base}}=\frac{10\sqrt3}{10}=\sqrt3
\displaystyle \tan A=\tan60^\circ
\displaystyle \therefore A=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find angle }x\text{:} \displaystyle \text{Answer:}
\displaystyle \text{Let the perpendicular from the top vertex to the base be }h.
\displaystyle \text{In the right-hand right-angled triangle,}
\displaystyle \tan60^\circ=\frac{30}{h}
\displaystyle \sqrt3=\frac{30}{h}
\displaystyle h=\frac{30}{\sqrt3}=10\sqrt3
\displaystyle \text{In the left-hand right-angled triangle,}
\displaystyle \sin x=\frac{10\sqrt3}{20}=\frac{\sqrt3}{2}
\displaystyle \sin x=\sin60^\circ
\displaystyle \therefore x=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find }AD\text{:} \displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{Since }BCDE\text{ is a rectangle,}
\displaystyle BE=CD=50\text{ m and }ED=BC=10\text{ m}.
\displaystyle \text{In right-angled }\triangle ABE,
\displaystyle \tan45^\circ=\frac{AE}{BE}
\displaystyle 1=\frac{AE}{50}
\displaystyle \therefore AE=50\text{ m}.
\displaystyle AD=AE+ED=50+10=60\text{ m}
\displaystyle \therefore AD=60\text{ m}.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \text{From the figure, }BC=AC\text{ and }\angle ACD=60^\circ.
\displaystyle \angle ACB=180^\circ-60^\circ=120^\circ
\displaystyle \text{Since }BC=AC,\quad \angle ABC=\angle BAC
\displaystyle \angle ABC=\frac{180^\circ-120^\circ}{2}=30^\circ
\displaystyle \text{Since }B,C,D\text{ are collinear, }\angle ABD=30^\circ.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \sin30^\circ=\frac{AD}{AB}
\displaystyle \frac{1}{2}=\frac{AD}{100}
\displaystyle \therefore AD=50\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Find the length of }AD.
\displaystyle \text{Given: }\angle ABC=60^\circ,\ \angle DBC=45^\circ\text{ and }BC=40\text{ cm}. \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AC}{BC}
\displaystyle \sqrt3=\frac{AC}{40}
\displaystyle \therefore AC=40\sqrt3\text{ cm}.
\displaystyle \text{In right-angled }\triangle DBC,
\displaystyle \tan45^\circ=\frac{DC}{BC}
\displaystyle 1=\frac{DC}{40}
\displaystyle \therefore DC=40\text{ cm}.
\displaystyle AD=AC-DC
\displaystyle =40\sqrt3-40=40(\sqrt3-1)\text{ cm}
\displaystyle \therefore AD=40(\sqrt3-1)\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Find the lengths of diagonals }AC\text{ and }BD.
\displaystyle \text{Given }AB=60\text{ cm and }\angle BAD=60^\circ. \displaystyle \text{Answer:}
\displaystyle \text{Let the diagonals }AC\text{ and }BD\text{ intersect at }O.
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \therefore OA=OC=\frac{1}{2}AC,\quad OB=OD=\frac{1}{2}BD
\displaystyle \text{Also, }AC\text{ bisects }\angle BAD.
\displaystyle \therefore \angle BAO=\frac{60^\circ}{2}=30^\circ
\displaystyle \text{In right-angled }\triangle AOB,
\displaystyle \cos30^\circ=\frac{OA}{AB}
\displaystyle \frac{\sqrt3}{2}=\frac{OA}{60}
\displaystyle \therefore OA=30\sqrt3\text{ cm}.
\displaystyle AC=2OA=2\times30\sqrt3=60\sqrt3\text{ cm}
\displaystyle \sin30^\circ=\frac{OB}{AB}
\displaystyle \frac{1}{2}=\frac{OB}{60}
\displaystyle \therefore OB=30\text{ cm}.
\displaystyle BD=2OB=2\times30=60\text{ cm}
\displaystyle \therefore AC=60\sqrt3\text{ cm and }BD=60\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{Find }AB. \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ACF,
\displaystyle \tan45^\circ=\frac{CF}{AC}=\frac{20}{AC}
\displaystyle 1=\frac{20}{AC}
\displaystyle \therefore AC=20.
\displaystyle \text{In right-angled }\triangle EDB,
\displaystyle \tan60^\circ=\frac{ED}{DB}=\frac{30}{DB}
\displaystyle \sqrt3=\frac{30}{DB}
\displaystyle \therefore DB=\frac{30}{\sqrt3}=10\sqrt3.
\displaystyle \text{Draw a perpendicular from }F\text{ to }DE\text{ meeting it at }G.
\displaystyle EG=ED-CF=30-20=10
\displaystyle \text{Also, }FG=CD.
\displaystyle \text{In right-angled }\triangle EFG,\quad \angle FEG=60^\circ,
\displaystyle \tan60^\circ=\frac{FG}{EG}=\frac{CD}{10}
\displaystyle \sqrt3=\frac{CD}{10}
\displaystyle \therefore CD=10\sqrt3.
\displaystyle AB=AC+CD+DB
\displaystyle =20+10\sqrt3+10\sqrt3
\displaystyle =20+20\sqrt3
\displaystyle \therefore AB=20(1+\sqrt3).
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{In trapezium }ABCD,\text{ as shown, }AB\parallel DC,
\displaystyle AD=DC=BC=20\text{ cm and }\angle A=60^\circ.\text{ Find:}
\displaystyle \text{(i) length of }AB
\displaystyle \text{(ii) distance between }AB\text{ and }DC. \displaystyle \text{Answer:}
\displaystyle \text{Draw }DE\perp AB\text{ and }CF\perp AB.
\displaystyle \text{Since }AD=BC,\ ABCD\text{ is an isosceles trapezium.}
\displaystyle \therefore \angle B=\angle A=60^\circ.

\displaystyle \text{(i) In right-angled }\triangle ADE,
\displaystyle \cos60^\circ=\frac{AE}{AD}
\displaystyle \frac{1}{2}=\frac{AE}{20}
\displaystyle \therefore AE=10\text{ cm}.
\displaystyle \text{Similarly, }BF=10\text{ cm}.
\displaystyle EF=DC=20\text{ cm}
\displaystyle AB=AE+EF+FB=10+20+10=40\text{ cm}
\displaystyle \therefore AB=40\text{ cm}.

\displaystyle \text{(ii) In right-angled }\triangle ADE,
\displaystyle \sin60^\circ=\frac{DE}{AD}
\displaystyle \frac{\sqrt3}{2}=\frac{DE}{20}
\displaystyle \therefore DE=10\sqrt3\text{ cm}.
\displaystyle \therefore \text{Distance between }AB\text{ and }DC=10\sqrt3\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Use the information given to find the length of }AB. \displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle AQP,
\displaystyle \tan30^\circ=\frac{AQ}{AP}
\displaystyle \frac{1}{\sqrt3}=\frac{10}{AP}
\displaystyle \therefore AP=10\sqrt3\text{ cm}.
\displaystyle \text{In right-angled }\triangle PBR,
\displaystyle \tan45^\circ=\frac{PB}{BR}
\displaystyle 1=\frac{PB}{8}
\displaystyle \therefore PB=8\text{ cm}.
\displaystyle AB=AP+PB
\displaystyle =10\sqrt3+8\text{ cm}
\displaystyle \therefore AB=(10\sqrt3+8)\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Find the length of }AB. \displaystyle \text{Answer:}
\displaystyle \text{Since }CDEB\text{ is a rectangle, }DE=CB=30\text{ cm}.
\displaystyle \text{In right-angled }\triangle DEB,
\displaystyle \tan60^\circ=\frac{EB}{DE}
\displaystyle \sqrt3=\frac{EB}{30}
\displaystyle \therefore EB=30\sqrt3\text{ cm}.
\displaystyle \text{In right-angled }\triangle DEA,
\displaystyle \tan45^\circ=\frac{AE}{DE}
\displaystyle 1=\frac{AE}{30}
\displaystyle \therefore AE=30\text{ cm}.
\displaystyle AB=AE+EB
\displaystyle =30+30\sqrt3=30(1+\sqrt3)\text{ cm}
\displaystyle \therefore AB=30(1+\sqrt3)\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{In the given figure, }AB\text{ and }EC\text{ are parallel to each other.}
\displaystyle \text{Sides }AD\text{ and }BC\text{ are }2\text{ cm each and are perpendicular to }AB.
\displaystyle \text{Given that }\angle AED=60^\circ\text{ and }\angle ACD=45^\circ;\text{ calculate:}
\displaystyle \text{(i) }AB\qquad\text{(ii) }AC\qquad\text{(iii) }AE \displaystyle \text{Answer:}
\displaystyle \text{Given }AD=BC=2\text{ cm}.

\displaystyle \text{(i) In right-angled }\triangle ADC,
\displaystyle \tan45^\circ=\frac{AD}{DC}
\displaystyle 1=\frac{2}{DC}
\displaystyle \therefore DC=2\text{ cm}.
\displaystyle \text{Since }ABCD\text{ is a rectangle, }AB=DC.
\displaystyle \therefore AB=2\text{ cm}.

\displaystyle \text{(ii) In right-angled }\triangle ADC,
\displaystyle \sin45^\circ=\frac{AD}{AC}
\displaystyle \frac{1}{\sqrt2}=\frac{2}{AC}
\displaystyle \therefore AC=2\sqrt2\text{ cm}.

\displaystyle \text{(iii) In right-angled }\triangle AED,
\displaystyle \sin60^\circ=\frac{AD}{AE}
\displaystyle \frac{\sqrt3}{2}=\frac{2}{AE}
\displaystyle AE=\frac{4}{\sqrt3}=\frac{4\sqrt3}{3}\text{ cm}
\displaystyle \therefore AE=\frac{4\sqrt3}{3}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{In the given figure, }\angle B=60^\circ,\ AB=8\text{ cm}
\displaystyle \text{and }BC=25\text{ cm. Calculate:}
\displaystyle \text{(i) }BE\qquad\text{(ii) }AC \displaystyle \text{Answer:}
\displaystyle \text{Given }AE\perp BC.

\displaystyle \text{(i) In right-angled }\triangle ABE,
\displaystyle \cos60^\circ=\frac{BE}{AB}
\displaystyle \frac{1}{2}=\frac{BE}{8}
\displaystyle \therefore BE=4\text{ cm}.

\displaystyle \text{(ii) In right-angled }\triangle ABE,
\displaystyle \sin60^\circ=\frac{AE}{AB}
\displaystyle \frac{\sqrt3}{2}=\frac{AE}{8}
\displaystyle \therefore AE=4\sqrt3\text{ cm}.
\displaystyle EC=BC-BE=25-4=21\text{ cm}
\displaystyle \text{In right-angled }\triangle AEC,
\displaystyle AC^2=AE^2+EC^2
\displaystyle =(4\sqrt3)^2+21^2=48+441=489
\displaystyle \therefore AC=\sqrt{489}\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{Find:}
\displaystyle \text{(i) }BC\qquad\text{(ii) }AD\qquad\text{(iii) }AC \displaystyle \text{Answer:}
\displaystyle \text{Given }AB=12\text{ cm},\ \angle C=30^\circ\text{ and }\angle B=90^\circ.

\displaystyle \text{(i) In right-angled }\triangle ABC,
\displaystyle \tan30^\circ=\frac{AB}{BC}
\displaystyle \frac{1}{\sqrt3}=\frac{12}{BC}
\displaystyle \therefore BC=12\sqrt3\text{ cm}.

\displaystyle \text{(ii) }\angle A=180^\circ-90^\circ-30^\circ=60^\circ
\displaystyle \text{Since }D\text{ lies on }AC,\quad \angle BAD=60^\circ.
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \cos60^\circ=\frac{AD}{AB}
\displaystyle \frac{1}{2}=\frac{AD}{12}
\displaystyle \therefore AD=6\text{ cm}.

\displaystyle \text{(iii) In right-angled }\triangle ABC,
\displaystyle \sin30^\circ=\frac{AB}{AC}
\displaystyle \frac{1}{2}=\frac{12}{AC}
\displaystyle \therefore AC=24\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{In right-angled triangle }ABC,\ \angle B=90^\circ.\text{ Find the magnitude of angle }A,\text{ if:}
\displaystyle \text{(i) }AB\text{ is }\sqrt3\text{ times of }BC.
\displaystyle \text{(ii) }BC\text{ is }\sqrt3\text{ times of }AB.
\displaystyle \text{Answer:}

\displaystyle \text{(i) }AB=\sqrt3\,BC
\displaystyle \tan A=\frac{BC}{AB}
\displaystyle =\frac{BC}{\sqrt3\,BC}=\frac{1}{\sqrt3}
\displaystyle \tan A=\tan30^\circ
\displaystyle \therefore A=30^\circ.

\displaystyle \text{(ii) }BC=\sqrt3\,AB
\displaystyle \tan A=\frac{BC}{AB}
\displaystyle =\frac{\sqrt3\,AB}{AB}=\sqrt3
\displaystyle \tan A=\tan60^\circ
\displaystyle \therefore A=60^\circ.
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{A ladder is placed against a vertical tower. If the ladder makes an angle of }
\displaystyle 30^\circ\text{ with the ground and reaches up to a height of }15\text{ m of the tower, find the}
\displaystyle \text{length of the ladder.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the length of the ladder }=l\text{ m}.
\displaystyle \sin30^\circ=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{15}{l}
\displaystyle \frac{1}{2}=\frac{15}{l}
\displaystyle \therefore l=30\text{ m}.
\displaystyle \therefore \text{The length of the ladder is }30\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{A kite is attached to a }100\text{ m long string. Find the greatest height reached}
\displaystyle \text{by the kite when its string makes an angle of }60^\circ\text{ with the level ground.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the greatest height reached by the kite }=h\text{ m}.
\displaystyle \sin60^\circ=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{h}{100}
\displaystyle \frac{\sqrt3}{2}=\frac{h}{100}
\displaystyle h=50\sqrt3\text{ m}
\displaystyle =50\times1.732=86.6\text{ m}
\displaystyle \therefore \text{The greatest height reached by the kite is }86.6\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{Find }AB\text{ and }BC,\text{ if:} \displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{Let }BC=x\text{ cm. Then }BD=(x+20)\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan45^\circ=\frac{AB}{BC}
\displaystyle 1=\frac{AB}{x}
\displaystyle \therefore AB=x
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}
\displaystyle \frac{1}{\sqrt3}=\frac{x}{x+20}
\displaystyle \sqrt3x=x+20
\displaystyle x(\sqrt3-1)=20
\displaystyle x=\frac{20}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}
\displaystyle =10(\sqrt3+1)\text{ cm}
\displaystyle \therefore BC=10(\sqrt3+1)\text{ cm}
\displaystyle \text{and }AB=10(\sqrt3+1)\text{ cm}.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \text{Let }BC=x\text{ cm. Then }BD=(x+20)\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}
\displaystyle \sqrt3=\frac{AB}{x}
\displaystyle \therefore AB=\sqrt3x
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan30^\circ=\frac{AB}{BD}
\displaystyle \frac{1}{\sqrt3}=\frac{\sqrt3x}{x+20}
\displaystyle x+20=3x
\displaystyle 2x=20
\displaystyle \therefore x=10\text{ cm}.
\displaystyle \therefore BC=10\text{ cm}
\displaystyle \text{and }AB=10\sqrt3\text{ cm}.

\displaystyle \text{(iii)}
\displaystyle \text{Answer:}
\displaystyle \text{Let }BC=x\text{ cm. Then }BD=(x+20)\text{ cm}.
\displaystyle \text{In right-angled }\triangle ABC,
\displaystyle \tan60^\circ=\frac{AB}{BC}
\displaystyle \sqrt3=\frac{AB}{x}
\displaystyle \therefore AB=\sqrt3x
\displaystyle \text{In right-angled }\triangle ABD,
\displaystyle \tan45^\circ=\frac{AB}{BD}
\displaystyle 1=\frac{\sqrt3x}{x+20}
\displaystyle \sqrt3x=x+20
\displaystyle x(\sqrt3-1)=20
\displaystyle x=\frac{20}{\sqrt3-1}\times\frac{\sqrt3+1}{\sqrt3+1}
\displaystyle =10(\sqrt3+1)\text{ cm}
\displaystyle \therefore BC=10(\sqrt3+1)\text{ cm}.
\displaystyle AB=\sqrt3\times10(\sqrt3+1)
\displaystyle =10(3+\sqrt3)\text{ cm}
\displaystyle \therefore AB=10(3+\sqrt3)\text{ cm}.
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{Find }PQ,\text{ if }AB=150\text{ m},\ \angle P=30^\circ\text{ and }\angle Q=45^\circ. \displaystyle \text{(i)}
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABP,
\displaystyle \tan30^\circ=\frac{AB}{PB}
\displaystyle \frac{1}{\sqrt3}=\frac{150}{PB}
\displaystyle \therefore PB=150\sqrt3\text{ m}.
\displaystyle \text{In right-angled }\triangle ABQ,
\displaystyle \tan45^\circ=\frac{AB}{BQ}
\displaystyle 1=\frac{150}{BQ}
\displaystyle \therefore BQ=150\text{ m}.
\displaystyle PQ=PB+BQ
\displaystyle =150\sqrt3+150=150(\sqrt3+1)\text{ m}
\displaystyle \therefore PQ=150(\sqrt3+1)\text{ m}.

\displaystyle \text{(ii)}
\displaystyle \text{Answer:}
\displaystyle \text{In right-angled }\triangle ABP,
\displaystyle \tan30^\circ=\frac{AB}{PB}
\displaystyle \frac{1}{\sqrt3}=\frac{150}{PB}
\displaystyle \therefore PB=150\sqrt3\text{ m}.
\displaystyle \text{In right-angled }\triangle ABQ,
\displaystyle \tan45^\circ=\frac{AB}{BQ}
\displaystyle 1=\frac{150}{BQ}
\displaystyle \therefore BQ=150\text{ m}.
\displaystyle PQ=PB-BQ
\displaystyle =150\sqrt3-150=150(\sqrt3-1)\text{ m}
\displaystyle \therefore PQ=150(\sqrt3-1)\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{If }\tan x^\circ=\frac{5}{12},\ \tan y^\circ=\frac{3}{4}\text{ and }AB=48\text{ m},
\displaystyle \text{find the length of }CD. \displaystyle \text{Answer:}
\displaystyle \text{Let }CD=h\text{ m}.
\displaystyle \text{In right-angled }\triangle ACD,
\displaystyle \tan x^\circ=\frac{CD}{AC}
\displaystyle \frac{5}{12}=\frac{h}{AC}
\displaystyle \therefore AC=\frac{12h}{5}.
\displaystyle \text{In right-angled }\triangle BCD,
\displaystyle \tan y^\circ=\frac{CD}{BC}
\displaystyle \frac{3}{4}=\frac{h}{BC}
\displaystyle \therefore BC=\frac{4h}{3}.
\displaystyle AB=AC-BC
\displaystyle 48=\frac{12h}{5}-\frac{4h}{3}
\displaystyle 48=\frac{36h-20h}{15}=\frac{16h}{15}
\displaystyle h=48\times\frac{15}{16}=45
\displaystyle \therefore CD=45\text{ m}.
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The perimeter of a rhombus is }96\text{ cm and an obtuse angle is }120^\circ.
\displaystyle \text{Find the lengths of its diagonals.}
\displaystyle \text{Answer:}
\displaystyle \text{Length of each side}=\frac{96}{4}=24\text{ cm}.
\displaystyle \text{Let the diagonals intersect at }O.
\displaystyle \text{The diagonals of a rhombus bisect each other at right angles.}
\displaystyle \text{They also bisect the angles of the rhombus.}
\displaystyle \therefore \text{Half of the }120^\circ\text{ angle}=60^\circ.
\displaystyle \text{In the right-angled triangle formed by a side and the half-diagonals,}
\displaystyle \cos60^\circ=\frac{\text{Half of one diagonal}}{24}
\displaystyle \frac{1}{2}=\frac{\text{Half of one diagonal}}{24}
\displaystyle \therefore \text{Half of one diagonal}=12\text{ cm}.
\displaystyle \therefore \text{One diagonal}=24\text{ cm}.
\displaystyle \sin60^\circ=\frac{\text{Half of the other diagonal}}{24}
\displaystyle \frac{\sqrt3}{2}=\frac{\text{Half of the other diagonal}}{24}
\displaystyle \therefore \text{Half of the other diagonal}=12\sqrt3\text{ cm}.
\displaystyle \therefore \text{Other diagonal}=24\sqrt3\text{ cm}.
\displaystyle \therefore \text{The lengths of the diagonals are }24\text{ cm and }24\sqrt3\text{ cm}.
\displaystyle \\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.