Question : $\triangle PQR$ is right angled at Q. If $m\angle R=60^{\circ}$, what is the length of PR(in cm), If RQ = $4\sqrt3$ cm?
Option 1: $8$
Option 2: $4$
Option 3: $\frac{8}{\sqrt3}$
Option 4: $8\sqrt3$
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Correct Answer: $8\sqrt3$
Solution : In $\triangle$PQR, $\cos R$ = $\frac{QR}{PR}$ $\cos 60^{\circ}$ = $\frac{\text{Base}}{\text{Hypotenuse}}$ ⇒ $\frac{1}{2}$ = $\frac{4\sqrt3}{x}$ ⇒ $x$ = $8\sqrt3$ ⇒ PR = $8\sqrt3$ cm Hence, the correct answer is $8\sqrt3$.
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Question : If $\triangle{PQR} \cong \triangle{STR}, \angle {Q}=50^{\circ}$ and $\angle {P}=70^{\circ}$ and ${PQ}$ is $8 {~cm}$. Which of the following options is NOT correct?
Option 1: $\angle {TSR}=80^{\circ}$
Option 2: $\angle {PRT}=60^{\circ}$
Option 3: ${PR}={RS}$
Option 4: ${TR}={RQ}$
Question : In a right-angled $\triangle PQR$, PQ = 5 cm, QR = 13 cm and $\angle P=90^{\circ}$. Find the value of $\tan Q-\tan R$.
Option 1: $\frac{5}{14}$
Option 2: $\frac{119}{60}$
Option 3: $\frac{60}{119}$
Option 4: $\frac{14}{5}$
Question : $\triangle PQR$ is right angled at Q. If PQ = 12 cm and PR = 13 cm, find $\tan P+\cot R$.
Option 1: $\frac{9}{10}$
Option 2: $0$
Option 3: $\frac{10}{12}$
Option 4: $\frac{12}{10}$
Question : $\triangle PQR$ is a right-angled triangle. $\angle Q = 90^\circ$, PQ = 12 cm, and QR = 5 cm. What is the value of $\operatorname{cosec}P+\sec R$?
Option 1: $\frac{26}{5}$
Option 2: $\frac{13}{5}$
Option 3: $\frac{18}{5}$
Question : It is given that $\triangle \mathrm{PQR} \cong \triangle \mathrm{MNY}$ and $PQ=8\ \mathrm{cm}, \angle Q = 55°$ and $\angle P = 72°$. Which of the following is true?
Option 1: $\mathrm{NY}=8 \mathrm{~cm}, \angle \mathrm{Y}=72^{\circ}$
Option 2: $\mathrm{NM}=8 \mathrm{~cm}, \angle \mathrm{M}=53^{\circ}$
Option 3: $\mathrm{NM}=8 \mathrm{~cm}, \angle \mathrm{Y}=53^{\circ}$
Option 4: $\mathrm{NY}=8 \mathrm{~cm}, \angle \mathrm{N}=55^{\circ}$
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