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: $\frac{18}{5}$
Option 4: $\frac{14}{5}$
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Correct Answer: $\frac{26}{5}$
Solution : As $PR^2=PQ^2+QR^2$ ⇒ $PR^2=12^2+5^2$ ⇒ $PR^2=144+25$ ⇒ $PR^2=169$ ⇒ $PR=\sqrt{169}$ ⇒ $PR=13$ cm Now, $\operatorname{cosec} \ P=\frac{PR}{QR}=\frac{13}{5}$ and $\sec R=\frac{PR}{QR}=\frac{13}{5}$ $\therefore \operatorname{cosec} \ P+\sec R=\frac{13}{5}+\frac{13}{5}=\frac{26}{5}$ Hence, the correct answer is $\frac{26}{5}$.
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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: $\frac{60}{119}$
Question : $\triangle $KLM is a right-angled triangle. $\angle$M = 90$^{\circ}$, KM = 12 cm and LM = 5 cm. What is the value of $\sec$ L?
Option 1: $\frac{13}{12}$
Option 2: $\frac{5}{12}$
Option 3: $\frac{12}{5}$
Option 4: $\frac{13}{5}$
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 : If in a right-angled $\triangle P Q R$, $\tan Q=\frac{5}{12}$, then what is the value of $\cos Q$?
Option 1: $\frac{12}{5}$
Option 3: $\frac{12}{13}$
Option 4: $\frac{5}{13}$
Question : Let ABC be a triangle right-angled at B. If $\tan A = \frac{12}{5}$, then find the values of $\operatorname{cosec A}$ and $\sec A$, respectively.
Option 1: $\frac{13}{10}, \frac{5}{13}$
Option 2: $\frac{13}{12},\frac{13}{5}$
Option 3: $\frac{10}{13}, \frac{5}{13}$
Option 4: $\frac{12}{13}, \frac{5}{13}$
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