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: $\frac{13}{5}$
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Correct Answer: $\frac{13}{5}$
Solution : By Pythagoras theorem, KL2 = KM2 + ML2 ⇒ KL = $\sqrt{12^2 + 5^2}$ = $\sqrt {144+69}$ = $\sqrt{169}$ = 13 $\therefore\sec$ L = $\frac{\text{KL}}{\text{ML}}$ = $\frac{13}{5}$ Hence, the correct answer is $\frac{13}{5}$.
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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: $\frac{14}{5}$
Question : $\triangle{ABC}$ is a right angled triangle. $\angle \mathrm{C}=90°$, AB = 25 cm and BC = 20 cm. What is the value of $\mathrm{sec}\; A$?
Option 1: $\frac{5}{3}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{4}{3}$
Option 4: $\frac{5}{4}$
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 : 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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