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: $\frac{14}{5}$
New: SSC CHSL Tier 2 answer key released | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
New: Unlock 10% OFF on PTE Academic. Use Code: 'C360SPL10'
Correct Answer: $\frac{119}{60}$
Solution : Given: PQ = 5 cm QR = 13 cm $\angle P=90^{\circ}$ In $\triangle PQR$, $RQ^2=PQ^2+PR^2$ ⇒ $13^2=5^2+PR^2$ ⇒ $PR^2=169-25$ ⇒ $PR^2=144$ $\therefore PR = 12$ Now, $\tan Q-\tan R$ $=\frac{12}{5}-\frac{5}{12}$ $=\frac{144-25}{60}$ $=\frac{119}{60}$ Hence, the correct answer is $\frac{119}{60}$.
Candidates can download this e-book to give a boost to thier preparation.
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Question : $\triangle PQR$ is a right-angled triangle. $\angle Q = 90^\circ$, PQ = 12 cm, and QR = 5 cm. What is the value of $\operatorname{cosec}P+\sec R$?
Question : $\triangle PQR$ is right angled at Q. If PQ = 12 cm and PR = 13 cm, find $\tan P+\cot R$.
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?
Question : If in a right-angled $\triangle P Q R$, $\tan Q=\frac{5}{12}$, then what is the value of $\cos Q$?
Question : $\triangle ABC$ is a right angled triangle, $\angle B$ = 90°, $AB$ = 12 cm, $BC$ = 5 cm. What is the value of $\cos A + \sin C$?
Regular exam updates, QnA, Predictors, College Applications & E-books now on your Mobile