Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

State True or False.
In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. The values of sin P = $513$ ,
cos P = $1213$ and tan P = $512$ .

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.

An Intiative by Sri Chaitanya

True

False

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Given that in ∆ PQR, right-angled at Q,
PR + QR = 25 cm ……(1)
and PQ = 5 cm.
We know that according to Pythagoras theorem,

h y p o t e n u s e 2 = b a s e 2 + p e r p e n d i c u l a r 2

In ∆ PQR, it is given that it is right angled at Q, so the hypotenuse is PR.

⇒ P R 2 = P Q 2 + Q R 2

⇒ P Q 2 = P R 2 − Q R 2

Given that QR + PR = 25,

⇒

25 = (PR + QR)(PR – QR)

⇒

PR – QR = 1 ……(2)
Add equation(1) and (2) for PR and QR.
We get,

⇒

PR + QR +PR – QR = 25+1

⇒

2PR = 26

⇒

PR = 13
Substitute PR in equation (2).

⇒

13 – QR = 1

⇒

QR = 12
So, PR= 13cm and PQ = 12cm.
Triangle PQR can depicted as,
Question Image

The values of sinP, cosP and tanP are,

sin P = P H = Q R P R sin P = 513 cos P = B H = P Q P R cos p = 1213 tan P = P B = Q R P Q tan P = 512

Hence the statement is true i.e, the values are sin P=

513

, cos P =

1213

and tan P =

512

.
Therefore, the correct option is 1.

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Online Course for IIT JEE

Best Online Course for NEET

Best Online Course for CBSE

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!