 ×     #### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
In ∆ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P

```
one month ago Anand Kumar Pandey
1372 Points
```							Dear StudentIn a given triangle PQR, right angled at Q, the following measures are PQ = 5 cmPR + QR = 25 cmNow let us assume, QR = xPR = 25-QRPR = 25- xAccording to the Pythagorean Theorem,PR^2= PQ^2+ QR^2Substitute the value of PR as x(25- x) ^2= 52+ x^225^2+ x^2–50x = 25 + x^2625 + x^2-50x -25–x^2= 0-50x = -600x= -600/-50x = 12 = QRNow, find the value of PRPR = 25- QRSubstitute the value of QRPR = 25-12PR = 13Now, substitute the value to the given problemThanks
```
one month ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Algebra

View all Questions »  ### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  ### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions