English

In ΔАВС, ∠ACB < 90°. The correct relation is ______. - Mathematics

Advertisements
Advertisements

Question

In ΔАВС, ∠ACB < 90°. The correct relation is ______.

Options

  • AB2 = BC2 + AC2

  • BC2 < AC2 + AB2

  • AB2 < BC2 + AC2

  • AB2 > BC2 + AC2

MCQ
Fill in the Blanks
Advertisements

Solution

In ΔАВС, ∠ACB < 90°. The correct relation is AB2 < BC2 + AC2.

Explanation:

By the cosine rule,

AB2 = BC2 + AC2 – 2·(BC)(AC)·cos(∠ACB). 

If ∠ACB < 90°. 

Then cos(∠ACB) > 0. 

So, the last term is positive and AB2 is less than BC2 + AC2.

shaalaa.com
  Is there an error in this question or solution?
Chapter 10: Pythagoras Theorem - Exercise 10B [Page 212]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 10 Pythagoras Theorem
Exercise 10B | Q 8. | Page 212
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×