In a triangle a 2 + b 2 + c 2 = c a + a b √ 3 , then triangle is Options: A . &nbsp Equilateral B . &nbsp Right angled and isosceles C . &nbsp Right angled D . &nbsp None of these

Answer: Option C : C Given, a 2 + b 2 + c 2 = c a + a b √ 3 a 2 + b 2 + c 2 − c a − a b √ 3 = 0 ⇒ ( a √ 3 2 − b ) 2 + ( a 2 − c ) 2 = 0 ⇒ a = 2 b √ 3 a n d a = c 2 W e c a n s e e , a 2 + b 2 = c 2 s i n B = b a = √ 3 2 ≠ 1 ⇒ N o t i s o s c e l e s ∴ G i v e n t r i a n g l e i s a R i g h t a n g l e d t r i a n g l e w h i c h i s n o t i s o s c e l e s . Option C is correct.

In a triangle a2+b2+c2=ca+ab√3, then triangle is

Discussion Forum : Solution Of Triangles

Question -

In a triangle $a^{2} + b^{2} + c^{2} = c a + a b \sqrt{3}$ , then triangle is

Options:

A . Equilateral

B . Right angled and isosceles

C . Right angled

D . None of these

Answer: Option C
:
C
Given,

a^{2} + b^{2} + c^{2} = c a + a b \sqrt{3} a^{2} + b^{2} + c^{2} - c a - a b \sqrt{3} = 0 \Rightarrow {(\frac{a \sqrt{3}}{2} - b)}^{2} + {(\frac{a}{2} - c)}^{2} = 0 \Rightarrow a = \frac{2 b}{\sqrt{3}} a n d a = \frac{c}{2} W e c a n s e e, a^{2} + b^{2} = c^{2} s i n B = \frac{b}{a} = \frac{\sqrt{3}}{2} \neq 1 \Rightarrow N o t i s o s c e l e s ∴ G i v e n t r i a n g l e i s a R i g h t a n g l e d t r i a n g l e w h i c h i s n o t i s o s c e l e s

.
Option C is correct.

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