Question

In a triangle the angles are in A.P. and the lengths of the two larger sides are 10 and 9 respectively. Then the length of the third side can be

• A. 5 ±$\sqrt { 6 }$
• B. 0.7
• C. 3 –$\sqrt { 6 }$
• D. None

Answer 1

Answer: (A)

5 ±$\sqrt { 6 }$

Answer 2

Let the angles be A = x – d, B = x, C = x + d. Then x – d + x + x + d = 180⁰

⇒ x = 60⁰ ⇒ two larger angles are B = 60⁰ and C.

Hence b = 9 and c = 10.

Now cosB =

⇒ $\frac { 1 } { 2 } = \frac { 100 + a ^ { 2 } - 81 } { 20 a }$ ⇒ a² – 10a + 19 = 0 ⇒ a = 5 ±$\sqrt { 6 }$