Question

If two vertices of a triangle are (1, 3) & (4, –1) and area of triangle is 5 sq. units, then angle at third vertex lies in

• A. $\left( 0,2\tan^{- 1}\frac{5}{4} \right\rbrack$
• B. $\left( 0,\tan^{- 1}\frac{5}{4} \right)$
• C. $\left( 2\tan^{- 1}\frac{5}{4},2 \right)$
• D. None of these

Answer 1

Answer: (A)

$\left( 0,2\tan^{- 1}\frac{5}{4} \right\rbrack$

Answer 2

Length of base AB = 5

\ length of altitude CM = 2

Max. ŠACB = 2 tan^–1$\frac{5}{4}$

\ ŠACB Ī$\left( 0,2\tan^{- 1}\frac{5}{4} \right\rbrack$