IN A TRIANGLE ABC WITH ALTITUDE AD, ANGLE BAC=45 DEGREES,DB=3 AND CD=2

THE AREA OF TRIANGLE ABC IS:

1) 6

2) 15

3) 15/4

4) 12

Tan x+y = tan x+ tan y / 1 - tan x tan yA = 45° let AD = k , ANGLE DAB= x and ANGLE DAC= y....such that x + y =45°Tan x = 3/k and tan y = 2/kPut the values in the formula...-> 5/k = k² - 6 / k²-> k² - 5k - 6 = 0-> k=6, -1 k acnnot be -1Hence area of triangle ABC = 1/2 * 6 * 5 = 15 sq units