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

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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

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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

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

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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

IN A TRIANGLE ABC WITH ALTITUDE AD, ANGLE BAC=45 DEGREES,DB=3 AND CD=2THE AREA OF TRIANGLE ABC IS:1) 6 2) 153) 15/44) 12

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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