In known triangle ABC, ab = 5, BC = 6, AC = 7, find the value of tanb and Tanc

For high ad on the edge of BC, it can be obtained from Pythagorean theorem: ab ^ 2 -- BD ^ 2 = ad ^ 2 = AC ^ 2 -- DC ^ 2,25 -- (6 -- DC) ^ 2 = 49 -- DC ^ 2, DC ^ 2 -- (6 -- DC) ^ 2 = 24,6 (2dc -- 6) = 24,2dc -- 6 = 4, DC = 5, BD = 1, ad = 2 radical 6, so tanb = ad / BD = 2 radical 6, Tanc = ad / DC = (2 radical 6) / 5

In the triangle ABC, ab = AC, BC = 6, tanb = 4 / 3, find the area of the triangle ABC

Ad ⊥ BC is made by crossing point a, and BC is made by crossing point D,
∵AB=AC,BC=6,tanB=4/3 ,
∴AD=4,AB=5；
∴S△ABC=1/2 BC*AD=1/2*6*4=12.

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