Question

in figure 39,AD andCF are respectively perpendicular to sides BC andAB of ABC .if FCD=50 find BAD​

Answer 1

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Name the point of intersection of AD and CE as O

angle ECD= 50 ( given)

By angle sum property in triangle OCD

angle OCD +angle ADC+angle DOC=180

50 + 90 + X = 180

140 +X= 180

X=40

therefore angle DOC= 40

and angle EOA =40 ( vertically oppo angles)

So similarly in triangle AOE

angle E + angle EOA+angle EAO =180

90 + 40 + x= 180

130+x = 180

X= 50

Therefore, angle BAD = 50

Answer 2

Answer:

Total sum of all angles of a triangle is 180

so ,  FCB + CBF + BFC = 180

    50 + CBF + 90 = 180

    140 + CBF = 180

    CBF = 180 -140

             = 40

In triangle ABD ,

       ABD + DAB + BDA = 180

BAD = 180 - 40 - 90

      = 180 - 130

BAD = 50

Step-by-step explanation:

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