ABC is a right angled triangle with right angle at B . From B a perpendicular is drawn to AC . If angle BAC = 35° , find angle ABD , angle CBD , angle DCB
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Answer:
/_ABD=
as we know
/_BAD+/_ABD+/_ADB =180°
35°+90°+/_ABD = 180°
/_ABD= 180°-125°
/_ABD = 55° ----(1)
/_CBD=
we know,
/_ABD & /_CBD are complementary angles
so
/_ABD+/_CBD = 90°
/_CBD= 90°-55°
/_CBD = 35° ----(2)
/_DCB =
/_ BDC+/_CBD+/_DCB = 180°
90°+35°+/_DCB=180°
/_DCB = 55° -----(3)
hope it helps...
mark brainliest....
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