please please help me to solve this maths question
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Answered by
6
Hellllo..
in this question I just give u hint and I hope u will find that
OK
first of all
in ∆ABC
we have given two angles then we can easily find the third angle
u know sum of ∆=180°
76+50+angleA= 180°
116+angle A=180°
angle A =180-116
= 64°
OK now AD is bisector of angleA
then angle BAD=64/2=32°
IN ∆ABD
76+32 +angle BDA=180°
108+ angle BDA=180°
AngleBDA=180-108=72°
in this question I just give u hint and I hope u will find that
OK
first of all
in ∆ABC
we have given two angles then we can easily find the third angle
u know sum of ∆=180°
76+50+angleA= 180°
116+angle A=180°
angle A =180-116
= 64°
OK now AD is bisector of angleA
then angle BAD=64/2=32°
IN ∆ABD
76+32 +angle BDA=180°
108+ angle BDA=180°
AngleBDA=180-108=72°
chaithanya313:
76+50=? Is iT 116 or 126 plz reply me..
126
is the right answer
180-126=?
Leave it........
54
am i right?
Ya 54
okk
:)
Answered by
4
Angle A+76°+50°=180°(angle sum property)
Angle A=54
AD bisects angle A
Therefore, angle BAD=27°
Consider ΔABD
76°+27°+angleADB=180°
Angle ADB=77
Angle A=54
AD bisects angle A
Therefore, angle BAD=27°
Consider ΔABD
76°+27°+angleADB=180°
Angle ADB=77
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