in the given figure, find the value of k.
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Answered by
12
hey mate...
here your answer is
in triangle ABC
A=34
B=x
C=30
by angle sum property A+B+C=180
34+x+30=180
64+x=180
x=116
angle abc=116
B=abc-kbd=180..........(linear pair)
=116-kbd=180
kbd=180-116
64
kbd=64
in triangle BED
B= y
B=64
D=45
by angle sum property B+B+D=180
B=71
bed+ckd=180.......(linear pair)
71+ckd=180
ckd=180-71
109
or
k=109
please make me a brainliest
here your answer is
in triangle ABC
A=34
B=x
C=30
by angle sum property A+B+C=180
34+x+30=180
64+x=180
x=116
angle abc=116
B=abc-kbd=180..........(linear pair)
=116-kbd=180
kbd=180-116
64
kbd=64
in triangle BED
B= y
B=64
D=45
by angle sum property B+B+D=180
B=71
bed+ckd=180.......(linear pair)
71+ckd=180
ckd=180-71
109
or
k=109
please make me a brainliest
Answered by
0
Answer:
119° is the correct answer.
Given:
∠CAB = 34°
∠BCA = 30°
∠EDB = 45°
To find:
the value of ∠CED= K
Solution:
In ∆ABC,
∠BCA + ∠CAB + ∠ABC = 180° (angle sum property)
34° + 30° + ∠ABC = 180°
∠ABC = 180- 74
∠ABC = 106°
now, in ∆BED,
∠EBD = 180° -ABC (linear pair)
∠EBD = 180° - 106°
∠EBD = 74°
now, k = ∠EBD + ∠BDE (angles opposite to exterior angle)
k = 45° + 74°
k = 119°
Hence, 119° is the correct answer.
#SPJ5
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