Question

Please solve number 8A for me someone

Answer 1

Answer:

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Answer 2

AnsWer :

The value of unknown ( a ) be 58°

and ( b ) = 127°.

Solution :

In ∆DEF,

Sum of all angle of triangle be 180°.

$\tt \longmapsto a + \angle FDE + \angle DEF = 180°$

$\tt \longmapsto a + 40 + 82 = 180°.$

$\tt\longmapsto a = 180 - 122.$

$\tt \longmapsto a = 58°.$

$\rule{90}1$

Note : ADF and BDE is straight line,

• So, Angle EDF = Angle ADB ( Vertically Opposite Angle of Triangle )

$\rule{120}1$

In ∆ ADB,

$\tt \longmapsto \angle DBA + \angle DAB + \angle ADB = 180°.$

$\tt \longmapsto \angle DBA + 45° + 82° = 180°.$

$\tt \longmapsto \angle DBA + 127° = 180°.$

$\tt \longmapsto \angle DBA = 53°.$

$\rule{90}1$

Note :

• ABC is a Straight line.
• Sum of Angle DBA + Angle DBC = 180°.

$\rule{90}1$

Now, For b , b = Angle DBC.

$\tt \longmapsto \angle DBA + \angle DBC = 180°.$

$\tt \longmapsto 53° + \angle DBC = 180°.$

$\tt \longmapsto \angle DBC = 127°.$

$\tt \longmapsto b = 127°.$

Therefore, the of a is ( 58° ) and b is 127°.