After
drawing
the Dotted Line from D to B, if you take the Triangle ABD, if AB = BD then it is an Isosceles Triangle. In that case when ̸͟ DAB is 55° given, then ̸͟ ADB should also be 55°. Therefore
̸͟
ABD will be 70° (180° – 55° – 55°) applying the rule of the sum of the angles of a triangle is 180°. As ̸͟ CDA is given as 100° and inasmuch as ̸͟ ADB is 55°, ̸͟ BDC will be 45° only.
In the Triangle DBC as BC=BD it is also an Isosceles Triangle. When ̸͟ BDC is=45° (since deduced) , ̸͟ BCD should also be 45°. Therefore ̸͟ DBC will be 90° (180°-45°-45°) applying the rule of the sum of the angles of a triangle is 180°.
Now ̸͟ ABD + ̸͟ CBD = 70° + 90° = 160°.
In the circumstances the exterior angle ABC will be
360° - 160°=200°
On 26 May 2014 22:14, Shabbir Taherali Dadabhai <dadabhai.shabbirtaherali0@gmail.com> wrote:
Dear Friends. Can you assist to solve the attached Maths Question for my son giving the method
Thanks
Shabbir Taherali Dadabhai
Cellphone no +254723024308 Sent from my BlackBerry®
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