In ?ACE
?A + ?C + ?E = 180°
Similarly in ?DFB
?D + ?F + ?B = 180°
? (?A + ?C + ?E) + (?D + ?F + ?B) = 360°
?A + ?B + ?C + ?D + ?E + ?F = 360°
Draw a figure as per given question,
In a Right triangle ADC, Use the (Pythagoras Theorem)
AC = ?DC2 - AD2
? AC = ? 152 - 9 2
? AC = ? 225 - 81
? AC = ? 144
? AC = 12 cm
In a Right triangle BCE, Use the formula
CB = ? CD2 - BE2
? CB = ?15 2 - 122
? CB = ?225 - 144
? CB = ? 81
? CB = 9 m
? Width of the street (AC + BC) = AB = 12 + 9 = 21 m.
Let, the angle be A
? Its complement angle = 90° ? A
According to the question
(90 ? A) = A + 60°
? 90 - 60 = A + A
? 2A = 30°
? A = 15°
?DCK = ?FDG = 55° (corr. ?s)
? ?ACE = ?DCK = 55° (vert. opp. ?s)
So, ?AEC = 180° ? (40° + 55°) = 85°
? ?HAB = ?AEC = 85° (corr. ?s)
Hence, x = 85°.
Since OP bisects ?BOC,
? ?BOC = 2?POC
Again, OQ bisects ?AOC,
? ?AOC = 2?QOC
Since ray OC stands on line AB, ?,
?AOC + ?BOC = 180°
? 2?QOC + 2?POC = 180°
? 2?QOC + ?POC = 180°
? ?QOC + ?POC = 90°
? ?POQ = 90°.
The above sum can also be restated as follows; The angle between the bisectors of a linear pair of angles is a right angle.
According to question,
?B = ?C = 55° , ?D = 25°
We can say ,
AB = AC ( ? ?B = ?C = 55° )
In Triangle ABC,
?A + ?B + ?C = 180°
? ?A + 55° + 55° = 180°
? ?A + 110° = 180°
? ?A = 180° - 110°
? ?A = 70° ..........................(1)
As per given figure,
?ACD + ?ACB = 180° ( ?ACB = ?C = 55°)
? ?ACD + 55° = 180°
? ?ACD = 180° - 55°
? ?ACD = 125° ....................... (2)
Now in Triangle ACD,
?CAD + ?ACD + ?CDA = 180°
? ?CAD + 125° + 25° = 180°
? ?CAD + 150° = 180°
? ?CAD = 30° ...........................(3)
( In a ?, greater angle has longer side opposite to it )
From the equation (1) , (2) and (3);
?B < ?A and ?CAD > ?D ;
? BC > CA and CA < CD
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