The sides AB and AC of ?ABC have been produced to D and E respectively. The bisectors of ?CBD and ?BCE meet at O. If ?A = 40°, then ?BOC is equal to:

Since A, B and C are the angles of a ?,
? A + B + C = 180° .......................................... (1)
According to question,
A ? B = 15° ;
? A = B + 15°....................................................(2)
B ? C = 30°;
? B = C + 30°;...................................................(3)
Put the value of B from equation (2) in Equation (1), we will get
? A = B + 15°
A = C + 30° + 15°
A = C + 45° ........................................................(4)
From a equation,
? A + B + C = 180°
? (C + 45°) + (C + 30°) + C = 180°
? 3C + 45° + 30° = 180°
? 3C = 180° ? 75° = 105°
? C = 35° ..........................................................(5)
From equation (4)
A = C + 45°
Put the value of C from equation (5) , we will get
? ?A = 35° + 45° = 80°.

?DAC = ?B + ?C
(Exterior angle prop. of a ? ABC)
According to question,
130° = 2A + 3A
5A = 130°
A = 26°
? ?B = 52°; ?C = 78°

CD || AB (Given)
Produce RQ to meet AB in S
?CRS = ?PSR (at. int. ?s)
But ?CRS = 55°
? ?PSR = 55°
Now in QSP
?QSP + ?QPS + ?PQS = 180°
55° + 38° + ?SQP = 180°
? ?SQP = 180° ? 93° = 87°
But angle a and ?PQS are linear
?a = 180° ? 87°
?a = 93°

Let, the measure of the required angled be A°.
Then, measure of its complement = ( 90 ? A )° measure of its supplement = (180 ? A)°
According to question,
6(90° ? A) = 2(180° ? A) ?12°
? 540° ? 6A = 360° ? 2A ? 12°
? 4A = 192°
? A = 48°.

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.