In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of ?BEE and ?DFE respectively, then ?EGF is equal to :

Through O, draw a line l parallel to both AB and CD. Then
?1 = 45° (alt. ?S)
and ?2 = 30° (alt. ?S)
? ?BOC = ?1 + ?2 = 45° + 30° = 75°
So, X = 360° ? ?BOC = 360° ? 75° = 285°
Hence X = 285°.

complement of 30°20? = 90° ? ( 30°20? ) = 90° ? ( 30° + 20? )
= (89° ? 30°) + (1° ? 20?)
= 59° + 60? ? 20? [ ? 1° = 60°?]
= 59° + 40? = 59°40?.

As we know that the angles are supplementary so sum of angles will be 180 degree.
Let us assume that the ratio factor is r.
According to question,
Angles are supplementary and have a ratio of 1:4.
r + 4r = 180
? 5r = 180
? r = 180/5
? r = 36

Series pattern
The elements of the given series are the numbers formed by joining together consecutive add numbers in order
i.e, 1 and 3, 3 and 5, 5 and 7, 7 and 9, 9 and 11,...
? Missing tern = Number formed by joining 11 and 13 = 1113

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°.

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°

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

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°.