?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°.
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°.
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°.
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