Mathematics Questions & Answers - Free Online Practice

1,216.

The position of three ships P,Q and R at sea are illustrated in the diagram. The arrows indicated the North direction. The bearing of Q from P is 050o and < PQR = 72o. Calculate the bearing of R and Q

155^o

80^o

158^o

91^o

View answer & explanation

Correct answer is C

< NPQ = < PQB = 50^o(alt. < s)

< PQB = 50^o

< PQR = < PQR < PQB + < QBR = 72^o

< QBR = < PQR - < PQB = 72^o - 50^o

= 22^o

< NQR = 180 - < QBR = 180^o - 22 = 158^o, the bearing of R from Q = 158^o

1,217.

Is the diagram, MN, PQ and RS are three intersecting straight lines. Which of the following statements is/are true? i. t = y ii. x + y + z + m = 180o iii. x + m + n = 180o iv. x + n = m + z

i and iv

iii

View answer & explanation

Correct answer is C

m + y + x = 180o(sum of < s on straight line)

y = n(vertically opp. angle)

m + n + x = 180o

1,218.

In the diagram, MQ//RS, < TUV = 70^o and < RLV = 30^o. Find the value of x

150^o

110^o

100^o

95^o

View answer & explanation

Correct answer is C

L + 30^o - 180^o(Sum of < s on straight line)

L = 180^o - 30^o = 150^o

L = q = 150^o(opposite < s are equal)

y = b = 30^o(alt. < s)

b + c = 180^o(sum of < s on str. line)

30^o + c 180

c = 180 - 30

c = 150^o

b = a = 30^o (opp < s are equal)

c = d = 150^o (opp < s are equal)

a + k + 70^o = 180^o (sum of < s on \(\bigtriangleup\))

30^o + k + 70^o = 180

k + 100^o = 180

k = 180 - 100

k = 80^o

x + 80^o = 180(sum of < s on straight line)

x = 180^o - 80^o

x = 100^o