Waec Mathematics Questions
Question 1336:
In the diagram, < WOX = 60o, < YOE = 50o and < OXY = 30o. What is the bearing of x from y?
View Answer & ExplanationQuestion 1337:
A = {2, 4, 6, 8}, B = {2, 3, 7, 9} and C = {x : 3 < x < 9} are subsets of the universal set U = {2, 3, 4, 5, 6, 7, 8, 9}. Find
(a) \(A \cap (B' \cap C')\) ;
(b) \((A \cup B) \cap (B \cup C)\).
View Answer & Explanation(a) \(A \cap (B' \cap C')\) ;
(b) \((A \cup B) \cap (B \cup C)\).
Question 1338:
(a) The angle of depression of a boat from the mid-point of a vertical cliff is 35°. If the boat is 120m from the foot of the cliff, calculate the height of the cliff.
(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
View Answer & Explanation(b) Towns P and Q are x km apart. Two motorists set out at the same time from P to Q at steady speeds of 60 km/h and 80 km/h. The faster motorist got to Q 30 minutes earlier than the other. Find the value of x.
Question 1339:
(a)
In the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r.
(b) In the diagram TS is a tangent to the circle at A. AB // CE, < AEC = 5x°, < ADB = 60° and < TAE = x. Find the value of x.
View Answer & ExplanationIn the diagram, < PQR = 125°, < QRS = r, < RST = 80° and < STU = 44°. Calculate the value of r.
(b) In the diagram TS is a tangent to the circle at A. AB // CE, < AEC = 5x°, < ADB = 60° and < TAE = x. Find the value of x.
Question 1340:
(a)

Curved Surface Area = \(\pi rl\)
\(115.5 = \frac{22}{7} \times r \times 10.5\)
\(115.5 = 33r\)
\(r = \frac{115.5}{33} = 3.5 cm\)
(b)

\(\therefore h^{2} + (3.50)^{2} = (10.5)^{2}\)
\(h^{2} = 10.5^{2} - 3.5^{2}\)
\(h^{2} = 98 \implies h = \sqrt{98}\)
\(h = 9.8994 cm \approxeq 9.90 cm\)
(c) Volume of a cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 9.90\)
= \(\frac{23.1 \times 11}{2}\)
= \(127.05 cm^{3} \approxeq 127 cm^{3}\)
View Answer & ExplanationCurved Surface Area = \(\pi rl\)
\(115.5 = \frac{22}{7} \times r \times 10.5\)
\(115.5 = 33r\)
\(r = \frac{115.5}{33} = 3.5 cm\)
(b)
\(\therefore h^{2} + (3.50)^{2} = (10.5)^{2}\)
\(h^{2} = 10.5^{2} - 3.5^{2}\)
\(h^{2} = 98 \implies h = \sqrt{98}\)
\(h = 9.8994 cm \approxeq 9.90 cm\)
(c) Volume of a cone = \(\frac{1}{3} \pi r^{2} h\)
= \(\frac{1}{3} \times \frac{22}{7} \times \frac{7}{2} \times \frac{7}{2} \times 9.90\)
= \(\frac{23.1 \times 11}{2}\)
= \(127.05 cm^{3} \approxeq 127 cm^{3}\)