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WAEC Mathematics Past Questions & Answers - Page 122

606.

Given that cos x^o = $\frac{1}{r}$ , express tan x in terms of r

$\frac{1}{\sqrt{r}}$

$\sqrt{r}$

$\sqrt{r^2 + 1}$

$\sqrt{r^2 - 1}$

View answer & explanation

Correct answer is D

cos x^o = $\frac{1}{r}$ ; $\sqrt{r^2 - 1}$

By Pythagoras r² = 1² + x² - 1

x = $\sqrt{r^2 - 1}$

tan x^o = $\sqrt{r^2 - 1}$

= $\sqrt{r^2 - 1}$

607.

Solve for x in the equation; $\frac{3}{5}$ (2x - 1) = $\frac{1}{4}$ (5x - 3)

zero

View answer & explanation

Correct answer is D

$\frac{3}{5}$ (2x - 1) = $\frac{1}{4}$ (5x - 3)

$\frac{6x}{5} - \frac{3}{5} = \frac{5x}{4} - \frac{3}{4}$

$\frac{6x}{5} - \frac{5x}{4} = \frac{3}{5} - \frac{3}{4}$

$\frac{24x - 25x}{20} = \frac{12 - 15}{20}$

$\frac{-x}{20} = \frac{-3}{20}$

-20x = -60

x = $\frac{-60}{-20}$

x = 3

608.

If N112.00 exchanges for D14.95, calculate the value of D1.00 in naira

0.13

7.49

8.00

13.00

View answer & explanation

Correct answer is B

D14.95 = N112.00

D1.00 = $\frac{N112}{D14.95} \times$ D1.00

= 7.49

609.

In the diagram, < WOX = 60^o, < YOE = 50^o and < OXY = 30^o. What is the bearing of x from y?

300^o

240^o

190^o

150^o

View answer & explanation

Correct answer is A

The bearing of x from y = 270^o + $\theta$

where $\theta$ + 50^o = y

in $\bigtriangleup$ OXY

O + X + Y = 180^o

Where O = 40^o + 30^o = 70^o

70^o + 30^o + y = 180^o

y + 100^o = 180^o

y = 180^o - 100^o = 30^o

$\theta$ + 50^o = 80^o

80^o - 50^o = 30^o

The bearing of x from y = 270^o + 30^o = 300^o

610.

In the diagram, 0 is the centre of the circle. Find the value x

View answer & explanation

Correct answer is D

POQ in a straight line

Hence, < POQ + < QOR = 180^o

56^o + < QOR = 180^o

< QOR = 180^o - 56^o

= 124^o

Now, in $\bigtriangleup$ QOR OR = OQ = Radius

< ORQ = < OQR = 2x (Base angles of an Isosceles $\bigtriangleup$ )

2x + 124 + 2x = 180^o

4x + 124 = 180

4x = 180 - 124

4x = 56

x = $\frac{56}{4}$

x = 14^o