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WAEC Mathematics Past Questions & Answers - Page 101

501.

An object is 6m away from the base of a mast. The angle of depression of the object from the top pf the mast is 50^o, Find, correct to 2 decimal places, the height of the mast

8.60m

7.51m

7.15m

1.19m

View answer & explanation

Correct answer is C

Tan 50^o = $\frac{h}{6}$

h = 6 tan 50

= 6 x 1.1917

= 7.1505

= 7.15

502.

If sin x = $\frac{5}{13}$ and 0^o $\leq$ x $\leq$ 90^o, find the value of (cos x - tan x)

$\frac{7}{13}$

$\frac{12}{13}$

$\frac{79}{156}$

$\frac{209}{156}$

View answer & explanation

Correct answer is C

Sin x = $\frac{5}{13}$

0^o $\leq$ x $\leq$ 90^o, (cos x - tan x)

AC² = AB² + BC²

13² = 5² + BC²

169 - 25 + BC²

169 - 25 = BC²

144 = BC²

Cos x = $\frac{Adj}{Hyp}$ = $\frac{12}{13}$

BC = $\sqrt{144}$

BC = 12

tan x = $\frac{opp}{adj} = \frac{5}{12}$

BC = 12

cos x - tan x = $\frac{12}{13} - \frac{5}{12}$

$\frac{144 - 65}{156} = \frac{79}{156}$

503.

A cube and cuboid have the same base area. The volume of the cube is 64cm $^3$ while that of the cuboid is 80cm $^3$ . Find the height of the cuboid

1cm

3cm

5cm

6cm

View answer & explanation

Correct answer is C

Volume of a cube with side a cm = a $^3$

a $^3$ = 64 $\implies$ a = 4cm

Base area of a cube = a $^2$

= 4 $^2$

= 16 cm $^2$

$\implies$ Base area of the cuboid = 16 cm $^2$

Volume of cuboid = Base area x height

80 = 16 x h

h = $\frac{80}{16}$

= 5 cm

504.

A chord is 2cm from the centre of a circle. If the radius of the circle is 5cm, find the length of the chord

2 $\sqrt{21}$ cm

$\sqrt{42}$ cm

2 $\sqrt{19}$ cm

$\sqrt{21}$ cm

View answer & explanation

Correct answer is A

From $\bigtriangleup$ OMQ find /MQ/ by Pythagoras OQ² = OM² + MQ²

5² = 2² + MQ²

25 = 4 + MQ²

25 - 4 = MQ²

21 - MQ²

MQ² = 21

MQ² = $\sqrt{21}$

Length of chord = 2 x $\sqrt{21}$ = 2 $\sqrt{21}$ cm

505.

Given that p^{$\frac{1}{3}$} = $\frac{3\sqrt{q}}{r}$ , make q the subject of the equation

q = p $\sqrt{r}$

q = p³r

q = pr³

q = pr^{$\frac{1}{3}$}

View answer & explanation

Correct answer is D

p^{$\frac{1}{3}$} = $\frac{3\sqrt{q}}{r}$ (cross multiply)

3 $\sqrt{q}$ = r x 3 $\frac{\sqrt{q}}{r}$ (cross multiply)

3 $\sqrt{q}$ = r x 3 $\sqrt{p}$ cube root both side

q = 3 $\sqrt{r}$ x p

q = r^{$\frac{1}{3}$}p = pr^{$\frac{1}{3}$}