JAMB Mathematics Past Questions & Answers

1,081.

If tan $\theta$ = $\frac{m^2 - n^2}{2mn}$ find sec $\theta$

$\frac{m^2 + n^2}{(m^2 - n^2)}$

$\frac{m^2 + n^2}{2mn}$

$\frac{mn}{2(m^2 + n^2)}$

$\frac{m^2n^2}{2(m^2 - n^2)}$

Correct answer is B

Tan $\theta$ = $\frac{m^2 - n^2}{2mn}$

$\frac{\text{Opp}}{\text{Adj}}$ by pathagoras theorem

= Hyp² = Opp² + Adj²

Hyp² = (m² - n²) + (2mn)²

Hyp² = m⁴ - 2m²n⁴ - 4m² - n²

Hyp² = m⁴ + 2m² + n²n

Hyp² = (m² - n²)²

Hyp² = $\frac{m^2 + n^2}{2mn}$

1,082.

The sine, cosine and tangent of 210o are respectively

$\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ , $\frac{\sqrt{3}}{2}$

$\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ , $\frac{\sqrt{3}}{3}$

$\frac{1}{2}$ , $\frac{\sqrt{3}}{2}$ , $\frac{\sqrt{3}}{2}$

$\frac{-1}{2}$ , $\frac{\sqrt{-3}}{2}$ , $\frac{\sqrt{3}}{3}$

View answer & explanation

Correct answer is D

210o = 180o - 210o = - 30o

From ratio of sides, sin -30o = - $\frac{1}{2}$

Cos 210o = 180o - 210o = -30o

= cos -30o = $\frac{-3}{2}$

But tan 30o = $\frac{1}{\sqrt{3}}$ , rationalizing this

= $\frac{1}{\sqrt{3}}$ x $\frac{\sqrt{3}}{\sqrt{3}}$ = $\frac{\sqrt{3}}{3}$

∴ = $\frac{-1}{2}$ , $\frac{\sqrt{-3}}{2}$ , $\frac{\sqrt{3}}{3}$

1,083.

Two chords QR and NP of a circle intersect inside the circle at x. If RQP = 37o, RQN = 49o and QPN = 35o, find PRQ

35^o

37^o

49^o

59^o

View answer & explanation

Correct answer is D

In PNO, ONP

= 180 - (35 + 86)

= 180 - 121

= 59°

PRQ = QNP = 59°(angles in the same segment of a circle are equal)

1,084.

Three angles of a nonagon are equal and the sum of six other angles is 1110o. Calculate the size of one of the equal angles

210^o

150^o

105^o

50^o

View answer & explanation

Correct answer is D

Sum of interior angles of any polygon is (2n - 4) right angle; n angles of the Nonagon = 9

Where 3 are equal and 6 other angles = 1110o

(2 x 9 - 4)90o = (18 - 4)90o

14 x 90o = 1260o

9 angles = 1260°; 6 angles = 110o

Remaining 3 angles = 1260o - 1110o = 150o

Size of one of the 3 angles $\frac{150}{3}$ = 50o