ONE EIGHTH CONTEST 2025
Anlo SHS - 31 points
Mpraeso SHS - 35 points
Mfantsiman Girls' SHS - 27 points
ONE EIGHTH CONTEST 2025
Pope John SHS & Minor Seminary - 58 points
Ghana SHS, Tamale - 25 points
Awudome SHS - 23 points
ONE EIGHTH CONTEST 2025
Swedru Secondary School - 36 points
Kumasi High School - 31 points
Anlo SHS - 30 points
ROUND 2
SPEED RACE
TRIGONOMETRY
QUESTION
Given that $\sin A=\dfrac{3}{5}$ and $\cos B=-\dfrac{1}{2}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$.
ANSWER: $\dfrac{4+3\sqrt3}{10}$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant.
$\cos^2A=1-\sin^2A$
$\cos^2A=1-\dfrac{9}{25}$
$\cos^2A=\dfrac{16}{25}$
Cosine is negative in the second quadrant:
$\cos A=-\dfrac{4}{5}$
$\sin^2B=1-\cos^2B$
$\sin^2B=1-\dfrac{1}{4}$
$\sin^2B=\dfrac{3}{4}$
Sine is positive in the second quadrant:
$\sin B=\dfrac{\sqrt3}{2}$
Substitute:
$\cos(A-B)=\left(-\dfrac{4}{5}\right)\left(-\dfrac{1}{2}\right)+\left(\dfrac{3}{5}\right)\left(\dfrac{\sqrt3}{2}\right)$
$\cos(A-B)=\dfrac{4}{10}+\dfrac{3\sqrt3}{10}$
$\cos(A-B)=\dfrac{4+3\sqrt3}{10}$
PRACTICE QUESTIONS
1. Given that $\sin A=\dfrac{3}{5}$ and $\cos B=-\dfrac{1}{2}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.920$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{9}{25}=\dfrac{16}{25}$
$\cos A=-\dfrac{4}{5}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{1}{4}=\dfrac{3}{4}$
$\sin B=\dfrac{\sqrt{3}}{2}$
Substitute:
$\cos(A-B)=\left(-\dfrac{4}{5}\right)\left(-\dfrac{1}{2}\right)+\left(\dfrac{3}{5}\right)\left(\dfrac{\sqrt{3}}{2}\right)$
$\cos(A-B)\approx0.920$
2. Given that $\sin A=\dfrac{5}{13}$ and $\cos B=-\dfrac{3}{5}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.862$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{25}{169}=\dfrac{144}{169}$
$\cos A=-\dfrac{12}{13}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{9}{25}=\dfrac{16}{25}$
$\sin B=\dfrac{4}{5}$
Substitute:
$\cos(A-B)=\left(-\dfrac{12}{13}\right)\left(-\dfrac{3}{5}\right)+\left(\dfrac{5}{13}\right)\left(\dfrac{4}{5}\right)$
$\cos(A-B)\approx0.862$
3. Given that $\sin A=\dfrac{8}{17}$ and $\cos B=-\dfrac{5}{13}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.774$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{64}{289}=\dfrac{225}{289}$
$\cos A=-\dfrac{15}{17}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{25}{169}=\dfrac{144}{169}$
$\sin B=\dfrac{12}{13}$
Substitute:
$\cos(A-B)=\left(-\dfrac{15}{17}\right)\left(-\dfrac{5}{13}\right)+\left(\dfrac{8}{17}\right)\left(\dfrac{12}{13}\right)$
$\cos(A-B)\approx0.774$
4. Given that $\sin A=\dfrac{7}{25}$ and $\cos B=-\dfrac{3}{5}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.800$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{49}{625}=\dfrac{576}{625}$
$\cos A=-\dfrac{24}{25}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{9}{25}=\dfrac{16}{25}$
$\sin B=\dfrac{4}{5}$
Substitute:
$\cos(A-B)=\left(-\dfrac{24}{25}\right)\left(-\dfrac{3}{5}\right)+\left(\dfrac{7}{25}\right)\left(\dfrac{4}{5}\right)$
$\cos(A-B)\approx0.800$
5. Given that $\sin A=\dfrac{4}{5}$ and $\cos B=-\dfrac{1}{2}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.993$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{16}{25}=\dfrac{9}{25}$
$\cos A=-\dfrac{3}{5}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{1}{4}=\dfrac{3}{4}$
$\sin B=\dfrac{\sqrt{3}}{2}$
Substitute:
$\cos(A-B)=\left(-\dfrac{3}{5}\right)\left(-\dfrac{1}{2}\right)+\left(\dfrac{4}{5}\right)\left(\dfrac{\sqrt{3}}{2}\right)$
$\cos(A-B)\approx0.993$
6. Given that $\sin A=\dfrac{20}{29}$ and $\cos B=-\dfrac{4}{5}$, and $A$ and $B$ are both obtuse. Evaluate $\cos(A-B)$, to 3 decimal places.
ANSWER: $\approx0.993$
SOLUTION
$\cos(A-B)=\cos A\cos B+\sin A\sin B$
$A$ is obtuse, so it is in the second quadrant, where cosine is negative:
$\cos^2A=1-\sin^2A=1-\dfrac{400}{841}=\dfrac{441}{841}$
$\cos A=-\dfrac{21}{29}$
$B$ is obtuse, so it is in the second quadrant, where sine is positive:
$\sin^2B=1-\cos^2B=1-\dfrac{16}{25}=\dfrac{9}{25}$
$\sin B=\dfrac{3}{5}$
Substitute:
$\cos(A-B)=\left(-\dfrac{21}{29}\right)\left(-\dfrac{4}{5}\right)+\left(\dfrac{20}{29}\right)\left(\dfrac{3}{5}\right)$
$\cos(A-B)\approx0.993$