Question 8
Determine the size of angle $A$ in the triangle.
That’s it. Step through the hints below to see why.
Not quite. $61.4^\circ$ is what you get if you subtract $9.9^2$ instead of $8.8^2$ on top of the fraction. The side you subtract is the one opposite angle $A$, which is $BC=8.8$.
Not quite. $118.6^\circ$ is $180^\circ-61.4^\circ$. It comes from two slips together: subtracting $9.9^2$ instead of $8.8^2$, and flipping the sign of the cosine. The side opposite $A$ is $BC=8.8$, and $\cos A$ is positive here.
Not quite. $131.5^\circ$ is $180^\circ-48.5^\circ$. You get it if you rearrange to $\cos A=\dfrac{a^2-b^2-c^2}{2bc}$, which flips the sign of the cosine. You also get it if you find another angle first, use the sine rule for $A$ and take the obtuse answer, because $\sin131.5^\circ=\sin48.5^\circ$. Here $\cos A\approx0.6626$ is positive, so $A$ is acute.
The triangle closes: $C$ lands on the dashed circle, so $BC=8.8$. This is angle $A\approx48.5^\circ$, option (A).
This is $180^\circ-48.5^\circ=131.5^\circ$, option (D). Here $BC$ is about , far longer than 8.8. Sine is the same at $A$ and $180^\circ-A$, but cosine changes sign, so the cosine rule gives only one angle.
This is the angle in option (). Here $BC$ is about , longer than 8.8, so $C$ misses the dashed circle and the triangle doesn’t close.
Very close, but not equal: $BC$ is about , not 8.8. Use the first button below to land exactly on the circle.
$BC$ is about , shorter than 8.8, so $C$ is inside the dashed circle. Open angle $A$ wider.
$BC$ is about , longer than 8.8, so $C$ is outside the dashed circle. Close angle $A$ up.
Angle $A$ sits opposite side $BC=8.8$. The other two sides, 9.9 and 11.3, meet at $A$.
You know all three sides and no angles, so use the cosine rule.
$$a^2=b^2+c^2-2bc\cos A$$
Rearrange for $\cos A$. The side opposite the angle is the one you subtract.
$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$
Here $a=8.8$, $b=9.9$ and $c=11.3$.
$\displaystyle \cos A=\frac{9.9^2+11.3^2-8.8^2}{2(9.9)(11.3)}$$\displaystyle \approx0.6626$
$$A=\cos^{-1}(0.6626)\approx48.5^\circ$$
Check your calculator is in degree mode. A quick sense check: 8.8 is the shortest side, so $A$ is the smallest angle, and the smallest angle of a triangle is never more than $60^\circ$.
$$\cos A=\frac{b^2+c^2-a^2}{2bc}$$
$\displaystyle \cos A=\frac{9.9^2+11.3^2-8.8^2}{2(9.9)(11.3)}$$\displaystyle =\frac{148.26}{223.74}$$\displaystyle \approx0.6626$
$$A=\cos^{-1}(0.6626)\approx48.5^\circ$$
The cosine is positive, so $A$ is acute. Unlike the sine rule, the cosine rule never gives a second, obtuse answer.
Only mark this done when you could do it without help. Reading the solution does not count.
Question wording is from the 2020 QCAA Mathematical Methods external assessment, © State of Queensland (QCAA) 2020, licensed under CC BY 4.0, and has been adapted. Tangent Tuition is not affiliated with the QCAA.
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