Trigonometry: angles and triangles

Answer: $\frac{4\pi}{3}$

Adding or taking off whole turns of $2\pi$ does not move a point on the unit circle. $-\frac{2\pi}{3} + 2\pi = \frac{4\pi}{3}\mathclose{,}$ which lies in $[0, 2\pi)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took off one turn too few → $-\frac{2\pi}{3}$
• Took off a multiple of $\pi$ that is not a whole number of turns → $\frac{7\pi}{3}$

Answer: $60$

$\tan A = \dfrac{2\sqrt{3}}{2} = \sqrt{3}\mathclose{,}$ so $A = 60^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other acute angle of the triangle → $30$
• Gave the ratio instead of the angle → $\sqrt{3}$

Answer: $\dfrac{\sqrt{6} - \sqrt{2}}{4}$

$\cos(A + B) = \cos A\cos B - \sin A\sin B\mathclose{.}$ With $A$ and $B$ the two special angles: $\cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \dfrac{\sqrt{6} - \sqrt{2}}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the cosines of the two angles → $\dfrac{\sqrt{3} + \sqrt{2}}{2}$
• Used the wrong sign in the middle of the formula → $\dfrac{\sqrt{6} + \sqrt{2}}{4}$

Answer: $26$

Pythagoras: the square on the hypotenuse equals the sum of the squares on the legs. $10^2 + 24^2 = 676\mathclose{,}$ and $\sqrt{676} = 26\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the legs instead of their squares → $34$
• Subtracted the squares, which finds a LEG, not the hypotenuse → $2 \sqrt{119}$

Answer: $\frac{23\pi}{12}$

$180^\circ = \pi$ radians, so multiply by $\dfrac{\pi}{180}\mathclose{:}$ $345 \times \dfrac{\pi}{180} = \frac{23\pi}{12}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by 180 but left out $\pi$ → $\frac{23}{12}$
• Multiplied by $\frac{180}{\pi}$ instead of $\frac{\pi}{180}$ → $\frac{62100}{\pi}$

Answer: $120$

The cosine rule turned round: $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{36 + 100 - 196}{120} = \dfrac{-60}{120} = -\dfrac{1}{2}\mathclose{,}$ so $C = 120^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+2ab\cos C\mathclose{,}$ which turns the sign of $\cos C$ round → $60$
• Gave $\cos C$ instead of the angle → $-\dfrac{1}{2}$

Answer: $37$

The three angles of any triangle add to $180^\circ\mathclose{,}$ so the third is $180 - 72 - 71 = 37\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $360^\circ\mathclose{,}$ the angles round a POINT → $217$
• Subtracted only one of the two given angles → $108$

Answer: $\frac{\sqrt{3}}{2}$

Whole turns of $2\pi$ change nothing: $\frac{8\pi}{3}$ is at the same place as $\frac{2\pi}{3}\mathclose{,}$ so $\sin\left(\frac{8\pi}{3}\right) = \sin\left(\frac{2\pi}{3}\right) = \frac{\sqrt{3}}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value with the wrong sign → $- \frac{\sqrt{3}}{2}$
• Gave the cosine instead → $- \frac{1}{2}$

Answer: $\dfrac{24}{7}$

By Pythagoras the hypotenuse is $\sqrt{24^2 + 7^2} = 25\mathclose{.}$ Opposite $A$ is $24\mathclose{,}$ next to it $7\mathclose{:}$ $\tan A = \dfrac{24}{7}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Turned the ratio upside down → $\dfrac{7}{24}$
• Divided by the hypotenuse → $\dfrac{24}{25}$

Answer: $\dfrac{240}{289}$

Because $\alpha$ is acute, $\cos\alpha = \sqrt{1 - \sin^2\alpha} = \dfrac{15}{17}$ (an 8-15-17 triangle). Then $\sin 2\alpha = 2\sin\alpha\cos\alpha = 2 \cdot \dfrac{8}{17} \cdot \dfrac{15}{17} = \dfrac{240}{289}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Doubled $\sin\alpha$ instead of using the formula → $\frac{16}{17}$
• Left out the factor 2 → $\frac{120}{289}$

Answer: $8$

Rearrange Pythagoras: the missing leg squared is $10^2 - 6^2 = 64\mathclose{,}$ so the leg is $\sqrt{64} = 8\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the lengths instead of their squares → $4$
• Added the squares, which is the rule for the HYPOTENUSE → $2 \sqrt{34}$

Answer: $0$

$2\pi$ is a special angle, so the value is exact. Its reference value is $0\mathclose{,}$ and the quadrant fixes the sign: $\sin\left(2\pi\right) = 0\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the cosine instead → $1$

Answer: $7$

The cosine rule: $c^2 = a^2 + b^2 - 2ab\cos C\mathclose{,}$ and $\cos 120^\circ = -\dfrac{1}{2}\mathclose{.}$ So $c^2 = 9 + 25 + 15 = 49\mathclose{,}$ and $c = \sqrt{49} = 7\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+2ab\cos C$ instead of $-2ab\cos C$ → $\sqrt{19}$
• Forgot to take the square root → $49$

Answer: $-\dfrac{2}{7}$

$\pi - \alpha$ is the mirror image in the $y$-axis: the $x$-coordinate changes sign: $\cos(\pi - \alpha) = -\dfrac{2}{7}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Got the sign the wrong way round → $\dfrac{2}{7}$

Answer: $4\sqrt{3}$

The tangent of $60^\circ$ links those two sides: $\tan 60^\circ = \dfrac{\text{opposite}}{\text{adjacent}}\mathclose{,}$ so the opposite side is $4 \cdot \sqrt{3} = 4\sqrt{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the sine instead of the tangent → $2\sqrt{3}$
• Divided by $\tan 60^\circ$ instead of multiplying → $\dfrac{4\sqrt{3}}{3}$

Answer: $-\dfrac{4}{5}$

$\cos^2\alpha = 1 - \sin^2\alpha = 1 - \dfrac{9}{25} = \dfrac{16}{25}\mathclose{,}$ so $\cos\alpha = \pm\dfrac{4}{5}\mathclose{.}$ In that quadrant cosine is negative: $\cos\alpha = -\dfrac{4}{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took the positive root without looking at the quadrant → $\frac{4}{5}$
• Used $1 - \sin\alpha$ instead of $\sqrt{1 - \sin^2\alpha}$ → $\frac{2}{5}$

Answer: $(- \frac{\sqrt{3}}{2}, \frac{1}{2})$

The point at angle $\theta$ is $(\cos\theta, \sin\theta)\mathclose{:}$ $(- \frac{\sqrt{3}}{2}, \frac{1}{2})\mathclose{.}$

Common mistakes, and the answer each one gives:

• Swapped the coordinates: $(\sin, \cos)$ instead of $(\cos, \sin)$ → $(\frac{1}{2}, - \frac{\sqrt{3}}{2})$
• Gave the point of the first-quadrant angle, without the signs → $(\frac{\sqrt{3}}{2}, \frac{1}{2})$

Answer: $17$

Pythagoras: the square on the hypotenuse equals the sum of the squares on the legs. $8^2 + 15^2 = 289\mathclose{,}$ and $\sqrt{289} = 17\mathclose{.}$

Common mistakes, and the answer each one gives:

• Added the legs instead of their squares → $23$
• Subtracted the squares, which finds a LEG, not the hypotenuse → $\sqrt{161}$

Answer: $\frac{\pi}{3}$

$180^\circ = \pi$ radians, so multiply by $\dfrac{\pi}{180}\mathclose{:}$ $60 \times \dfrac{\pi}{180} = \frac{\pi}{3}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Divided by 180 but left out $\pi$ → $\frac{1}{3}$
• Multiplied by $\frac{180}{\pi}$ instead of $\frac{\pi}{180}$ → $\frac{10800}{\pi}$

Answer: $4 \sqrt{6}$

The sine rule: $\dfrac{b}{\sin B} = \dfrac{a}{\sin A}\mathclose{,}$ so $b = \dfrac{a\sin B}{\sin A} = \dfrac{8 \cdot \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = 4 \sqrt{6}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Turned the ratio upside down, $b = \frac{a\sin A}{\sin B}$ → $\frac{8 \sqrt{6}}{3}$
• Used the angles themselves instead of their sines → $\frac{64}{3}$

Answer: $30$

$\sin A = \dfrac{5}{10} = \dfrac{1}{2}\mathclose{,}$ so $A = 30^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the other acute angle of the triangle → $60$
• Gave the ratio instead of the angle → $\frac{1}{2}$

Answer: $\dfrac{\sqrt{6} + \sqrt{2}}{4}$

$\cos(A - B) = \cos A\cos B + \sin A\sin B\mathclose{.}$ With $A$ and $B$ the two special angles: $\cos\left(\frac{\pi}{12}\right) = \frac{1}{2} \cdot \frac{\sqrt{2}}{2} + \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{2}}{2} = \dfrac{\sqrt{6} + \sqrt{2}}{4}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the cosines of the two angles → $\dfrac{1 - \sqrt{2}}{2}$
• Used the wrong sign in the middle of the formula → $\dfrac{\sqrt{2} - \sqrt{6}}{4}$

Answer: $114$

The three angles of any triangle add to $180^\circ\mathclose{,}$ so the third is $180 - 46 - 20 = 114\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $360^\circ\mathclose{,}$ the angles round a POINT → $294$
• Subtracted only one of the two given angles → $134$

Answer: $\frac{2\pi}{3}$

Adding or taking off whole turns of $2\pi$ does not move a point on the unit circle. $\frac{14\pi}{3} - 4\pi = \frac{2\pi}{3}\mathclose{,}$ which lies in $[0, 2\pi)\mathclose{.}$

Common mistakes, and the answer each one gives:

• Took off one turn too few → $\frac{8\pi}{3}$
• Took off a multiple of $\pi$ that is not a whole number of turns → $-\frac{\pi}{3}$

Answer: $4$

Rearrange Pythagoras: the missing leg squared is $5^2 - 3^2 = 16\mathclose{,}$ so the leg is $\sqrt{16} = 4\mathclose{.}$

Common mistakes, and the answer each one gives:

• Subtracted the lengths instead of their squares → $2$
• Added the squares, which is the rule for the HYPOTENUSE → $\sqrt{34}$

Answer: $0$

$0$ is a special angle, so the value is exact. Its reference value is $0\mathclose{,}$ and the quadrant fixes the sign: $\sin\left(0\right) = 0\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the cosine instead → $1$

Answer: $120$

The cosine rule turned round: $\cos C = \dfrac{a^2 + b^2 - c^2}{2ab} = \dfrac{81 + 225 - 441}{270} = \dfrac{-135}{270} = -\dfrac{1}{2}\mathclose{,}$ so $C = 120^\circ\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used $+2ab\cos C\mathclose{,}$ which turns the sign of $\cos C$ round → $60$
• Gave $\cos C$ instead of the angle → $-\dfrac{1}{2}$

Answer: $\dfrac{4}{5}$

By Pythagoras the hypotenuse is $\sqrt{4^2 + 3^2} = 5\mathclose{.}$ Opposite $A$ is $4\mathclose{,}$ next to it $3\mathclose{:}$ $\sin A = \dfrac{4}{5}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Used the side next to $A$ instead of the opposite one → $\dfrac{3}{5}$
• Divided by the other short side instead of the hypotenuse → $\dfrac{4}{3}$

Answer: $\dfrac{527}{625}$

Because $\alpha$ is acute, $\cos\alpha = \sqrt{1 - \sin^2\alpha} = \dfrac{24}{25}$ (a 7-24-25 triangle). Then $\cos 2\alpha = 1 - 2\sin^2\alpha = 1 - 2 \cdot \left(\dfrac{7}{25}\right)^{2} = \dfrac{527}{625}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Doubled $\cos\alpha$ instead of using the formula → $\frac{48}{25}$
• Used $1 - \sin^2\alpha\mathclose{,}$ leaving out the 2 → $\frac{576}{625}$

Answer: $- \frac{1}{2}$

Whole turns of $2\pi$ change nothing: $-\frac{4\pi}{3}$ is at the same place as $\frac{2\pi}{3}\mathclose{,}$ so $\cos\left(-\frac{4\pi}{3}\right) = \cos\left(\frac{2\pi}{3}\right) = - \frac{1}{2}\mathclose{.}$

Common mistakes, and the answer each one gives:

• Gave the value with the wrong sign → $\frac{1}{2}$
• Gave the sine instead → $\frac{\sqrt{3}}{2}$