Geometry and measurement: trigonometric ratios (Mathematics I)

In a right triangle with sides in the ratio 1:2:√3, the side opposite the 30° angle is half the hypotenuse, so sin30°=1/2. The value √3/2 is cos30°, √2/2 is sin45°, and √3/3 is tan30°; these are easy to mix up, so it is best to learn the whole table.

In a right isosceles triangle with sides 1:1:√2 we get cos45°=1/√2=√2/2. The value 1/2 is cos60° and √3/2 is cos30°. A distinctive feature of 45° is that sin and cos take the same value, √2/2.

In a right triangle with sides 1:2:√3, the side opposite the 60° angle is √3 and the adjacent side is 1, so tan60°=√3. The value √3/3 is tan30°, which is what you get by swapping the opposite and adjacent sides. The value √3/2 is sin60°.

sin120°=sin(180°−60°)=sin60°=√3/2. Since sin stays positive for 90°<θ<180°, attaching a minus sign to give −√3/2 is wrong. The value 1/2 is sin30°, which equals cos60°.

cos135°=−cos45°=−√2/2. Note that cos changes sign under 180°−θ. Forgetting the sign and answering √2/2 is the commonest error. The value −√3/2 is cos150°.

tan150°=−tan30°=−√3/3. The tangent is negative for obtuse angles. Forgetting the sign gives √3/3, and confusing tan30° with tan60° gives −√3.

Sine is (opposite)/(hypotenuse). The hypotenuse is the longest side, 5, and the side opposite A is 3, so sinA=3/5. The value 4/5 is cosA and 3/4 is tanA. Identify which side is opposite and which is adjacent on a diagram before forming the ratio.

Tangent is (opposite)/(adjacent). The side opposite A is 3 and the adjacent side is 4, so tanA=3/4. Turning it upside down as 4/3 is a common error. The value 3/5 is sinA and 4/5 is cosA.

cos²θ=1−sin²θ=1−1/9=8/9. Since θ is acute, cosθ=√(8/9)=2√2/3. Answering 8/9 forgets to take the square root, and √2/3 comes from miscomputing √8=2√2.

sin²θ=1−cos²θ=1−1/16=15/16. For 90°<θ<180° we have sinθ>0, so sinθ=√15/4. Making sin negative just because cos is negative is the typical error, and 15/16 forgets the square root.

cos²θ=1−4/9=5/9, and since the angle is obtuse, cosθ=−√5/3. Then tanθ=sinθ/cosθ=(2/3)÷(−√5/3)=−2/√5=−2√5/5. Taking cos as positive leads to the wrong answer 2√5/5.

By the identity sin(90°−θ)=cosθ, we have sin55°=sin(90°−35°)=cos35°. This says that looking at the same ratio of sides from the two acute angles of a right triangle (35° and 55°) interchanges sine and cosine.

cos150°=cos(180°−30°)=−cos30°=−√3/2. Under 180°−θ the cosine changes sign. The value −1/2 is cos120° and −√2/2 is cos135°, so be careful which acute angle you base it on.

From 1+tan²θ=1/cos²θ we get 1/cos²θ=1+4=5, so cos²θ=1/5. Since θ is acute, cosθ=1/√5=√5/5. The value 1/5 forgets the square root, and 2√5/5 is sinθ.

Squaring both sides gives sin²θ+2sinθcosθ+cos²θ=1/4. Substituting sin²θ+cos²θ=1 gives 1+2sinθcosθ=1/4, so sinθcosθ=(1/4−1)/2=−3/8. Dropping the sign gives 3/8, and forgetting to divide by 2 gives −3/4.

By the law of sines, 2R=a/sinA=4÷(1/2)=8, so R=4. Answering 8 reports the value of 2R instead. The rule of thumb is to use the law of sines whenever you have a matching pair of a side and its opposite angle.

From the law of sines a/sinA=b/sinB we get b=a·sinB/sinA=√2·(√3/2)÷(√2/2)=√2·√3/√2=√3. The value √6/2 is what you get by forgetting to divide by sinA and writing b=a·sinB. Two angles and one side call for the law of sines.

By the law of cosines, a²=9+25−2·3·5·cos60°=34−15=19, so a=√19. Forgetting the term −2bc·cosA gives √34, and taking it with a plus sign gives a²=49 and the wrong answer 7.

cosA=(b²+c²−a²)/2bc=(25+9−49)/(2·5·3)=−15/30=−1/2, so A=120°. Getting the sign of the numerator wrong gives 1/2 and hence 60°. Since cos135°=−√2/2 and cos150°=−√3/2, take care not to confuse these values.

cosA=(b²+c²−a²)/2bc=(4+9−7)/(2·2·3)=6/12=1/2, so A=60°. Do not confuse this with cos30°=√3/2 or cos45°=√2/2. When all three sides are known, use the law of cosines in its cosine form.

By the law of sines, sinA=a·sinB/b=2·(√3/2)/√6=√3/√6=1/√2=√2/2. Since A is acute, A=45°. Getting sinA=1/2 by mistake would give 30°. Rationalise √3/√6 carefully.

a²=4+16−2·2·4·cos120°=20−16·(−1/2)=20+8=28, so a=2√7. Taking cos120°=−1/2 as positive gives a²=12 and 2√3, while forgetting the correction term gives a²=20 and 2√5.

The largest angle faces the longest side, 4. cosθ=(2²+3²−4²)/(2·2·3)=(4+9−16)/12=−3/12=−1/4. Since it is negative, the largest angle is obtuse. The values 7/8 and 11/16 are the cosines of the other two angles, which appear if you pick the wrong longest side.

The angle θ opposite the longest side 7 satisfies cosθ=(9+25−49)/(2·3·5)=−1/2, so θ=120°. By the law of sines, 2R=7/sin120°=7÷(√3/2)=14/√3, hence R=7/√3=7√3/3. Do not report the value of 2R, namely 14√3/3, as R.

Finding the remaining side from two sides and the included angle is the classic use of the law of cosines, a²=b²+c²−2bc·cosA. The law of sines is for when a matching pair of a side and its opposite angle is known, and the Pythagorean theorem applies only to right triangles.

S=(1/2)·5·8·sin60°=20·(√3/2)=10√3. Forgetting to multiply by 1/2 gives 20√3. Mistaking sin60°=√3/2 for 1/2, which is sin30°, gives 10.

S=(1/2)·7·8·sinθ=28sinθ=14, so sinθ=1/2. Since θ is acute, θ=30°. Do not go from sinθ=1/2 to 60° by confusing it with sin60°=√3/2; know the standard values exactly.

S=(1/2)·4·3√2·sin135°=(1/2)·4·3√2·(√2/2)=(1/2)·12=6. Simply use sin135°=√2/2, which is positive. The option 6√2 forgets to multiply by sin135°, 12 forgets the factor (1/2), and 3√2 comes from taking sin135° to be 1/2.

By the law of cosines, cosA=(36+49−25)/(2·6·7)=60/84=5/7, so sinA=√(1−25/49)=2√6/7. Then S=(1/2)·6·7·(2√6/7)=6√6. Heron's formula checks this: s=9 and S=√(9·4·3·2)=√216=6√6.

From S=(1/2)r(a+b+c) we get 6√6=(1/2)r·18=9r, so r=6√6/9=2√6/3. Forgetting to halve the perimeter 18 and writing 18r gives √6/3. The standard route is to find the area first and then r.

The area is S=(1/2)·3·4=6. From S=(1/2)r(3+4+5)=6r we get r=1. For a right triangle you can also use r=(a+b−c)/2=(3+4−5)/2=1, where c is the hypotenuse.

The height is h=12·tan30°=12·(√3/3)=4√3, about 6.9 m. Confusing this with tan60°=√3 gives 12√3. The basic pattern for elevation problems is height = horizontal distance × tan(angle of elevation).

The foot of the perpendicular from the apex to the base is the centroid of the equilateral base, and the distance from a base vertex to that centroid is 6/√3=2√3. The height is h=√(6²−(2√3)²)=√(36−12)=√24=2√6. Do not confuse the distance to the centroid with the length of a median, 3√3.

The base area is (√3/4)·36=9√3 and the height is 2√6, so V=(1/3)·9√3·2√6=6√18=18√2. Forgetting the factor 1/3 gives 54√2. The formula V=(√2/12)a³=(√2/12)·216=18√2 confirms it.

By the law of sines, a:b:c=sinA:sinB:sinC=7:5:3. Putting a=7k, b=5k and c=3k gives cosA=(25+9−49)k²/(2·15k²)=−1/2, so A=120°. The crucial point is that a ratio of sines can be turned into a ratio of sides.