Exercises — Trig ratios of standard angles — 0°, 30°, 45°, 60°, 90° (derive, don't memorize blindly)

Read them straight off the derived table (each is a triangle ratio, not a guessed number).

• $\sin 30°=\frac{1}{2}$ — in the $30$–$60$–$90$ triangle the side opposite the small angle is the short leg $1$, over hypotenuse $2$.
• $\cos 60°=\frac{1}{2}$ — for $60°$ the adjacent side is that same short leg $1$, over hypotenuse $2$.
• $\tan 45°=1$ — the $45$–$45$–$90$ triangle has equal legs, so rise $=$ run.

Answers: $\sin 30°=\frac{1}{2},\cos 60°=\frac{1}{2},\tan 45°=1.$

$\tan \theta =\frac{\sin \theta }{\cos \theta }$. It blows up (is undefined) exactly when the denominator $\cos \theta =0$. Among our five angles, only $\cos 90°=0$. Geometrically the adjacent side has shrunk to length $0$: the ray points straight up, so "forward reach" is nothing and steepness is infinite.

Answer: $\theta =90°$, because $\cos 90°=0$ (division by zero).

What connects the pieces? We know the adjacent side (base $=6$) and want the opposite side (height). The ratio linking opposite to adjacent is tangent: $\tan 30°=\frac{\text{opposite}}{\text{adjacent}}=\frac{h}{6}.$ Substitute $\tan 30°=\frac{1}{\sqrt{3}}$: $\frac{1}{\sqrt{3}}=\frac{h}{6}\Rightarrow h=\frac{6}{\sqrt{3}}.$ Rationalize (clear the radical from the bottom, per Rationalization) by multiplying top and bottom by $\sqrt{3}$: $h=\frac{6}{\sqrt{3}}\cdot \frac{\sqrt{3}}{\sqrt{3}}=\frac{6\sqrt{3}}{3}=2\sqrt{3}\text{m}\approx 3.46\text{m}.$

Answer: $h=2\sqrt{3}\text{m}\approx 3.46\text{m}.$

The string is the hypotenuse ($20$), the height is opposite the $60°$ angle. Opposite over hypotenuse is sine: $\sin 60°=\frac{h}{20},\sin 60°=\frac{\sqrt{3}}{2}.$ $\frac{\sqrt{3}}{2}=\frac{h}{20}\Rightarrow h=20\cdot \frac{\sqrt{3}}{2}=10\sqrt{3}\text{m}\approx 17.32\text{m}.$

Answer: $h=10\sqrt{3}\text{m}\approx 17.32\text{m}.$

Substitute the derived values $\tan 45°=1,\tan 60°=\sqrt{3},\tan 30°=\frac{1}{\sqrt{3}}$: $\frac{1+\sqrt{3}}{\frac{1}{\sqrt{3}}}.$ Dividing by a fraction $=$ multiplying by its reciprocal ${\left(\frac{1}{\sqrt{3}}\right)}^{-1}=\sqrt{3}$: $(1+\sqrt{3})\cdot \sqrt{3}=\sqrt{3}+\sqrt{3}\cdot \sqrt{3}=\sqrt{3}+3.$

Answer: $3+\sqrt{3}(\approx 4.73).$

Substitute $\sin 60°=\frac{\sqrt{3}}{2}$, $\cos 60°=\frac{1}{2}$ (here ${\sin }^2\theta$ means $(\sin \theta {)}^2$): ${\left(\frac{\sqrt{3}}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2=\frac{3}{4}+\frac{1}{4}=1.✓$ Why it must be $1$: in the Unit Circle the point at angle $\theta$ is $(\cos \theta ,\sin \theta )$ and lies on a circle of radius $1$. The Pythagorean Theorem on the little right triangle from the centre gives ${\cos }^2\theta +{\sin }^2\theta =1^2$. It's Pythagoras wearing a disguise.

Answer: equals $1$; it is the Pythagorean theorem on the unit circle.

Numerator, plug in $\sin 30°=\frac{1}{2},\cos 60°=\frac{1}{2},\cos 30°=\frac{\sqrt{3}}{2},\sin 60°=\frac{\sqrt{3}}{2}$: $\frac{1}{2}\cdot \frac{1}{2}+\frac{\sqrt{3}}{2}\cdot \frac{\sqrt{3}}{2}=\frac{1}{4}+\frac{3}{4}=1.$ (Nice check: this numerator is $\sin (30°+60°)=\sin 90°=1$ by the angle-sum rule — see Symmetry in Trigonometry.) Denominator $\cos 45°=\frac{\sqrt{2}}{2}$. So $\frac{1}{\frac{\sqrt{2}}{2}}=\frac{2}{\sqrt{2}}=\frac{2}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{2}=\sqrt{2}.$

Answer: $\sqrt{2}(\approx 1.414).$

$\tan 30°\cdot \tan 60°=\frac{1}{\sqrt{3}}\cdot \sqrt{3}=\frac{\sqrt{3}}{\sqrt{3}}=1.$ Interpretation: $30°+60°=90°$, so they are complementary. For any pair adding to $90°$, one angle's opposite side is the other's adjacent side — the triangle is the same, just viewed from the other corner. Hence $\tan (90°-\theta )=\frac{1}{\tan \theta }$ (this is the cotangent), and their product is $1$.

Answer: product $=1$; complementary angles give reciprocal tangents.

Let the foot of the tower be $F$, with horizontal distance $x=BF$ (from the nearer point $B$). Two right triangles share the same vertical height $h$.

From $B$ (angle $60°$, adjacent $=x$): \tan 60° = \frac{h}{x}\;\Rightarrow\; h = x\tan 60° = x\sqrt3. \tag{1} From $A$ (angle $30°$, adjacent $=x+30$): \tan 30° = \frac{h}{x+30}\;\Rightarrow\; h = (x+30)\tan 30° = \frac{x+30}{\sqrt3}. \tag{2} Set (1) $=$ (2): $x\sqrt{3}=\frac{x+30}{\sqrt{3}}.$ Multiply both sides by $\sqrt{3}$: $3x=x+30\Rightarrow 2x=30\Rightarrow x=15\text{m}.$ Substitute into (1): $h=15\sqrt{3}\text{m}\approx 25.98\text{m}.$

Answer: $h=15\sqrt{3}\text{m}\approx 25.98\text{m}.$

Note $\cos 30°=\sin 60°=\frac{\sqrt{3}}{2}$. So $\frac{\cos 30°}{1+\sin 60°}=\frac{\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}}=\frac{\frac{\sqrt{3}}{2}}{\frac{2+\sqrt{3}}{2}}=\frac{\sqrt{3}}{2+\sqrt{3}}.$ Rationalize by the conjugate $2-\sqrt{3}$ (this kills the radical: $(2+\sqrt{3})(2-\sqrt{3})=4-3=1$): $\frac{\sqrt{3}}{2+\sqrt{3}}\cdot \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{\sqrt{3}(2-\sqrt{3})}{1}=2\sqrt{3}-3.$ Hmm — that gives $2\sqrt{3}-3\approx 0.464$. Check the target numerically: $\frac{0.8660}{1+0.8660}=0.464$, and $2-\sqrt{3}\approx 0.268\ne 0.464$. So the true simplified value is $2\sqrt{3}-3$; the stated $2-\sqrt{3}$ was a decoy. Always verify numerically!

Answer: $\frac{\cos 30°}{1+\sin 60°}=2\sqrt{3}-3(\approx 0.464)$. The claimed $2-\sqrt{3}$ is false.

Why can every value on this page be rebuilt in under a minute? ::: Because each is a side-ratio of the $45$–$45$–$90$ or $30$–$60$–$90$ Special Triangles, whose sides come from the Pythagorean Theorem — draw, label, divide.