What are Standard Angles? Although a trigonometric ratio exists for every angle, five angles — $0^{\circ}$, $30^{\circ}$, $45^{\circ}$, $60^{\circ}$, $90^{\circ}$ — appear again and again because their ratios come out as clean surds. Memorising this one table unlocks almost every numerical question in the chapter.
The standard table.
| Ratio | $0^{\circ}$ | $30^{\circ}$ | $45^{\circ}$ | $60^{\circ}$ | $90^{\circ}$ |
|---|
| $\sin$ | $0$ | $\tfrac12$ | $\tfrac{1}{\sqrt2}$ | $\tfrac{\sqrt3}{2}$ | $1$ |
| $\cos$ | $1$ | $\tfrac{\sqrt3}{2}$ | $\tfrac{1}{\sqrt2}$ | $\tfrac12$ | $0$ |
| $\tan$ | $0$ | $\tfrac{1}{\sqrt3}$ | $1$ | $\sqrt3$ | $\infty$ |
| $\operatorname{cosec}$ | $\infty$ | $2$ | $\sqrt2$ | $\tfrac{2}{\sqrt3}$ | $1$ |
| $\sec$ | $1$ | $\tfrac{2}{\sqrt3}$ | $\sqrt2$ | $2$ | $\infty$ |
| $\cot$ | $\infty$ | $\sqrt3$ | $1$ | $\tfrac{1}{\sqrt3}$ | $0$ |
An easy way to remember the sine row. Write $0,1,2,3,4$ under $0^{\circ},30^{\circ},45^{\circ},60^{\circ},90^{\circ}$, divide each by $4$ and take the square root: $\sqrt{0},\sqrt{\tfrac14},\sqrt{\tfrac24},\sqrt{\tfrac34},\sqrt{1}$, which gives $0,\tfrac12,\tfrac{1}{\sqrt2},\tfrac{\sqrt3}{2},1$. The cosine row is the same list reversed, and $\tan\theta=\dfrac{\sin\theta}{\cos\theta}$ fills in the third row.
Where the surds come from. The $45^{\circ}$ values use an isosceles right triangle with equal legs $1$ and hypotenuse $\sqrt2$, so $\sin45^{\circ}=\cos45^{\circ}=\tfrac{1}{\sqrt2}$. The $30^{\circ}$ and $60^{\circ}$ values come from an equilateral triangle of side $2$ cut in half, giving a right triangle with sides $1,\sqrt3,2$.
Notice the special "not defined" entries. $\tan90^{\circ}$ and $\sec90^{\circ}$ are undefined because they would need division by $\cos90^{\circ}=0$. Likewise $\cot0^{\circ}$ and $\operatorname{cosec}0^{\circ}$ are undefined since $\sin0^{\circ}=0$. Never write these as a finite number.
Reading printed tables for non-standard angles. For an angle like $37^{\circ}24'$, comprehensive "natural sine/cosine/tangent" tables list whole degrees down the side and minutes ($1^{\circ}=60'$) across the top; you read the four-decimal value at the intersection and add the small "mean-difference" correction. In Class 10 problems we usually engineer angles so that only standard values are needed.
Some Applications of Trigonometry — Heights and Distances. The same ratios let us measure heights and distances that we cannot reach directly. The key vocabulary:
- Line of sight: the straight line from the observer’s eye to the object being viewed.
- Angle of elevation: the angle the line of sight makes above the horizontal when the object is higher than the eye (looking up at a tower top).
- Angle of depression: the angle the line of sight makes below the horizontal when the object is lower than the eye (looking down from a lighthouse at a ship). Because the horizontal lines at the two ends are parallel, the angle of depression at the top equals the angle of elevation measured from the bottom (alternate angles).
A reliable method for every heights-and-distances problem: (1) draw a neat figure and mark the right angle at the foot of the vertical object; (2) label the known and required lengths; (3) at each given angle choose the ratio that connects the known side to the unknown side — use $\tan$ when you know/need the base and height, $\sin$ or $\cos$ when the hypotenuse (line of sight) is involved; (4) substitute the standard value and solve. For two-triangle problems (two observation points, or an observer of height $h$), set up one equation per triangle and combine them.
Common mistakes. Confusing elevation with depression; forgetting to add the observer’s own height to the computed height; leaving an answer with a surd in the denominator (always rationalise, e.g. $\tfrac{1}{\sqrt3}=\tfrac{\sqrt3}{3}$); and measuring the angle from the vertical instead of from the horizontal.
Evaluate $\sin60^{\circ}\cos30^{\circ}+\sin30^{\circ}\cos60^{\circ}$.
Solution- Step 1: From the table $\sin60^{\circ}=\dfrac{\sqrt3}{2}$, $\cos30^{\circ}=\dfrac{\sqrt3}{2}$, $\sin30^{\circ}=\dfrac12$, $\cos60^{\circ}=\dfrac12$.
- Step 2: Substitute: $\dfrac{\sqrt3}{2}\cdot\dfrac{\sqrt3}{2}+\dfrac12\cdot\dfrac12$.
- Step 3: $=\dfrac{3}{4}+\dfrac{1}{4}$.
- Step 4: $=\dfrac{4}{4}=1$.
Answer: $1$.
Evaluate $2\tan^{2}45^{\circ}+\cos^{2}30^{\circ}-\sin^{2}60^{\circ}$.
Solution- Step 1: $\tan45^{\circ}=1$, $\cos30^{\circ}=\dfrac{\sqrt3}{2}$, $\sin60^{\circ}=\dfrac{\sqrt3}{2}$.
- Step 2: $2(1)^{2}+\left(\dfrac{\sqrt3}{2}\right)^{2}-\left(\dfrac{\sqrt3}{2}\right)^{2}$.
- Step 3: The last two terms are equal, so they cancel: $2+\dfrac34-\dfrac34$.
- Step 4: $=2$.
Answer: $2$.
Evaluate $\dfrac{\tan^{2}60^{\circ}+4\cos^{2}45^{\circ}+3\sec^{2}30^{\circ}}{\operatorname{cosec}^{2}30^{\circ}}$.
Solution- Step 1: $\tan60^{\circ}=\sqrt3$, $\cos45^{\circ}=\dfrac{1}{\sqrt2}$, $\sec30^{\circ}=\dfrac{2}{\sqrt3}$, $\operatorname{cosec}30^{\circ}=2$.
- Step 2: Numerator $=(\sqrt3)^{2}+4\left(\dfrac{1}{\sqrt2}\right)^{2}+3\left(\dfrac{2}{\sqrt3}\right)^{2}=3+4\cdot\dfrac12+3\cdot\dfrac43$.
- Step 3: $=3+2+4=9$. Denominator $=2^{2}=4$.
- Step 4: Quotient $=\dfrac{9}{4}$.
Answer: $\dfrac{9}{4}$.
If $\tan2A=\cot(A-18^{\circ})$ where $2A$ is acute, find $A$.
Solution- Step 1: Write $\cot(A-18^{\circ})=\tan\bigl(90^{\circ}-(A-18^{\circ})\bigr)=\tan(108^{\circ}-A)$.
- Step 2: So $\tan2A=\tan(108^{\circ}-A)$, giving $2A=108^{\circ}-A$.
- Step 3: $3A=108^{\circ}$.
- Step 4: $A=36^{\circ}$.
Answer: $A=36^{\circ}$.
Solve for $x$:\ $x\,\tan45^{\circ}\,\sin30^{\circ}=\cos60^{\circ}\,\tan60^{\circ}$.
Solution- Step 1: $\tan45^{\circ}=1$, $\sin30^{\circ}=\dfrac12$, $\cos60^{\circ}=\dfrac12$, $\tan60^{\circ}=\sqrt3$.
- Step 2: LHS $=x\cdot1\cdot\dfrac12=\dfrac{x}{2}$; RHS $=\dfrac12\cdot\sqrt3=\dfrac{\sqrt3}{2}$.
- Step 3: $\dfrac{x}{2}=\dfrac{\sqrt3}{2}$.
- Step 4: Multiply both sides by $2$: $x=\sqrt3$.
Answer: $x=\sqrt3$.
A tower stands vertically on the ground. From a point $30$ m from its foot, the angle of elevation of the top is $30^{\circ}$. Find the height of the tower.
Solution- Step 1: Let the height be $h$. The base ($30$ m) and the height are the two legs of the right triangle, so use tangent.
- Step 2: $\tan30^{\circ}=\dfrac{h}{30}$.
- Step 3: $\dfrac{1}{\sqrt3}=\dfrac{h}{30}\Rightarrow h=\dfrac{30}{\sqrt3}$.
- Step 4: Rationalise: $h=\dfrac{30}{\sqrt3}\cdot\dfrac{\sqrt3}{\sqrt3}=\dfrac{30\sqrt3}{3}=10\sqrt3$ m $\approx17.32$ m.
Answer: $10\sqrt3\ \text{m}\approx17.32\ \text{m}$.
A kite is flying at a height of $60$ m above the ground. The string makes an angle of $60^{\circ}$ with the horizontal. Assuming the string is straight, find its length.
Solution- Step 1: The string is the hypotenuse; the height $60$ m is the side opposite the $60^{\circ}$ angle, so use sine.
- Step 2: $\sin60^{\circ}=\dfrac{60}{\ell}$ where $\ell$ is the string length.
- Step 3: $\dfrac{\sqrt3}{2}=\dfrac{60}{\ell}\Rightarrow \ell=\dfrac{60\times2}{\sqrt3}=\dfrac{120}{\sqrt3}$.
- Step 4: Rationalise: $\ell=\dfrac{120\sqrt3}{3}=40\sqrt3$ m $\approx69.28$ m.
Answer: $40\sqrt3\ \text{m}\approx69.28\ \text{m}$.
A ladder $15$ m long leans against a wall, making an angle of $60^{\circ}$ with the ground. How high up the wall does the ladder reach?
Solution- Step 1: The ladder is the hypotenuse ($15$ m); the height on the wall is opposite the $60^{\circ}$ angle, so use sine.
- Step 2: $\sin60^{\circ}=\dfrac{h}{15}$.
- Step 3: $\dfrac{\sqrt3}{2}=\dfrac{h}{15}\Rightarrow h=15\cdot\dfrac{\sqrt3}{2}=\dfrac{15\sqrt3}{2}$.
- Step 4: $h=\dfrac{15\sqrt3}{2}\approx12.99$ m.
Answer: $\dfrac{15\sqrt3}{2}\ \text{m}\approx12.99\ \text{m}$.
From the top of a $75$ m high lighthouse, the angle of depression of a ship is $30^{\circ}$. Find the horizontal distance of the ship from the foot of the lighthouse.
Solution- Step 1: The angle of depression equals the angle of elevation at the ship, so the angle at the ship is $30^{\circ}$.
- Step 2: Height $75$ m is opposite the $30^{\circ}$ angle and the horizontal distance $d$ is adjacent, so use tangent.
- Step 3: $\tan30^{\circ}=\dfrac{75}{d}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{75}{d}$.
- Step 4: $d=75\sqrt3$ m $\approx129.9$ m.
Answer: $75\sqrt3\ \text{m}\approx129.9\ \text{m}$.
A tree breaks in a storm and the broken part bends so that its top touches the ground, making an angle of $30^{\circ}$ with the ground at a distance of $8$ m from the foot. Find the original height of the tree.
Solution- Step 1: Let the unbroken stump be $h$ and the broken (slanting) part be $\ell$ (the hypotenuse). The horizontal distance from the foot to the top is $8$ m.
- Step 2: $\tan30^{\circ}=\dfrac{h}{8}\Rightarrow h=8\tan30^{\circ}=\dfrac{8}{\sqrt3}=\dfrac{8\sqrt3}{3}$ m.
- Step 3: $\cos30^{\circ}=\dfrac{8}{\ell}\Rightarrow \ell=\dfrac{8}{\cos30^{\circ}}=\dfrac{8}{\sqrt3/2}=\dfrac{16}{\sqrt3}=\dfrac{16\sqrt3}{3}$ m.
- Step 4: Original height $=h+\ell=\dfrac{8\sqrt3}{3}+\dfrac{16\sqrt3}{3}=\dfrac{24\sqrt3}{3}=8\sqrt3$ m $\approx13.86$ m.
Answer: $8\sqrt3\ \text{m}\approx13.86\ \text{m}$.
The angle of elevation of the top of a tower from a point on the ground is $30^{\circ}$. On walking $40$ m towards the tower, the elevation becomes $60^{\circ}$. Find the height of the tower.
Solution- Step 1: Let the height be $h$ and the distance from the nearer point to the foot be $x$. From the nearer point: $\tan60^{\circ}=\dfrac{h}{x}\Rightarrow \sqrt3=\dfrac{h}{x}\Rightarrow x=\dfrac{h}{\sqrt3}$.
- Step 2: From the farther point ($x+40$ away): $\tan30^{\circ}=\dfrac{h}{x+40}\Rightarrow \dfrac{1}{\sqrt3}=\dfrac{h}{x+40}\Rightarrow x+40=h\sqrt3$.
- Step 3: Substitute $x=\dfrac{h}{\sqrt3}$: $\dfrac{h}{\sqrt3}+40=h\sqrt3\Rightarrow 40=h\sqrt3-\dfrac{h}{\sqrt3}=\dfrac{3h-h}{\sqrt3}=\dfrac{2h}{\sqrt3}$.
- Step 4: $h=\dfrac{40\sqrt3}{2}=20\sqrt3$ m $\approx34.64$ m.
Answer: $20\sqrt3\ \text{m}\approx34.64\ \text{m}$.
A $1.5$ m tall observer is $28.5$ m from a tower. The angle of elevation of the top of the tower from his eyes is $45^{\circ}$. Find the height of the tower.
Solution- Step 1: Let the part of the tower above the observer’s eye level be $H$. The horizontal distance is $28.5$ m.
- Step 2: $\tan45^{\circ}=\dfrac{H}{28.5}\Rightarrow 1=\dfrac{H}{28.5}\Rightarrow H=28.5$ m.
- Step 3: This $H$ is measured from the observer’s eye, which is $1.5$ m above the ground, so add the observer’s height.
- Step 4: Total height of tower $=28.5+1.5=30$ m.
Answer: $30\ \text{m}$.