Notes for SSC CGL examinations — turn every angle into a ratio, every identity into a shortcut, and every line of sight into a solvable triangle.
An angle records rotation from an initial ray to a terminal ray. Counter-clockwise rotation is positive; clockwise rotation is negative. One complete revolution is \(360^\circ\).
Acute: \(0^\circ\lt\theta\lt90^\circ\); right: \(90^\circ\); obtuse: \(90^\circ\lt\theta\lt180^\circ\).
One radian is the central angle subtended by an arc equal in length to the radius.
Arc length follows \(s=r\theta\) only when \(\theta\) is in radians.
Relative to an acute angle \(\theta\): opposite lies across from \(\theta\), adjacent touches it but is not the hypotenuse, and hypotenuse faces the right angle.
Here \(P\) is perpendicular/opposite, \(B\) is base/adjacent, and \(H\) is hypotenuse. If the chosen angle changes, opposite and adjacent swap; hypotenuse never changes.
Ratios are dimensionless: matching side units cancel. For a fixed acute angle, all similar right triangles give the same six values.
Hence \(\csc\theta=1/\sin\theta\), not \(1/\cos\theta\).
These let you rewrite an expression in only sine and cosine. They also explain why tangent is undefined when \(\cos\theta=0\).
If \(\tan\theta=3/4\) and \(\theta\) is acute, draw opposite \(3k\), adjacent \(4k\), so hypotenuse \(5k\).
For a non-acute angle, use quadrant signs after finding magnitudes.
| Ratio | \(0^\circ\) | \(30^\circ\) | \(45^\circ\) | \(60^\circ\) | \(90^\circ\) |
|---|---|---|---|---|---|
| \(\sin\theta\) | \(0\) | \(1/2\) | \(1/\sqrt2\) | \(\sqrt3/2\) | \(1\) |
| \(\cos\theta\) | \(1\) | \(\sqrt3/2\) | \(1/\sqrt2\) | \(1/2\) | \(0\) |
| \(\tan\theta\) | \(0\) | \(1/\sqrt3\) | \(1\) | \(\sqrt3\) | undefined |
| \(\csc\theta\) | undefined | \(2\) | \(\sqrt2\) | \(2/\sqrt3\) | \(1\) |
| \(\sec\theta\) | \(1\) | \(2/\sqrt3\) | \(\sqrt2\) | \(2\) | undefined |
| \(\cot\theta\) | undefined | \(\sqrt3\) | \(1\) | \(1/\sqrt3\) | \(0\) |
Rationalised and unrationalised equivalent forms represent the same value: \(1/\sqrt3=\sqrt3/3\), and \(2/\sqrt3=2\sqrt3/3\).
A \(45^\circ-45^\circ-90^\circ\) triangle has side ratio \(1:1:\sqrt2\). A \(30^\circ-60^\circ-90^\circ\) triangle has opposite-side ratio \(1:\sqrt3:2\).
A ratio is undefined when its denominator is zero.
Do not write “infinity” as an ordinary real value unless a limit is explicitly being discussed.
In a right triangle, the two acute angles total \(90^\circ\). Switching to the complementary angle swaps opposite and adjacent, so each function changes to its co-function.
Reverse forms are equally valid: \(\csc(90^\circ-\theta)=\sec\theta\).
Use the reference angle for magnitude, then the quadrant for sign:
Example: \(\sin150^\circ=\sin30^\circ=1/2\), but \(\cos150^\circ=-\cos30^\circ=-\sqrt3/2\).
Rewrite unfamiliar ratios, factor before expanding, and look for a fundamental identity.
State denominator restrictions; cancellation is valid only where the original expression is defined.
Conjugates exploit \(1-sin^2\theta=\cos^2\theta\) or \(\sec^2\theta- an^2\theta=1\).
Because multiplying numerator and denominator by \(\sec\theta-\tan\theta\) makes the denominator \(1\).
For acute angles, common comparisons can be solved from monotonic change: sine rises from \(0\) to \(1\), cosine falls from \(1\) to \(0\), and tangent rises from \(0\) upward.
Do not extend an acute-angle conclusion blindly to all real angles.
Horizontals at the observer and object are parallel, so depression angle equals the corresponding elevation angle by alternate interior angles.
After drawing the right triangle, choose the ratio containing the known and required sides. Most problems use tangent because vertical height and horizontal distance are involved.
For line-of-sight length \(L\): \(h=L\sin\theta\), \(d=L\cos\theta\).
If the observer’s eye is \(e\) above ground, tangent gives height above eye level, not total object height.
If viewing a point below eye level, subtract the vertical drop as appropriate.
Sun elevation \(\theta\), object height \(h\), shadow length \(s\):
If two vertical objects are under the same sun angle, their heights are proportional to their shadow lengths.
From top of height \(h\), angle of depression to a ground point is \(\theta\). The ground point’s angle of elevation to the top is also \(\theta\).
This equality comes from parallel horizontals, not from complementarity.
Keep the calculator out: SSC standard-angle problems are designed to collapse using \(1/\sqrt3,1,\sqrt3\) and familiar right-triangle triples.
For \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\):
Cosine uses the same sequence backwards: \(\sqrt{4,3,2,1,0}/2\).
If sine and cosine values are remembered, do not separately memorise tangent:
Then get cotangent by flipping. This reconstructs the full six-row table from only two rows.
Whenever you see \(90^\circ-\theta\), swap within these pairs:
The angle returns to \(\theta\).
Write numerator opposite, denominator adjacent; use Pythagoras for hypotenuse. It is faster and safer than algebraic substitutions.
Memorise only \(\sin^2+\cos^2=1\). Divide by the square of the denominator you want.
If one factor equals \(x\), its conjugate factor equals \(1/x\).
These apply when observer height is zero or already adjusted.
At the same moment, the sun angle is common, so similar triangles eliminate trigonometry entirely.
Cross-multiply; no standard angle is required.
They depend on the chosen acute angle. When the angle switches, these two sides swap; the hypotenuse remains fixed.
At \(90^\circ\), \(\cos=0\), so tangent and secant divide by zero. They are undefined, not zero.
\(30\) radians is not \(30^\circ\). Attach the unit before conversion, and use \(s=r\theta\) only with radian \(\theta\).
\(\sin(90^\circ-\theta)\) becomes cosine, not sine. The \(90^\circ\) trigger forces a co-function swap.
The plus identities are \(1+\tan^2=\sec^2\) and \(1+\cot^2=\csc^2\). Rearranged differences keep the larger square first.
Angle of depression is between the downward line of sight and a horizontal through the observer—not between line of sight and the vertical.
Tangent from the observer’s eye gives height above eye level. Add observer height for the total object height.
The sloping line is hypotenuse; the ground distance is adjacent. They are equal only in a degenerate case, never in an ordinary right triangle.
You may cancel common factors, not separate terms. Factor first.
Do not “cancel sine” only from the first term while leaving an altered denominator.
Sine = Opposite / Hypotenuse; Cosine = Adjacent / Hypotenuse; Tangent = Opposite / Adjacent.
At \(90^\circ-\theta\), friends exchange seats: sine–cosine, tangent–cotangent, secant–cosecant.
Moving anticlockwise through quadrants I–IV: All, Sine, Tangent, Cosine are positive.
Across standard angles from \(0^\circ\) to \(90^\circ\), sine climbs \(0\to1\); cosine slides \(1\to0\).
The mother identity \(\sin^2+cos^2=1\) produces tangent–secant by dividing with cosine and cotangent–cosecant by dividing with sine.
Both angles begin at an eye-level horizontal. The direction of the line of sight decides the name.
Vertical height over horizontal ground distance is tangent, so tower questions often begin with \(\tan\theta=h/d\).
The eye-level triangle gives only the top portion. Put the observer’s height back at the end.