10th Class Math Chapter 7 Notes PDF 2026 – Trigonometry

Rule

Quadrant I (All): all six ratios positive.

Quadrant II (Sin): only $sin, csc$ positive.

Quadrant III (Tan): only $tan, cot$ positive.

Quadrant IV (Cos): only $cos, sec$ positive.

Example 1 — sign of a ratio

Find the sign of $cos 14 5^{\circ}$ .

$14 5^{\circ}$ is in Quadrant II (between $9 0^{\circ}$ and $18 0^{\circ}$ ).
In Quadrant II, only sin/cosec are positive.

Answer: Negative

Rule

$sin (18 0^{\circ} - θ) = sin θ$ , $cos (18 0^{\circ} - θ) = - cos θ$ , $tan (18 0^{\circ} - θ) = - tan θ$

$sin (9 0^{\circ} + θ) = cos θ$ , $cos (9 0^{\circ} + θ) = - sin θ$

Example 2 — an exact value beyond $9 0^{\circ}$

Find the exact value of $sin 15 0^{\circ}$ .

$15 0^{\circ} = 18 0^{\circ} - 3 0^{\circ}$ .
$sin (18 0^{\circ} - 3 0^{\circ}) = sin 3 0^{\circ}$ .

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

$\frac{a}{sin α} = \frac{b}{sin β} = \frac{c}{sin γ}$ — use when you know two angles and a side, or two sides and a non-included angle.

Example 3 — using the sine rule

In $△ A B C$ , $b = 8$ cm, $α = 9 2^{\circ}$ , $β = 7 7^{\circ}$ . Find $a$ .

$\frac{a}{sin 9 2 ^{\circ}} = \frac{8}{sin 7 7 ^{\circ}}$ .
$a = \frac{8 sin 9 2 ^{\circ}}{sin 7 7 ^{\circ}}$ .

Answer: $a \approx 8.20$ cm

$a^{2} = b^{2} + c^{2} - 2 b c cos α$ — use when you know all three sides, or two sides and the included angle.

Example 4 — using the cosine rule

In $△ A B C$ , $b = 5.2$ cm, $c = 5$ cm, $γ = 4 8^{\circ}$ (find angle $β$ using the sine rule after finding side $a$ , illustrative outline).

With two sides and a non-included angle, first use the sine rule: $\frac{sin β}{b} = \frac{sin γ}{c}$ .
$sin β = \frac{b sin γ}{c} = \frac{5.2 sin 4 8 ^{\circ}}{5}$ .
Solve for $β$ , then the third angle $= 18 0^{\circ} - β - γ$ , then side $a$ by the sine rule again.

Answer: Solved step by step using the sine rule twice

Area $= \frac{1}{2} ab sin γ$ (two sides and the included angle)

Heron's formula (all three sides known): $s = \frac{a + b + c}{2}$ , Area $= s (s - a) (s - b) (s - c)$

Example 5 — area from two sides and an angle

Find the area of a triangle with $a = 7$ cm, $b = 8$ cm, $γ = 3 8^{\circ}$ .

Area $= \frac{1}{2} (7) (8) sin 3 8^{\circ}$ .

Answer: $\approx 17.24$ cm $^{2}$

Definition

A bearing is measured clockwise from North, always given as three digits (e.g. $04 5^{\circ}$ , $27 0^{\circ}$ ). Angle-of-elevation/depression and bearing problems usually need the sine or cosine rule once the given information is turned into a triangle.

Example 6 — elevation angle

A surveyor $60$ m from a tower sees its top at an elevation angle of $2 8^{\circ}$ . Find the tower's height.

$tan 2 8^{\circ} = \frac{h}{60}$ .
$h = 60 tan 2 8^{\circ}$ .

Answer: $h \approx 31.9$ m

Quick Recap

●ASTC: All (Q1), Sin (Q2), Tan (Q3), Cos (Q4) positive
● $sin (18 0^{\circ} - θ) = sin θ$ ; $cos (18 0^{\circ} - θ) = - cos θ$ ; use these to get exact values past $9 0^{\circ}$
●Sine rule $\frac{a}{s i n α} = \frac{b}{s i n β} = \frac{c}{s i n γ}$ : two angles+a side, or two sides+non-included angle
●Cosine rule $a^{2} = b^{2} + c^{2} - 2 b c cos α$ : three sides, or two sides+included angle
●Area $= \frac{1}{2} ab sin γ$ , or Heron's formula from three sides
●Bearings: clockwise from North, three digits

10th Class Math Chapter 7 Notes

Signs of Trigonometric Ratios (ASTC)

Reduction Formulas

The Sine Rule and Solving Triangles

The Cosine Rule

Area of a Triangle

Bearings and Applications

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