Applications of Trigonometric Functions (Grade 12 NSC Matric Mathematics): Revision Notes

Worked Example: Triangle calculation using multiple rules

Question: Given triangle ABC with $a = 14$ cm, $c = 17$ cm, and angle $B = 19°$, find:

1. Length b
2. Angle C
3. Area of triangle ABC

Solution:

Step 1: Find side b using cosine rule

Since we have two sides and the included angle, we use the cosine rule:

• $b^2 = a^2 + c^2 - 2ac \cos B$
• $b^2 = (14)^2 + (17)^2 - 2(14)(17) \cos 19°$
• $b^2 = 196 + 289 - 476 \cos 19°$
• $b^2 = 485 - 476 \cos 19°$
• $b^2 ≈ 34.93$
• $b ≈ 5.91$ cm

Step 2: Find angle C using sine rule

Now we can apply the sine rule. To avoid rounding errors, we keep the unrounded value of $b$ in the calculator memory:

• $\frac{\sin C}{c} = \frac{\sin B}{b}$
• $\frac{\sin C}{17} = \frac{\sin 19°}{b}$
• $\sin C = \frac{17 × \sin 19°}{b}$
• $\sin C ≈ 0.9364$

This gives an acute solution of $C ≈ 69.5°$, but we must check whether the obtuse solution applies.

Reasoning for the obtuse case: In any triangle, the largest side lies opposite the largest angle. Since $c = 17$ is the longest side ($c > a > b$), angle $C$ must be the largest angle in the triangle. If $C$ were the acute value $69.5°$, then $A = 180° - 19° - 69.5° = 91.5°$ would be larger than $C$ — a contradiction. Therefore $C$ must be obtuse:

• $C = 180° - 69.5° = 110.5°$

Step 3: Calculate the area

Using the area rule with the two sides and the included angle that were given:

• Area = $\frac{1}{2}ac \sin B$
• Area = $\frac{1}{2}(14)(17) \sin 19°$
• Area ≈ $38.7$ cm²

Worked Example: Height of a pole

Question: A pole with top T has its base F in the same horizontal plane as points A and B. The angle of elevation from B to T is 25°, AB = 120 m, angle FAB = 40°, and angle FBA = 30°. Calculate the height h of the pole.

Solution:

Step 1: Analyse the given information

We have two triangles to consider: △FAB (horizontal) and △TFB (vertical).

The link between them is side FB.

Step 2: Find FB using triangle FAB

In △FAB, we know two angles and one side, so we use the sine rule:

Angle F = 180° - 40° - 30° = 110°

• $\frac{FB}{\sin A} = \frac{AB}{\sin F}$
• $\frac{FB}{\sin 40°} = \frac{120}{\sin 110°}$
• $FB = \frac{120 × \sin 40°}{\sin 110°}$
• $FB ≈ 82.1$ m

Step 3: Find height FT using triangle TFB In △TFB:

• Angle F = 90° (vertical pole)
• Angle B = 25° (given)
• FB = 82.1 m (from step 2)

Using trigonometry:

• $\tan B = \frac{FT}{FB}$
• $\tan 25° = \frac{FT}{82.1}$
• $FT = 82.1 × \tan 25°$
• $FT ≈ 38.3$ m

Therefore, the height of the pole is approximately 38 m.

Worked Example: Height of a building using angles

Question: A vertical building stands with its foot at point A on horizontal ground. D is the top of the building, so DA is vertical and DA = h. Points B and C also lie on the horizontal ground, forming triangle ABC. Given BC = b, angle DBA = α (the angle of elevation from B to D), angle DBC = β, and angle DCB = θ, prove that $h = \frac{b \sin α \sin θ}{\sin(β + θ)}$.

Solution:

Step 1: Identify the triangles

Because A is the foot of the building, triangle DBA is right-angled at A, with DA = h as the vertical side. We will work in the slanted triangle BCD to find BD, then use △DBA to extract the height.

Step 2: Express BD using triangle BCD

In △BCD, angle BDC = 180° - (β + θ)

Using the sine rule, and noting that $\sin(180° - (β + θ)) = \sin(β + θ)$:

$\frac{BD}{\sin θ} = \frac{BC}{\sin(\text{angle BDC})} = \frac{b}{\sin(β + θ)}$

Therefore: $BD = \frac{b \sin θ}{\sin(β + θ)}$

Step 3: Find height h using right-angled triangle DBA

Since △DBA is right-angled at A:

$\sin α = \frac{h}{BD}$

Substituting our expression for BD:

$h = BD \sin α = \frac{b \sin α \sin θ}{\sin(β + θ)}$

This proves the required relationship.