Introduction to Trigonometry

Step-by-Step Calculation

1. \[\begin{aligned}AC &= \sqrt{[(4k)² + (3k)²]} \\ & = \sqrt{(16k² + 9k²)} \\ &= \sqrt{(25k²)} \\ & = 5k\end{aligned}\]
2. \[\begin{aligned}\sin A &= \dfrac{BC}{AC}\\ &= \dfrac{4k}{5k}\\ &= \dfrac{4}{5}\end{aligned}\]
3. \[\begin{aligned}\cos A &= \dfrac{AB}{AC} \\&= \dfrac{3k}{5k} \\&= \dfrac{3}{5}\end{aligned}\]
4. \[\tan A = 4/3 \text{ (given)}\]
5. \[\begin{aligned}\text{cosec }A &= \dfrac{1}{\sin A}\\&= \dfrac{1}{4/5}\\&=\dfrac{5}{4}\end{aligned}\]
6. \[\begin{aligned}\sec A &=\dfrac{1}{\cos A} \\&= \dfrac{1}{3/5}\\&= \dfrac{5}{3}\end{aligned}\]
7. \[\begin{aligned}\cot A &=\dfrac{1}{\tan A}\\&=\dfrac{1}{4/3}\\&= \dfrac{3}{4}\end{aligned}\]

Finding Missing Side

1. Given

   Triangle ACB is right-angled at C
   AB = 29 (Hypotenuse), BC = 21 (Base), ∠ABC = θ

2. Find Missing Side

   AC² = AB² − BC² = 29² − 21²
   = (29 + 21)(29 − 21) = 50 × 8 = 400
   ⇒ AC = 20

   Corrected: Earlier mistake avoided (AC was incorrectly labeled as AB).

3. Trigonometric Ratios

   \[\begin{aligned}\sin θ &= \dfrac{AC}{AB} \\&= \dfrac{20}{29} \\\\ \cos θ &= \dfrac{BC}{AB} \\&= \dfrac{21}{29}\end{aligned}\]

4. Part (i): Evaluate cos²θ + sin²θ

   \[ \begin{aligned} \cos^2θ + \sin^2θ&= \left(\dfrac{21}{29}\right)^2 + \left(\dfrac{20}{29}\right)^2\\ &= \dfrac{441 + 400}{841}\\ &= \dfrac{841}{841} \\&= 1 \end{aligned} \]

   Verified Identity: sin²θ + cos²θ = 1

5. Part (ii): Evaluate cos²θ − sin²θ

   \[ \begin{aligned} \cos^2 \theta - \sin^2 \theta &= \left(\dfrac{21}{29}\right)^2 - \left(\dfrac{20}{29}\right)^2 \\ &= \dfrac{(21 + 20)(21 - 20)}{29^2} \\ &= \dfrac{41 \times 1}{841} \\ &= \dfrac{41}{841} \end{aligned} \]

given

To Find

1. Form Equation
2. \[OQ − PQ = 1 ⇒ OQ = PQ + 1\]
3. Apply Pythagoras Theorem
4. \[\begin{aligned}OQ^2 &= OP^2 + PQ^2\\ (PQ + 1)^2 &= 7^2 + PQ^2\end{aligned}\]
5. \[\begin{aligned}PQ^2 + 2PQ + 1 &= 49 + PQ^2\\ 2PQ &= 48 \\⇒ PQ &= 24\end{aligned}\]
6. Find Hypotenuse
7. \[\begin{aligned}OQ &= PQ + 1 \\&= 24 + 1 \\&= 25\end{aligned}\]
8. Trigonometric Ratios
9. \[\begin{aligned}\sin Q &= \dfrac{\text{Opposite}}{\text{Hypotenuse}} \\&= \dfrac{OP}{OQ} \\&= \dfrac{7}{25}\end{aligned}\]
10. \[\begin{aligned}\cos Q &= \dfrac{\text{Adjacent}}{\text{Hypotenuse}} \\&= \dfrac{PQ}{OQ} \\&= \dfrac{24}{25}\end{aligned}\]

Solution

Given

To Find

1. Find Hypotenuse (AC)
2. \[ \begin{aligned} \sin 30° &= AB / AC\\ 1/2 &= 5 / AC\\ ⇒ AC &= 10 cm \end{aligned} \]
3. Find Base (BC)
4. Using Pythagoras:\[ \begin{aligned} BC^2 &= AC^2 − AB^2\\ &=10^2 − 5^2 \\&= 100 − 25 = 75\\ ⇒ BC &= \sqrt{75} \\&= 5\sqrt{3} cm \end{aligned} \]

Solution

1. Given

   Triangle PQR is right-angled at Q
   PQ = 3 cm, PR = 6 cm

2. To Find

   ∠QPR (angle at P) and ∠PRQ (angle at R)

3. Find \(\angle R\)
4. \[\begin{aligned} \sin R &= \dfrac{\text{Opposite}}{\text{Hypotenuse}} \\&= PQ / PR\\ &= \dfrac{3}{6} \\&= \dfrac{1}{2}\\ ⇒ \angle R &= 30° \end{aligned} \]
5. Find \(\angle P\)
6. Using angle sum property: \[\begin{aligned} \angle P + \angle Q + \angle R &= 180°\\ \angle P + 90° + 30° &= 180°\\ ⇒ \angle P &= 60° \end{aligned} \]
7. Alternative Method (Direct Recognition)

8. Since:

   PQ : PR = 3 : 6 = 1 : 2

   This matches standard 30° triangle ratio:

   Opposite : Hypotenuse = 1 : 2 ⇒ angle = 30°

   Hence:

   \[\begin{aligned}\angle R = 30°\\\angle P = 60° \end{aligned} \]

9. Final Answer

   ∠QPR = 60°,   ∠PRQ = 30°

Solution