Solutions: Applications of Trigonometry - Test 1

Q1. (a) \(10\sqrt{3}\) m

Let height be \(h\). Distance \(d = 30\) m. Angle \(\theta = 30^\circ\).
\(\tan 30^\circ = \frac{h}{30} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{30} \Rightarrow h = \frac{30}{\sqrt{3}} = 10\sqrt{3}\) m.

Q2. (a) \(60^\circ\)

\(\tan \theta = \frac{\text{Height}}{\text{Shadow}} = \frac{6}{2\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\).
\(\tan \theta = \sqrt{3} \Rightarrow \theta = 60^\circ\).

Q3. (b) is decreasing

As the sun goes lower in the sky (angle of elevation decreases), the length of the shadow increases.

Q4. (b) Both A and R are true but R is not the correct explanation of A.

Assertion is true: \(\tan \theta = \frac{h}{l} = 1 \Rightarrow \theta = 45^\circ\).
Reason is true (Pythagoras theorem), but the explanation for the angle relies on trigonometric ratios, not just Pythagoras theorem.

Q5. Find height of pole.

Length of rope (hypotenuse) = 20 m. Angle = \(30^\circ\).
\(\sin 30^\circ = \frac{\text{Height}}{20} \Rightarrow \frac{1}{2} = \frac{h}{20} \Rightarrow h = 10\) m.

Q6. Find length of string.

Height = 60 m. Angle = \(60^\circ\).
\(\sin 60^\circ = \frac{60}{\text{Length}} \Rightarrow \frac{\sqrt{3}}{2} = \frac{60}{L} \Rightarrow L = \frac{120}{\sqrt{3}} = 40\sqrt{3}\) m.

Q7. Find height of chimney.

Let height of chimney above observer's eye level be \(h\).
\(\tan 45^\circ = \frac{h}{28.5} \Rightarrow 1 = \frac{h}{28.5} \Rightarrow h = 28.5\) m.
Total height = \(h + \text{Observer's height} = 28.5 + 1.5 = 30\) m.

Q8. Find distance between ships.

Let the lighthouse be \(h = 100\) m.
Ship 1 distance \(x\): \(\tan 30^\circ = \frac{100}{x} \Rightarrow x = 100\sqrt{3}\) m.
Ship 2 distance \(y\): \(\tan 45^\circ = \frac{100}{y} \Rightarrow y = 100\) m.
Total distance = \(x + y = 100\sqrt{3} + 100 = 100(\sqrt{3} + 1)\) m.

Q9. Find height of tree.

Let broken part be \(AC\) (hypotenuse) and standing part be \(AB\) (perpendicular). Base \(BC = 8\) m.
\(\cos 30^\circ = \frac{8}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{AC} \Rightarrow AC = \frac{16}{\sqrt{3}}\).
\(\tan 30^\circ = \frac{AB}{8} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{8} \Rightarrow AB = \frac{8}{\sqrt{3}}\).
Total height = \(AB + AC = \frac{24}{\sqrt{3}} = 8\sqrt{3}\) m.

Q10. Find height of tower.

Let building height \(AB = 7\) m. Tower \(CD = h\).
From top of building A, angle of depression of foot C is \(45^\circ\).
Horizontal distance \(BC = 7 \cot 45^\circ = 7\) m.
From A, angle of elevation of top D is \(60^\circ\). Let horizontal line from A meet CD at E.
\(AE = BC = 7\) m. \(CE = AB = 7\) m.
In \(\triangle AED\), \(\tan 60^\circ = \frac{DE}{AE} \Rightarrow \sqrt{3} = \frac{DE}{7} \Rightarrow DE = 7\sqrt{3}\) m.
Height of tower \(h = CE + DE = 7 + 7\sqrt{3} = 7(\sqrt{3} + 1)\) m.

(i) \(\tan \theta = \frac{\text{Height}}{\text{Distance}} = \frac{42}{42} = 1 \Rightarrow \theta = 45^\circ\).

(ii) \(\tan 60^\circ = \frac{42}{d} \Rightarrow \sqrt{3} = \frac{42}{d} \Rightarrow d = \frac{42}{\sqrt{3}} = 14\sqrt{3}\) m.

(iii) \(\tan 60^\circ = \frac{h}{20} \Rightarrow \sqrt{3} = \frac{h}{20} \Rightarrow h = 20\sqrt{3}\) m.