Solutions: Applications of Trigonometry - Vardaan Learning Institute

1. Ratio of pole to shadow $\sqrt{3}:1$. Angle of elevation is:

$\tan \theta = P/B = \sqrt{3}/1 = \sqrt{3} \Rightarrow \theta = 60^\circ$.

Correct Option: (c) $60^\circ$

2. Ladder makes $60^\circ$ with ground. Foot 2m away. Length of ladder:

$\cos 60^\circ = Base/Hyp = 2/L$.
$1/2 = 2/L \Rightarrow L = 4$ m.

Correct Option: (a) 4 m

3. Depression of car $30^\circ$, 150m high tower. Distance from tower:

$\tan 30^\circ = 150/x \Rightarrow 1/\sqrt{3} = 150/x \Rightarrow x = 150\sqrt{3}$ m.

Correct Option: (b) $150\sqrt{3}$ m

4. 20m away from foot, elevation $60^\circ$. Height of tower:

$\tan 60^\circ = H/20 \Rightarrow \sqrt{3} = H/20 \Rightarrow H = 20\sqrt{3}$.

Correct Option: (b) $20\sqrt{3}$ m

5. Pole 6m high, shadow $2\sqrt{3}$m. Sun's elevation:

$\tan \theta = 6/2\sqrt{3} = 3/\sqrt{3} = \sqrt{3} \Rightarrow \theta = 60^\circ$.

Correct Option: (a) $60^\circ$

6. Elevation angles complementary, dist a and b. Height of tower:

$\tan \theta = h/a$. $\tan(90-\theta) = h/b \Rightarrow \cot \theta = h/b$.
$\tan \theta \cdot \cot \theta = (h/a)(h/b) \Rightarrow 1 = h^2/ab \Rightarrow h = \sqrt{ab}$.

Correct Option: (b) $\sqrt{ab}$

7. If the length of the shadow of a tower is increasing, then the angle of elevation...

As shadow length increases (denominator in $H/L$), the value of $\tan \theta$ decreases, so $\theta$ decreases.

Correct Option: (b) Decreases

8. A kite is flying at 50m. String 100m. Inclination?

$\sin \theta = \text{Height}/\text{String} = 50/100 = 1/2$.
$\theta = 30^\circ$.

Correct Option: (a) $30^\circ$

9. Elevation when shadow equals height:

$\tan \theta = H/H = 1 \Rightarrow \theta = 45^\circ$.

Correct Option: (b) $45^\circ$

10. Assertion: Both height and distance increased by 10%...

New $\tan \theta' = (1.1H)/(1.1B) = H/B = \tan \theta$. Angle unchanged.

Correct Option: (a) Both A and R are true and R is the correct explanation of A.

11. Kite height 60m, string $60^\circ$. Length of string.

$\sin 60^\circ = 60/L \Rightarrow \sqrt{3}/2 = 60/L \Rightarrow L = 120/\sqrt{3} = 40\sqrt{3}$.

$40\sqrt{3}$ m

12. Broken tree hits ground at $30^\circ$, 8m from base. Height of tree.

Let top pt be C, break pt B, foot A.
$AC = 8$. In $\triangle BAC$, $\tan 30 = AB/8 \Rightarrow AB = 8/\sqrt{3}$.
$\cos 30 = 8/BC \Rightarrow \sqrt{3}/2 = 8/BC \Rightarrow BC = 16/\sqrt{3}$.
Total Height = $AB+BC = 24/\sqrt{3} = 8\sqrt{3}$ m.

$8\sqrt{3}$ m

13. Shadow 40m longer at $30^\circ$ than $60^\circ$. Height of tower.

Let height h, shadow x at 60. Shadow x+40 at 30.
$\tan 60 = h/x \Rightarrow h = x\sqrt{3}$.
$\tan 30 = h/(x+40) \Rightarrow 1/\sqrt{3} = x\sqrt{3}/(x+40)$.
$x+40 = 3x \Rightarrow 2x = 40 \Rightarrow x=20$.
$h = 20\sqrt{3}$.

$20\sqrt{3}$ m

14. Observer 1.5m, 28.5m away, angle $45^\circ$. Height of chimney.

Let Chimney be AB. Observer CD. Eyes E.
$\triangle ADE$ (where A is top relative to eyes). Base = 28.5.
$\tan 45 = AE/28.5 \Rightarrow AE = 28.5$.
Total Height = $AE + 1.5 = 28.5 + 1.5 = 30$ m.

30 m

15. From 7m building, tower top $60^\circ$, foot $45^\circ$. Height of tower.

Depression $45^\circ$ means distance between building and tower = 7 m.
Height of tower above building h'.
$\tan 60 = h'/7 \Rightarrow h' = 7\sqrt{3}$.
Total height = $7 + 7\sqrt{3} = 7(\sqrt{3}+1)$.

$7(\sqrt{3}+1)$ m

16. From 75m lighthouse, ships $30^\circ, 45^\circ$. Distance between ships.

$\tan 45 = 75/x \Rightarrow x = 75$.
$\tan 30 = 75/(x+d) \Rightarrow 1/\sqrt{3} = 75/(75+d)$.
$75+d = 75\sqrt{3} \Rightarrow d = 75(\sqrt{3}-1)$.

$75(\sqrt{3}-1)$ m

17. Balloon height 88.2m, girl 1.2m. Angle $60^\circ \to 30^\circ$. Distance travelled.

Effective height h = $88.2 - 1.2 = 87$ m.
$\tan 60 = 87/x \Rightarrow x = 87/\sqrt{3} = 29\sqrt{3}$.
$\tan 30 = 87/(x+d) \Rightarrow x+d = 87\sqrt{3}$.
$d = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}$.

$58\sqrt{3}$ m

18. Angles complementary, dist 4m, 9m. Prove height 6m.

Using formula from Q6: $h = \sqrt{ab}$.
$h = \sqrt{4 \times 9} = \sqrt{36} = 6$ m.

Proved

19. Two pillars, road 80m. Heights equal. angles 60, 30.

Let height H. Point P at x meter from one pillar.
$H/x = \tan 60 = \sqrt{3} \Rightarrow H = x\sqrt{3}$.
$H/(80-x) = \tan 30 = 1/\sqrt{3} \Rightarrow H = (80-x)/\sqrt{3}$.
$x\sqrt{3} = (80-x)/\sqrt{3} \Rightarrow 3x = 80-x \Rightarrow 4x = 80 \Rightarrow x = 20$.
Position: 20m from pole with $60^\circ$ elevation.
Height: $20\sqrt{3}$ m.

Height: $20\sqrt{3}$ m; Point: 20m from first pillar

20. Ship deck 10m high. Hill top 60, base 30. Dist and Height.

Let hill height H. Ship dist D.
Height of hill above deck level $= H-10$. depression angle 30 to base means angle of elevation from base to deck is 30.
$10/D = \tan 30 = 1/\sqrt{3} \Rightarrow D = 10\sqrt{3}$ m.
$(H-10)/D = \tan 60 = \sqrt{3} \Rightarrow H-10 = 10\sqrt{3} \cdot \sqrt{3} = 30$.
$H = 40$ m.

Distance: $10\sqrt{3}$ m; Height: 40 m

21. Cloud elevation $\alpha$, reflection depression $\beta$. Prove height formula.

Let cloud height be H. Observer height h. Dist d.
Cloud above observer $= H-h$. Reflection below lake surface $= H$. Reflection below observer $= H+h$.
$\tan \alpha = (H-h)/d \Rightarrow d = (H-h)\cot \alpha$.
$\tan \beta = (H+h)/d \Rightarrow d = (H+h)\cot \beta$.
$(H-h)\cot \alpha = (H+h)\cot \beta \Rightarrow (H-h)/\tan \alpha = (H+h)/\tan \beta$.
$H\tan\beta - h\tan\beta = H\tan\alpha + h\tan\alpha$.
$H(\tan\beta - \tan\alpha) = h(\tan\beta + \tan\alpha)$.
$H = \frac{h(\tan\beta + \tan\alpha)}{\tan\beta - \tan\alpha}$.

Proved

22. Car approaching tower. 30 deg to 60 deg in 6 sec. Time to reach foot.

Let tower height h. Distances $d_1$ (at 30) and $d_2$ (at 60).
$h/d_1 = \tan 30 \Rightarrow d_1 = h\sqrt{3}$.
$h/d_2 = \tan 60 \Rightarrow d_2 = h/\sqrt{3}$.
Distance covered in 6s: $d_1 - d_2 = h(\sqrt{3} - 1/\sqrt{3}) = 2h/\sqrt{3}$.
Speed $v = (2h/\sqrt{3})/6 = h/(3\sqrt{3})$.
Time to cover remaining distance $d_2$:
$t = d_2/v = (h/\sqrt{3}) / (h/3\sqrt{3}) = 3$ seconds.

3 seconds

23. Case Study: Skyscraper. Boy 100m dist, 30 elev. Girl on 20m building, 45 elev.

(i) Height of bird ($H$) from ground: In boy's triangle, hypotenuse is 100m.
$\sin 30 = H/100 \Rightarrow 1/2 = H/100 \Rightarrow H=50$ m.
(ii) Distance of bird from girl ($D$): Height relative to girl $= 50-20=30$.
$\sin 45 = 30/D \Rightarrow 1/\sqrt{2} = 30/D \Rightarrow D = 30\sqrt{2}$ m.
(iii) Horizontal distance: Boy dist $x_B = 100 \cos 30 = 50\sqrt{3}$. Girl dist $x_G = 30 \cot 45 = 30$.
Total distance (opposite sides) = $50\sqrt{3} + 30$ m.

(i) 50 m (ii) $30\sqrt{2}$ m (iii) $50\sqrt{3} + 30$ m