Applications of trigonometry

1. Ye topic hai kya?

Right triangle zameen pe: tower ki height, aankh se doori, kite ki dori. Angle horizontal se measure. Paper: 30°/45°/60° se h = d tan θ, do jagah se do elevations, lighthouse se do ships.

Yahan lock

Line of sight, elevation (up), depression (down), tan = height/distance, sin jab string/hyp, two-triangle, same-side vs opposite ships.

Yahan kya nahi

Sirf sin 30=1/2 bina figure (Intro). Pole 6 m, shadow 4 m, angle nahi (Triangles similar). Sihi 13 m, 5 m door, angle nahi (Pythagoras). Clock 5:20.

2. Angle of elevation — neeche se up

Observer ground (ya aankh) pe. Horizontal aankh se nikalti line. Object ka top us horizontal se upar dikhe to beech ka angle = elevation. Line of sight = aankh se top tak.

θ elevation. Tower vertical, ground horizontal → right triangle. tan θ = h/d.

Ground problems (aankh ground pe assume, jab tak height of eye na ho): tan θ = opposite/adjacent = h/d → h = d tan θ, d = h / tan θ = h cot θ.

Example

Tower se 24 m door, top ka elevation 30°. Height?

Solution

tan 30
h/24 = 1/√3 → h = 24/√3 = 8√3 m.
Trap
sin 30 se 24×1/2=12 — hyp nahi di. 60° mix se 24√3. Similar-shadow (6 m pole) yahan angle hai, ratio of shadows nahi.

Answer8√3 m

45° checkpoint: tan 45° = 1 → height = distance. Tower 40 m, elevation 45° → observer 40 m door.

3. Angle of depression — upar se neeche

Observer ooncha (lighthouse, building top). Horizontal aankh se. Object neeche, line of sight horizontal se neeche → depression.

α depression, dashed = eye-horizontal. Alternate angles: ground pe elevation bhi α. tan α = h/d.

Example

30 m building ke top se car ka depression 60°. Car kitni door?

Solution

tan 60
30/d = √3 → d = 30/√3 = 10√3 m.

Answer10√3 m

Horizontals parallel (eye-line aur ground) → depression (upar) = elevation (neeche wale point se top ki taraf). Figure banao, phir tan. “Depression 60 isliye elevation 30” mat ghumao.

4. Kaun sa ratio — tan vs sin

Diya	Maanga	Ratio
d, θ	height	tan θ = h/d
h, θ	distance	tan θ = h/d ya cot
string / slope length, θ	height	sin θ = h/hyp
string, θ	ground doori	cos θ = d/hyp

Example

Patang ki dori 80 m, elevation 60°. Patang ki height? (dori tight, straight.)

Solution

sin
hyp = 80. h = 80 sin 60° = 80·√3/2 = 40√3 m.
Nahi
tan 60 se 80√3 — 80 adjacent nahi, hypotenuse hai. Sihi 13 m / 5 m (angle-less Pythagoras) alag.

Answer40√3 m

5. Do elevations — walk towards / away

Do jagah se same tower. Doosra triangle chhota (paas) ya bada (door). Height same, base alag. Subtract/add the walk.

Paas aane pe elevation badhta. 30° door, 60° paas, beech 18 m.

Example

Pehle elevation 30°. Tower ki taraf 18 m chale, ab 60°. Height?

Solution

Do tan
Paas wali base x: h = x tan 60° = x√3. Door wali: h = (x+18) tan 30° = (x+18)/√3.
Equal
x√3 = (x+18)/√3 → 3x = x+18 → 2x = 18 → x = 9. h = 9√3 m.

Answer9√3 m

Ulta walk (door jaana): elevation girta, 60° phir 30°. Same algebra, + walk door wali base pe.

30°/60° swap: tan 30 aur tan 60 ulta lagaoge to x negative ya height galat. Figure: bada angle = chhoti base. Towards = paas wala 60°.

6. Window + do ships

Window: aankh building mein, saamne tower. Upar elevation (top), neeche depression (foot). Dono ka base same d. Total tower = neeche ka hissa + window height + (kabhi window ke upar extra).

Example

Khidki ground se 15 m. Saamne tower ke top ka elevation 30°, foot ka depression 45°. Tower ki height?

Solution

Neeche
tan 45° = 15/d → d = 15 m.
Upar
tan 30° = extra/15 → extra = 15/√3 = 5√3. Height = 15 + 5√3 m.

Answer15 + 5√3 m

Do ships, same side: lighthouse h, depressions 45° (paas) aur 30° (door). Distances h cot 45, h cot 30. Beech = far − near.

Example

60 m lighthouse. Do ships same side, depression 45° aur 30°. Unke beech doori?

Solution

Near / far
45° → d₁ = 60 cot 45° = 60. 30° → d₂ = 60 cot 30° = 60√3.
Gap
60√3 − 60 = 60(√3 − 1) m. Opposite sides hote to sum 60(√3+1).

Answer60(√3 − 1) m

7. Sawal — basic se pro

Q1 · Basic

Elevation kahan measure?

Horizontal se upar, line of sight tak
Vertical tower se
Ground se 90° hamesha
Clock hands jaisa

Solution

Def
A. C: 90 tab jab object zenith, yahan nahi.

AnswerA

Q2 · Basic

24 m door, elevation 30°. Height?

8√3 m
12 m
24√3 m
24 m

Solution

24/√3
8√3. B: sin 30. C: tan 60. D: tan 45.

Answer8√3 m

Q3 · Basic

Tower 40 m, elevation 45°. Observer kitni door?

40 m
40√3 m
40/√3 m
20 m

Solution

tan 45=1
d=h=40.

Answer40 m

Q4 · Basic

30 m top se depression 60°. Car doori?

10√3 m
30√3 m
15 m
30 m

Solution

30/√3
10√3. B: tan 30 se 30√3 (elevation 30 mix).

Answer10√3 m

Q5 · Medium

Dori 80 m, elevation 60°. Patang height?

40√3 m
80√3 m
40 m
80 m

Solution

sin 60
40√3. B: tan, 80 ko base maan liya.

Answer40√3 m

Q6 · Medium

30° phir 18 m towards, 60°. Height?

9√3 m
18√3 m
9 m
18 m

Solution

x=9
h=9√3. B: 18 ko x maan. D: walk ko height.

Answer9√3 m

Q7 · Medium

Khidki 15 m, elev 30°, dep 45°. Tower height?

15 + 5√3 m
15 m
5√3 m
15√3 m

Solution

d=15
+ 5√3. B: sirf window. C: sirf upar ka tukda.

Answer15 + 5√3 m

Q8 · Medium

60 m light, ships same side 45° aur 30° depression. Beech?

60(√3 − 1) m
60(√3 + 1) m
60 m
60√3 m

Solution

Far−near
60√3−60. B: opposite sides ka sum.

Answer60(√3 − 1) m

Q9 · Pro

60° elevation, 12 m away walk, ab 30°. Height?

6√3 m
12√3 m
6 m
18√3 m

Solution

Away
h = y√3 = (y+12)/√3 → 3y = y+12 → y=6. h=6√3. Towards wala 9√3 mat copy.

Answer6√3 m

Q10 · Pro

Dori 90 m, elevation 30°. Height?

45 m
90√3 m
45√3 m
30 m

Solution

90 sin 30
45. C: sin 60 mix. B: tan 60 × 90.

Answer45 m

Practice

Pehle khud, phir neeche kholo.

1 2 3 4 5 6 7 8 9 10

Depression kahan?

Oonchi jagah, horizontal se neeche
Ground se upar
Hamesha 45°
sin ka dusra naam

Solution

Def
A. B elevation.

AnswerA

d = 10 m, elevation 60°. h?

10√3 m
10/√3 m
5 m
10 m

Solution

10 tan 60
10√3. B tan 30.

Answer10√3 m

h = 20√3 m, elevation 60°. d?

20 m
20√3 m
60 m
10 m

Solution

tan 60
20√3 / d = √3 → d=20.

Answer20 m

40 m se depression 30°. Ship doori?

40√3 m
40/√3 m
20 m
40 m

Solution

cot 30=√3
40√3. B tan 30 se.

Answer40√3 m

Ground problems mein usually kaun sa?

tan θ = h/d
sin²+cos² table sum
Distance formula (x,y)
DE ∥ BC

Solution

Opp/adj
tan. B Intro chapter. C Coordinate. D Triangles BPT.

AnswerA

60 m light, ships opposite sides, depression 45° aur 30°. Beech?

60(√3+1) m
60(√3−1) m
60√3 m
120 m

Solution

Sum
Near 60, far 60√3, opposite ⇒ jod: 60+60√3 = 60(√3+1). B = same-side Q8 (subtract).

Answer60(√3+1) m

Dori 80 m, 60°. Ground pe doori (base)?

40 m
40√3 m
80 m
80/√3 m

Solution

cos 60
d = 80 · 1/2 = 40. Height 40√3 thi; base alag.

Answer40 m

Elevation 0° matlab?

Object aankh ke level, h=0 relative
Object infinitely far hamesha
tan undefined
90° tower

Solution

tan 0=0
h=0. C tan 90.

AnswerA

Ships opposite sides, 45° aur 30°, h=60. Beech?

60(√3+1) m
60(√3−1) m
0
60 m

Solution

Sum
60+60√3. B same-side subtract.

Answer60(√3+1) m

P10

Khidki 15 m, dep 45° foot. Tower foot kitni door?

15 m
15√3 m
5√3 m
30 m

Solution

tan 45=1
d=15. C upar wala extra; yahan sirf foot.

Answer15 m