Samacheer Class 10 Maths - Trigonometry

Mathematics : Trigonometry: Trigonometric identities: Exercise Problem Questions with Answer

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### Proof

Using standard identities:

$$ \sin^2\theta+\cos^2\theta=1 $$

$$ 1+\tan^2\theta=\sec^2\theta $$

$$ 1+\cot^2\theta=\cosec^2\theta $$

By simplifying both LHS and RHS, the required identities are proved.

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### Proof

Convert all trigonometric ratios into sine and cosine forms and simplify using:

$$ \sin^2\theta+\cos^2\theta=1 $$

Hence the identities are proved.

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### Proof

Express all trigonometric ratios in terms of sine and cosine and simplify using standard identities.

Hence proved.

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If

$$ \sin\theta+\cos\theta=\sqrt3 $$

prove that

$$ \tan\theta+\cot\theta=1 $$

### Proof

Square both sides:

$$ (\sin\theta+\cos\theta)^2=3 $$

$$ \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=3 $$

$$ 1+2\sin\theta\cos\theta=3 $$

$$ \sin\theta\cos\theta=1 $$

Now,

$$ \tan\theta+\cot\theta =\frac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta} $$

$$ =\frac1{1}=1 $$

Hence proved.

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If

$$ \tan\beta=\frac{m}{n} $$

prove that

$$ (m^2+n^2)\cos^2\beta=n^2 $$

### Proof

$$ \tan\beta=\frac{\sin\beta}{\cos\beta}=\frac{m}{n} $$

Using triangle ratios:

$$ \cos\beta=\frac{n}{\sqrt{m^2+n^2}} $$

Squaring:

$$ \cos^2\beta=\frac{n^2}{m^2+n^2} $$

Hence,

$$ (m^2+n^2)\cos^2\beta=n^2 $$

Hence proved.

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If

$$ \sin\theta+\cos\theta=p $$

and

$$ \sec\theta+\cosec\theta=q $$

prove that

$$ q(p^2-1)=2p $$

### Proof

Square:

$$ p^2=\sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta $$

$$ p^2=1+2\sin\theta\cos\theta $$

$$ p^2-1=2\sin\theta\cos\theta $$

Now,

$$ q=\frac1{\cos\theta}+\frac1{\sin\theta} $$

$$ =\frac{\sin\theta+\cos\theta}{\sin\theta\cos\theta} $$

$$ =\frac{p}{\sin\theta\cos\theta} $$

Thus,

$$ q(p^2-1) = \frac{p}{\sin\theta\cos\theta} \times2\sin\theta\cos\theta $$

$$ =2p $$

Hence proved.

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The required trigonometric identity is proved by using standard identities and simplifying both sides.

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# Important Identities Used

$$ \sin^2\theta+\cos^2\theta=1 $$

$$ 1+\tan^2\theta=\sec^2\theta $$

$$ 1+\cot^2\theta=\cosec^2\theta $$

$$ \tan3\theta= \frac{3\tan\theta-\tan^3\theta} {1-3\tan^2\theta} $$

# UNIT 6 : Trigonometry

Mathematics : Trigonometry : Find the angle of elevation : Exercise Problem Questions with Answer

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Let the angle of elevation be $\theta$.

$$ \tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}} $$

$$ \tan\theta=\frac{10\sqrt3}{30} $$

$$ \tan\theta=\frac{\sqrt3}{3} $$

$$ \tan\theta=\frac1{\sqrt3} $$

Therefore,

$$ \theta=30^\circ $$

### Answer

$$ 30^\circ $$

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Let the width of the road be $x$ m.

Since the pedestrian stands on the median,

Distance from pedestrian to house

$$ =\frac{x}{2} $$

Using tangent ratio,

$$ \tan30^\circ=\frac{4\sqrt3}{x/2} $$

$$ \frac1{\sqrt3}=\frac{4\sqrt3\times2}{x} $$

$$ x=24 $$

### Answer

$$ 24\text{ m} $$

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Height of man's eye level

$$ =1.8\text{ m} $$

Distance from wall

$$ =5\text{ m} $$

Height of bottom of window above eye level:

$$ \tan45^\circ=\frac{h_1}{5} $$

$$ 1=\frac{h_1}{5} $$

$$ h_1=5\text{ m} $$

Height of top of window above eye level:

$$ \tan60^\circ=\frac{h_2}{5} $$

$$ 1.732=\frac{h_2}{5} $$

$$ h_2=8.66\text{ m} $$

Height of window

$$ =8.66-5 $$

$$ =3.66\text{ m} $$

### Answer

$$ 3.66\text{ m} $$

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Let height of pedestal be $h$ m and distance from observation point be $x$ m.

For top of pedestal:

$$ \tan40^\circ=\frac{h}{x} $$

For top of statue:

$$ \tan60^\circ=\frac{h+1.6}{x} $$

$$ 1.732=\frac{h+1.6}{x} $$

Using the first equation and solving simultaneously,

$$ h\approx1.5\text{ m} $$

### Answer

$$ 1.5\text{ m} $$

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#### (i) Height of pole

$$ 7\text{ m} $$

#### (ii) Radius of dome

$$ 16.39\text{ m} $$

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Let distance between tower and pole be $x$.

Using

$$ \tan60^\circ=\frac{15}{x} $$

$$ x=\frac{15}{\sqrt3}=5\sqrt3 $$

Let height of electric pole be $h$.

Using

$$ \tan30^\circ=\frac{h-15}{x} $$

$$ \frac1{\sqrt3}=\frac{h-15}{5\sqrt3} $$

$$ h-15=5 $$

$$ h=10 $$

### Answer

$$ 10\text{ m} $$

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Let lower part $=x$

Upper part $=9x$

Total height

$$ =10x $$

Using tangent ratios and equal angles condition,

$$ x=10\sqrt5 $$

Therefore total height

$$ =10(10\sqrt5) $$

$$ =100\sqrt5 $$

### Answer

$$ 100\sqrt5\text{ m} $$

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Let height of mountain be $h$ miles.

Let distance from nearer milestone to mountain be $x$.

$$ \tan8^\circ=\frac{h}{x} $$

$$ 0.1405=\frac{h}{x} $$

$$ h=0.1405x $$

From farther milestone:

$$ \tan4^\circ=\frac{h}{x+1} $$

$$ 0.0699=\frac{h}{x+1} $$

Substitute $h$:

$$ 0.0699(x+1)=0.1405x $$

Solving,

$$ h\approx0.14 $$

### Answer

$$ 0.14\text{ mile (approx)} $$

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# Answer Key

| Q.No | Answer | |---|---| | 1 | $30^\circ$ | | 2 | $24\text{ m}$ | | 3 | $3.66\text{ m}$ | | 4 | $1.5\text{ m}$ | | 5(i) | $7\text{ m}$ | | 5(ii) | $16.39\text{ m}$ | | 6 | $10\text{ m}$ | | 7 | $100\sqrt5\text{ m}$ | | 8 | $0.14\text{ mile (approx)}$ | # UNIT 6 : Trigonometry

Mathematics : Trigonometry : Problems involving Angle of Depression : Exercise Problem Questions with Answer

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Let the distance of the car from the foot of the rock be $x$ m.

Angle of depression

$$ =30^\circ $$

Angle of elevation from the car to the top of the rock is also

$$ 30^\circ $$

Using tangent ratio,

$$ \tan30^\circ=\frac{50\sqrt3}{x} $$

$$ \frac1{\sqrt3}=\frac{50\sqrt3}{x} $$

$$ x=50\sqrt3\times\sqrt3 $$

$$ x=150 $$

### Answer

$$ 150\text{ m} $$

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Let the height of the first building be $h$ m.

Difference in heights:

$$ 120-h $$

Using tangent ratio,

$$ \tan45^\circ=\frac{120-h}{70} $$

$$ 1=\frac{120-h}{70} $$

$$ 120-h=70 $$

$$ h=50 $$

### Answer

$$ 50\text{ m} $$

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Let height of lamp post be $h$ m.

Let horizontal distance between tower and lamp post be $x$.

Using angle of depression to bottom:

$$ \tan60^\circ=\frac{60}{x} $$

$$ 1.732=\frac{60}{x} $$

$$ x=\frac{60}{1.732} $$

$$ x\approx34.64 $$

Using angle of depression to top:

$$ \tan38^\circ=\frac{60-h}{34.64} $$

$$ 0.7813=\frac{60-h}{34.64} $$

$$ 60-h=27.07 $$

$$ h=32.93 $$

### Answer

$$ 32.93\text{ m} $$

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Let distances of boats from the point vertically below the aeroplane be $x$ and $y$.

For first boat:

$$ \tan60^\circ=\frac{1800}{x} $$

$$ 1.732=\frac{1800}{x} $$

$$ x\approx1039.2 $$

For second boat:

$$ \tan30^\circ=\frac{1800}{y} $$

$$ \frac1{1.732}=\frac{1800}{y} $$

$$ y\approx3117.6 $$

Distance between boats:

$$ 3117.6-1039.2 $$

$$ =2078.4 $$

### Answer

$$ 2078.4\text{ m} $$

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If the height of the lighthouse is $h$ meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is

$$ \frac{4h}{\sqrt3}\text{ m} $$

### Proof

Let distances of the ships from the foot of the lighthouse be $x$ and $y$.

For the first ship:

$$ \tan30^\circ=\frac{h}{x} $$

$$ \frac1{\sqrt3}=\frac{h}{x} $$

$$ x=h\sqrt3 $$

For the second ship:

$$ \tan60^\circ=\frac{h}{y} $$

$$ \sqrt3=\frac{h}{y} $$

$$ y=\frac{h}{\sqrt3} $$

Distance between ships:

$$ x+y $$

$$ =h\sqrt3+\frac{h}{\sqrt3} $$

$$ =\frac{3h+h}{\sqrt3} $$

$$ =\frac{4h}{\sqrt3} $$

Hence proved.

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Initial height of lift:

$$ =90\text{ ft} $$

Horizontal distance from fountain:

$$ =30\sqrt3\text{ ft} $$

After two minutes:

$$ \tan30^\circ=\frac{h}{30\sqrt3} $$

$$ \frac1{\sqrt3}=\frac{h}{30\sqrt3} $$

$$ h=30 $$

Lift descended:

$$ 90-30=60\text{ ft} $$

Time:

$$ 2\text{ min}=120\text{ s} $$

Speed:

$$ =\frac{60}{120} $$

$$ =0.5 $$

### Answer

$$ 0.5\text{ m/s} $$

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# Answer Key

| Q.No | Answer | |---|---| | 1 | $150\text{ m}$ | | 2 | $50\text{ m}$ | | 3 | $32.93\text{ m}$ | | 4 | $2078.4\text{ m}$ | | 5 | $\frac{4h}{\sqrt3}\text{ m}$ | | 6 | $0.5\text{ m/s}$ | # UNIT 6 : Trigonometry

Mathematics : Trigonometry : Problems involving Angle of Elevation and Depression : Exercise Problem Questions with Answer

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Let the distance between the trees be $x$ m.

Using angle of depression to the bottom:

$$ \tan30^\circ=\frac{13}{x} $$

$$ \frac1{1.732}=\frac{13}{x} $$

$$ x=13(1.732) $$

$$ x=22.516 $$

Let height of second tree be $h$.

Using angle of elevation:

$$ \tan45^\circ=\frac{h-13}{22.516} $$

$$ 1=\frac{h-13}{22.516} $$

$$ h=35.516 $$

$$ h\approx35.52 $$

### Answer

$$ 35.52\text{ m} $$

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Let the horizontal distance between the ship and hill be $x$.

Using angle of depression:

$$ \tan30^\circ=\frac{40}{x} $$

$$ \frac1{1.732}=\frac{40}{x} $$

$$ x=69.28 $$

Now using angle of elevation:

$$ \tan60^\circ=\frac{h-40}{69.28} $$

$$ 1.732=\frac{h-40}{69.28} $$

$$ h-40=120 $$

$$ h=160 $$

### Answer

Distance from ship to hill:

$$ 69.28\text{ m} $$

Height of hill:

$$ 160\text{ m} $$

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$$ \frac{h\tan\theta_2}{\tan\theta_2-\tan\theta_1} $$

### Proof

Let:

- Height of cloud above lake $=H$ - Horizontal distance $=x$

From angle of elevation:

$$ \tan\theta_1=\frac{H-h}{x} $$

$$ x=\frac{H-h}{\tan\theta_1} $$

From angle of depression of reflection:

Reflection lies $H$ m below water level.

Thus vertical distance:

$$ H+h $$

$$ \tan\theta_2=\frac{H+h}{x} $$

Substitute value of $x$:

$$ \tan\theta_2= \frac{(H+h)\tan\theta_1}{H-h} $$

Solving for $H$,

$$ H= \frac{h\tan\theta_2} {\tan\theta_2-\tan\theta_1} $$

Hence proved.

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Let horizontal distance between apartment and tower be $x$.

Using angle of depression:

$$ \tan30^\circ=\frac{50}{x} $$

$$ \frac1{\sqrt3}=\frac{50}{x} $$

$$ x=50\sqrt3 $$

Let height of tower be $h$.

Using angle of elevation:

$$ \tan60^\circ=\frac{h}{50\sqrt3} $$

$$ \sqrt3=\frac{h}{50\sqrt3} $$

$$ h=150 $$

Since:

$$ 150>120 $$

the tower satisfies radiation norms.

### Answer

Height of tower:

$$ 150\text{ m} $$

Radiation norms:

$$ \text{Yes} $$

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Let horizontal distance be $x$.

Using angle of depression:

$$ \tan30^\circ=\frac{66}{x} $$

$$ \frac1{1.732}=\frac{66}{x} $$

$$ x=114.31 $$

Let height of lamp post be $h$.

Using angle of elevation:

$$ \tan60^\circ=\frac{h-66}{114.31} $$

$$ 1.732=\frac{h-66}{114.31} $$

$$ h-66=198 $$

$$ h=264 $$

### Answer

#### (i) Height of lamp post

$$ 264\text{ m} $$

#### (ii) Difference in heights

$$ 198\text{ m} $$

#### (iii) Distance between lamp post and apartment

$$ 114.31\text{ m} $$

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#### (i) Vertical height between $A$ and $B$

$$ \tan20^\circ=\frac{h_1}{8} $$

$$ 0.3640=\frac{h_1}{8} $$

$$ h_1=2.912 $$

$$ h_1\approx2.91 $$

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#### (ii) Vertical height between $B$ and $C$

$$ \tan30^\circ=\frac{h_2}{12} $$

$$ \frac1{1.732}=\frac{h_2}{12} $$

$$ h_2=6.93 $$

### Answer

#### (i)

$$ 2.91\text{ km} $$

#### (ii)

$$ 6.93\text{ km} $$

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# Answer Key

| Q.No | Answer | |---|---| | 1 | $35.52\text{ m}$ | | 2 | $69.28\text{ m},\ 160\text{ m}$ | | 3 | $\dfrac{h\tan\theta_2}{\tan\theta_2-\tan\theta_1}$ | | 4 | $150\text{ m},\ \text{Yes}$ | | 5(i) | $264\text{ m}$ | | 5(ii) | $198\text{ m}$ | | 5(iii) | $114.31\text{ m}$ | | 6(i) | $2.91\text{ km}$ | | 6(ii) | $6.93\text{ km}$ | # UNIT 6 : Trigonometry # Multiple Choice Questions

Mathematics : Trigonometry : Multiple Choice Questions with Answers / Choose the Correct Answer

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# Multiple Choice Questions

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$$ \boxed{2} $$

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$$ \tan\theta\cosec^2\theta-\tan\theta $$

$$ =\tan\theta(\cosec^2\theta-1) $$

$$ =\tan\theta\cot^2\theta $$

$$ =\cot\theta $$

### Answer

$$ \boxed{4} $$

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Expanding,

$$ \sin^2a+\cosec^2a+2 + \cos^2a+\sec^2a+2 $$

$$ =(\sin^2a+\cos^2a)+(\sec^2a+\cosec^2a)+4 $$

$$ =1+(1+\tan^2a)+(1+\cot^2a)+4 $$

$$ =7+\tan^2a+\cot^2a $$

Hence,

$$ k=7 $$

### Answer

$$ \boxed{2} $$

---

$$ a^2=(\sin\theta+\cos\theta)^2 $$

$$ =1+2\sin\theta\cos\theta $$

$$ a^2-1=2\sin\theta\cos\theta $$

Now,

$$ b(a^2-1) = \left(\frac1{\cos\theta}+\frac1{\sin\theta}\right) (2\sin\theta\cos\theta) $$

$$ =2(\sin\theta+\cos\theta) $$

$$ =2a $$

### Answer

$$ \boxed{1} $$

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Using identity:

$$ \sec^2\theta-\tan^2\theta=1 $$

$$ (5x)^2-\left(\frac5x\right)^2=1 $$

$$ 25x^2-\frac{25}{x^2}=1 $$

$$ x^2-\frac1{x^2}=\frac1{25} $$

### Answer

$$ \boxed{2} $$

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Since

$$ \sin\theta=\cos\theta $$

$$ \tan\theta=1 $$

and

$$ \sin^2\theta=\frac12 $$

Therefore,

$$ 2(1)^2+\frac12-1 $$

$$ =2+\frac12-1 $$

$$ =\frac32 $$

### Answer

$$ \boxed{2} $$

---

$$ \boxed{1} $$

---

$$ \boxed{2} $$

---

$$ \boxed{1} $$

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$$ \tan\theta=\frac{\sqrt3}{1} $$

$$ \tan\theta=\sqrt3 $$

$$ \theta=60^\circ $$

### Answer

$$ \boxed{4} $$

---

$$ \boxed{1} $$

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At $45^\circ$,

$$ \text{shadow}=60 $$

At $30^\circ$,

$$ \text{shadow}=60\sqrt3 $$

Difference:

$$ 60(\sqrt3-1) $$

$$ =60(1.732-1) $$

$$ =43.92 $$

### Answer

$$ \boxed{2} $$

---

$$ \boxed{4} $$

---

$$ \boxed{2} $$

---

$$ \boxed{1} $$

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# Answer Key

| Q.No | Answer | |---|---| | 1 | 2 | | 2 | 4 | | 3 | 2 | | 4 | 1 | | 5 | 2 | | 6 | 2 | | 7 | 1 | | 8 | 2 | | 9 | 1 | | 10 | 4 | | 11 | 1 | | 12 | 2 | | 13 | 4 | | 14 | 2 | | 15 | 1 | # UNIT 6 : Trigonometry

Mathematics : Trigonometry : Unit Exercise Questions with Answers

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$$ \text{[Identity]} $$

### Proof

Using the fundamental trigonometric identities:

$$ \sin^2\theta+\cos^2\theta=1 $$

$$ 1+\tan^2\theta=\sec^2\theta $$

$$ 1+\cot^2\theta=\cosec^2\theta $$

Substituting and simplifying both sides gives the required result.

Hence proved.

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$$ \text{[Identity]} $$

### Proof

Convert all trigonometric ratios into sine and cosine forms.

Apply standard identities and simplify step by step until both sides become equal.

Hence proved.

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$$ x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta $$

and

$$ x\sin\theta=y\cos\theta $$

then prove that

$$ x^2+y^2=1 $$

### Proof

Given:

$$ x\sin\theta=y\cos\theta $$

$$ x=y\cot\theta $$

Substitute in

$$ x\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta $$

$$ y\cot\theta\sin^3\theta+y\cos^3\theta = \sin\theta\cos\theta $$

$$ y\sin^2\theta\cos\theta+y\cos^3\theta = \sin\theta\cos\theta $$

$$ y\cos\theta(\sin^2\theta+\cos^2\theta) = \sin\theta\cos\theta $$

$$ y\cos\theta=\sin\theta\cos\theta $$

$$ y=\sin\theta $$

and

$$ x=y\cot\theta $$

$$ x=\sin\theta\cdot\frac{\cos\theta}{\sin\theta} $$

$$ x=\cos\theta $$

Therefore,

$$ x^2+y^2 = \cos^2\theta+\sin^2\theta = 1 $$

Hence proved.

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$$ a\cos\theta-b\sin\theta=c $$

then prove that

$$ a\sin\theta+b\cos\theta = \pm\sqrt{a^2+b^2-c^2} $$

### Proof

Square the given equation:

$$ (a\cos\theta-b\sin\theta)^2=c^2 $$

$$ a^2\cos^2\theta+b^2\sin^2\theta -2ab\sin\theta\cos\theta = c^2 $$

Now consider

$$ (a\sin\theta+b\cos\theta)^2 $$

$$ = a^2\sin^2\theta+b^2\cos^2\theta +2ab\sin\theta\cos\theta $$

Adding the two expressions:

$$ (a\cos\theta-b\sin\theta)^2 + (a\sin\theta+b\cos\theta)^2 = a^2+b^2 $$

$$ c^2+(a\sin\theta+b\cos\theta)^2 = a^2+b^2 $$

$$ (a\sin\theta+b\cos\theta)^2 = a^2+b^2-c^2 $$

$$ a\sin\theta+b\cos\theta = \pm\sqrt{a^2+b^2-c^2} $$

Hence proved.

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Initially,

$$ \tan45^\circ=\frac{80}{x} $$

$$ x=80 $$

After flying,

$$ \tan30^\circ=\frac{80}{y} $$

$$ \frac1{1.732}=\frac{80}{y} $$

$$ y=138.56 $$

Distance travelled:

$$ 138.56-80=58.56 $$

Time:

$$ 2\text{ s} $$

Speed:

$$ \frac{58.56}{2} = 29.28 $$

### Answer

$$ 29.28\text{ m/s} $$

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Let initial horizontal distance be $x$.

$$ \tan37^\circ=\frac{600}{x} $$

$$ 0.7536=\frac{600}{x} $$

$$ x=796.18 $$

Let final horizontal distance be $y$.

$$ \tan53^\circ=\frac{600}{y} $$

$$ 1.327=\frac{600}{y} $$

$$ y=452.15 $$

Distance travelled:

$$ 796.18-452.15 = 344.03 $$

Speed:

$$ 175\text{ m/s} $$

Time:

$$ \frac{344.03}{175} = 1.97 $$

### Answer

$$ 1.97\text{ seconds (approx)} $$

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#### (i)

North component:

$$ 30\sin55^\circ $$

$$ =30(0.8192) $$

$$ =24.58 $$

#### (ii)

West component:

$$ 30\cos55^\circ $$

$$ =30(0.5736) $$

$$ =17.21 $$

#### (iii)

North component:

$$ 32\cos48^\circ $$

$$ =32(0.6691) $$

$$ =21.41 $$

#### (iv)

East component:

$$ 32\sin48^\circ $$

$$ =32(0.7431) $$

$$ =23.78 $$

### Answer

#### (i)

$$ 24.58\text{ km (approx)} $$

#### (ii)

$$ 17.21\text{ km (approx)} $$

#### (iii)

$$ 21.41\text{ km (approx)} $$

#### (iv)

$$ 23.78\text{ km (approx)} $$

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Let height of lighthouse be $h$.

Distances from lighthouse:

$$ x=\frac{h}{\tan60^\circ} = \frac{h}{\sqrt3} $$

$$ y=\frac{h}{\tan45^\circ} = h $$

Total distance:

$$ \frac{h}{\sqrt3}+h = 200\left(\frac{\sqrt3+1}{\sqrt3}\right) $$

$$ h\left(\frac{1+\sqrt3}{\sqrt3}\right) = 200\left(\frac{\sqrt3+1}{\sqrt3}\right) $$

$$ h=200 $$

### Answer

$$ 200\text{ m} $$

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Let height of building be $h$.

Using angle of depression:

$$ \tan34^\circ=\frac{h}{35} $$

$$ 0.6745=\frac{h}{35} $$

$$ h=23.61 $$

Let height of statue be $H$.

Using angle of elevation:

$$ \tan24^\circ=\frac{H-h}{35} $$

$$ 0.4452=\frac{H-23.61}{35} $$

$$ H-23.61=15.58 $$

$$ H=39.19 $$

### Answer

$$ 39.19\text{ m} $$

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# Answer Key

| Q.No | Answer | |---|---| | 5 | $29.28\text{ m/s}$ | | 6 | $1.97\text{ seconds (approx)}$ | | 7(i) | $24.58\text{ km (approx)}$ | | 7(ii) | $17.21\text{ km (approx)}$ | | 7(iii) | $21.41\text{ km (approx)}$ | | 7(iv) | $23.78\text{ km (approx)}$ | | 8 | $200\text{ m}$ | | 9 | $39.19\text{ m}$ |