Chapter-8, INTRODUCTION TO TRIGONOMETRY AND ITS APPLICATIONS

Explanation:

Cos A = 4/5 provided

The value of tan A must be determined.

We are aware that cos x = adjacent/hypotenuse.

Adjoining = 4

(5) Hypotenuse

(Hypotenuse) Using the Pythagorean Theorem2 equals (opposite)2 + (nearby)2.² (5)2 = (contrary)² + (4)²

 25 = (contrary)²+ 16

(contrary)² = 25 - 16 

(contrary)² = 9

Using the square root,

Contrary = 3

We are aware that tan A = opposite/nearby tan A = 3/4.

As a result, tan A has a value of 3/4.

Explanation:

Provided that sin A = 0,

We need to determine what cot A is worth.

As we are aware, sin A = opposite and hypotenuse

Contrary = 1

2 Hypotenuse

(Hypotenuse) Using the Pythagorean Theorem2 = (contrary)²+ (nearby)² (2)² = (contiguous)² + (1)²

4 = (contiguous)2 + 1 (nearby)² = 4 - 1 (nearby)² = 3

Using the square root,

Adjoining = 3

We are aware that cot A equals next to or across from

Cot A = √3/1

Therefore, cot A has a value of 3.

Explanation:

The value of the formula [cosec (75° +  θ) - sec (15° - θ ) - tan (55° + θ ) + cot (35° - θ )] must be determined.

Cosec A is known to equal sec (90° - A).

Cot A also equals tan (90° - A).

Cosec (75° + θ ) now equals sec (90° - (75° +  θ)) equals sec (15° - θ ), and cot (35° - θ ) equals tan (90° - (35° - )) equals tan (55°+ ).

This means that [cosec (75° +  θ) - sec (15° -  θ) - tan (55° +  θ) + cot (35° -  θ)]

Tan (55° + θ ) + Tan (55° + θ ) = [sec (15° +  θ) - sec (15° - θ ) - tan (55° + θ )]

= 0

As a result, the formula [cosec (75° +  θ) - sec (15° - θ ) - tan (55° +  θ) + cot (35° - θ )] has no value.

Explanation:

Since sin = a/b,

We must ascertain the cos valueθ.

Knowing that sin A equals the reverse and hypotenuse

Contrary = a

b = hypotenuse

(Hypotenuse) Using the Pythagorean Theorem² = (contrary)² + (nearby)² (b)²= (contiguous)² + (a)² (contiguous)² = b² - a²

Using the square root,

Adjoining = (b² - a2), which

We are aware that adjacent/hypotenuse = cos A

cos = b² - a² / b

Cos is therefore equal to (b² - a²)/b.

Explanation:

Since cos(a+β) = 0,

We must determine the cost of sin (a- β).

According to the trigonometric ratios, cos(a+β) = 0 (a +β) = cos-1(0)

At 90°, cos is zero.

So, (α + β) = 90°

Now, α = 90° - β

Change the value of, and you get sin( - ) = sin(90° -β  -β ) = sin(90° - 2β).

Sin (90° - A) = cos A is the trigonometric ratio of complementary angles.

Sin (90° - 2β) = cos (2), then.

As a result, sin (90-A) = cos 2.

Explanation:

The ratio of (tan1°, tan2°, tan3°,... tan89°) must be determined.

Tan (90° - A) = cot A is a trigonometric ratio of complement angles.

You can write the formula as (tan1°, tan2°, tan3°,... tan 44°). Tan 45 degrees (tan(90°- 44°)) tan(90°- 43°) Tan (90°- 42°) = (tan1°, tan2°, tan3°,... tan 44°) Tan (90°- 1°) Tan 45° (cot 44°, cot 43°, cot 42°, cot 41°, cot 1°)

We are aware that tan cot 1

Thus, (tan1°x  cot1°) (tan2° xcot2°) (tan3° xcot3°) tan 45°..........(tan44° xcot44°) = 1 1 1 ................. 1 = 1

The value of (tan1°, tan2°, tan3°,..., tan89°) is therefore 1.

Explanation:

Provided that sin = cos 

Given, 9α < 90°

The value of tan5 must be determined.

9 is an acute angle because it is 90 degrees.

Sin (90° - A) = cos A is a trigonometric ratio of complementary angles.

Cos 9 = sin(90° - 9), then.

Consequently, sin(90° - 9) = sin 90° - 9 =

When simplified, + 9 = 90°

10α = 90°

α = 90°/10 α = 9° 

To locate tan5

Change the value to provide tan 5 = tan(5 9) = tan 45°.

Tan 45° = 1 using the trigonometric ratio of angles.

Explanation:

because angle C is 90 degrees and ABC is right-angled

thus,

A+B=180-C

A+B=180-90

A+B=90

Consequently, cos(A+B)=cos90. =0

Explanation:

SinA + Sin2 A = 1 given

Sin A = 1 - sin2 A can be used to represent it. First, we need to calculate the value of (cos2 A + cos4 A).

Cos2 A = 1 - Sin2 A is the trigonometric identity that is used. (2) The second equation is sin A = cos2 A.

The result is that (cos2 A + cos4 A) = (cos2 A + (sin A)2) = cos2 A + sin2 A.

Cos2 + Sin2 = 1 using the trigonometric identities.

(Cos2 A + Cos4 A) = 1 as a result.

Explanation:

Because sinα and cosβ are both halves,

We must determine the value of ( +β ).

The trigonometric ratios of angles can be used to:

Given that sinα = /

α = sin⁻¹(1/2)

α = 30°

Given that cosβ = 1/2 = cosβ 1(1/2) = 60 degrees

Now, (α + β) = 30° + 60° = 90°

As a result, the value of ( a+β ) equals 90°.

Explanation:

The solution is (B).

With that expression, sin222∘+sin268∘cos222∘/cos268∘+sin263∘+cos63∘ sin27∘=sin222∘/sin2(90∘−22∘)cos2(90∘−68∘)+cos268∘+sin263∘+cos63∘sin27∘=sin222∘+cos222∘sin268∘+co268∘+sin263∘+cos63∘cos63∘[cos(90)=sin and sin(90)=cos] = 11 + (sin263+cos263)    [∵sin2θ+cos2θ=1]=1+1=2.

Explanation:

Considering that 4 tanθ=3,

Finding the value of (4sinθ - cosθ)/(4sinθ + cosθ) is necessary.

So, tan = 3/4 now.

As we now know, tan A = opposite/adjacent

Contrary = 3

Adjoining = 4

(Hypotenuse) Using the Pythagorean Theorem2 = (contrary)2 + (nearby)² (hypotenuse)² = (3)² + (4)² (hypotenuse)(Hypotenuse) 2 = 9 + 16² = 25

Hypotenuse equals 5 after taking the square root.

Knowing that sin A equals the reverse and hypotenuse

So, sinθ = 3/5

We are aware that adjacent/hypotenuse = cos A

So, cosθ = 4/5

This means that (4sinθ - cosθ) = 4(3/5) - (4/5) = 12/5 - 4/5 = (12 - 4)/5 = 8/5.

Now, (4sinθ + cosθ) = 4(3/5) + (4/5) = 12 + 4/5 = 16

Thus, (8/5) / (16/5) = 8(5) / 16(5) = 8/16 = 1/2 (4sinθ - cosθ)/(4sin θ+ cosθ).

Explanation:

With sinθ(-cosθ) = 0

We must determine the value of sin4 + cos4, where sin = cos.

As we are aware, sinθ A/cosθ A = tanθ A.

Cos/Sin = Tanθ

Thus, tanθ ⁻¹= 1 = tan 1(1).

tan θ45° = 1

So, θ = 45°

So, sin⁴θ  = sin⁴θ 45° now.

According to the table above, sin 45° = 1/2.

So, sin⁴ 45° = (1/√2)⁴ = 1/4

Cos⁴θ  = cos⁴θ 45° now.

From the table above, cos 45° equals 1/2.

So, cos⁴ 45° = (1/√2)⁴ = 1/4

cos4θ + sin⁴θ = 1/4 + 1/4 = (1 + 1)/4 = 2/4 = 1/2

As a result, sin⁴ + cos⁴ = 1/2.

Explanation:

The value of sin (45° + θ ) must be determined. - cos (45° - θ)

cos (90° - A) = sin is the trigonometric ratio of the complementary angles. Cos (90° - (45° -  θ)) = A cos (45° - θ )

= sin(45°+ θ ) = cos(90° - 45° +  θ)

Cos(45°- θ) - Sin(45°+ θ) = Sin(45°+ θ) - Sin(45°+ θ) = 0.

Consequently, the supplied expression has a value of 0.

Explanation:

Since the pole's length is 6 meters,

A 2 to 3-m-long shadow is created on the ground by the pole.

The elevation of the sun must be determined.

Let the elevation angle be.

The figure shows,

tan = opposite/hypotenuse tan = 6/2√3 tan = 3/√3 tan = 3 tan = opposite/hypotenuse tan = adjacent tan = 2√3 m

Tan 60° = 3 using the trigonometric ratios of angles.

So, θ = 60°

Consequently, the elevation angle is 60°.

Explanation:

Assuming, tan 47°/cot 43°= 1,

The validity of the relationship mentioned above must be established.

Tan (90° - A) = cot is the trigonometric ratio of complementary angles. A tan 47° is equal to tan (90° - 43°) = cot 43°.

In other words, tan 47°/cot 43° = cot 43°/cot 43° = 1.

Because of this, tan 47°/cot 43° = 1.

Explanation:

 The formula is cos²  23° - sin²  67°.

Whether (cos² 23° - sin² 67°) is positive must be determined.

Algebraic identity yields (a² - b²) = (a - b)(a + b).

As a result, cos² 23° - sin² 67° equals (cos 23° - sin 67°)(cos 23° + sin 67°).

Using the trigonometric formula for the ratio of complementary angles, sin (90° - A) = cos A, sin (90° - 23°) = cos 23°, and so forth.

Cos² 23° - Sin² 67° = Cos² 23° - Cos23° + Cos23° = (0) Cos² 23° - Cos23° = 0

Thus, cos² 23° - sin² 67° equals 0.

Explanation:

Considering that the formula is sin 80° - cos 80°

Whether the expression is negative must be determined.

According to the trigonometric ratio of angles, sinθ grows asθ cosθ decreases asθ the angle increases θ from 0° to 90°.

In other words, (sin 80° - cos 80°) equals (growing value - declining value) = positive value.

(Sin 80° - Cos 80°) > 0 as a result.

Explanation:

 (1cos²).sec²θ =sin²θ×sec²θ

=sinθ×secθ

=sinθ/cosθ

=tanθ

So it is true.

Explanation:

Assume that cos A + cos² A = 1.

We must demonstrate that sin²A + sin⁴A = 1.

Sin² A = 1 - cos² A is the trigonometric identity.

Therefore, cos A = 1 - cos² A

sin² A = cos A

At this point, (sin² A + sin⁴ A) = (sin² A + (cos A)²) = sin² A + cos² A.

Cos² + Sin² = 1 using the trigonometric identities.

As a result, the expression (sin² A + sin⁴ A) has a value of 1.

Explanation:

The formula is as follows: (tanθ + 2) (2 tanθ + 1) = 5 tanθ + se²2.θ

We must assess whether or not the supplied expression is true.

LHS: (tan θ + 2) (2 tan θ + 1)

By virtue of the distributive and multiplicative properties, = 2tan² θ + tan  θ+ 4tan θ + 2 = 2tan² θ + 5tan θ + 2

Using the trigonometric ratio of complementary angles, 1 + tan² A = sec² A and tan² A = sec² A - 1 may be calculated.

In other words, 2(sec² θ - 1) + 5tan + 2 = 2sec² θ - 2 + 5tan + 2 = 2sec²  θ+ 5tan RHS = 5 tan + sec²  θ

RHS LHS

Explanation:

We must evaluate whether the length of a tower's shadow is growing in order to ascertain whether the sun's angular height is likewise growing.

If a pole with a height of 2√3 m casts a shadow of 2 m, the sun is at an elevation of.

According to the Pythagorean Theorem, tan = opposite or adjacent, tanθ = 2√3/2, tanθ = 3, and tan⁻¹ = (3).

We understand that tanθ 60° =2 3.

So, θ = 60°

When the shadow lengthens to 4 m, the angle of elevation becomes.

tan θ = 2√3/(2 + 4)

tan θ = 2√3/6

tan θ = √3/√3

Tan ⁻¹= 1/√3 equals Tan (1/√3)

We are aware that 30° tan = 1/√3

So, θ = 30°

The two examples show emphatically that as the length of the shadow rises, the angle of elevation decreases.

Explanation:

Assume that a man sees a cloud and its reflection in a lake from a platform three meters above the water level.

We must determine whether the cloud's elevation angle and its reflection's deflection angle are equal.

M should represent the stance of a guy standing on a platform.

The cloud is C.

MO is the distance between the platform and the lake's surface.

MO = 3 m

Let 1 represent the elevation angle.

Let 2 be the depression's angle.

The height of the cloud's reflection is equal to (h + 3) m.

Triangular MPC

tan θ₁=CM/PM

Tan θ₁ = PM/h

PM = tan/h θ₁ ------------------- (1)

MPC's triangle

Tanθ = C'M / PM

Tan θ equals (h + 3)/PM

(h + 3)/tan θ = PM ----------------- (2)

Based on (1) and (2),

(h + 3)/tanθ₂ = h/tan1

(h + 3) tanθ₁ = h(tanθ₂)

Therefore, tanθ₁ = h/(h + 3) tanθ₂

It is evident that 1 and 2

As a result, the angle of depression and elevation are not equal.

Explanation:

Assuming that an is a positive number, a 1.

We must establish whether 2sinθ can be a + (1/a).

A + (1/a) / 2 is the arithmetic mean of a and 1/a.

The geometric mean of a and 1/a is equal to 1 (a (1/a)).

As a result, [a + (1/a)]/2 > 1 [a + (1/a)] > 2 where [a + (1/a)] > geometric mean

Currently, 2sinθ > 2 sin > 1

Consequently, 2sinθ cannot equal a + (1/a).

Explanation:

Given that ab > 0, a and b are two separate numbers.

We must determine whether cosθ = (a2 + b2)/2ab.

Mathematical mean > Geometric Mean

As GM=ab and AM = (a + b)/2

Thus, we obtain (a²+ b²)/2 > a²b² (a² + b²)/2 > ab.

By cross multiplying (a² + b²)/2ab, 1 is obtained.

Given is that cos > 1.

We are aware that cosθ has a value between 0 and 1.

Hypotenuse/Base > 1

Hypotenuse > Base

Since cosθ cannot have a value larger than 1, the assertion is untrue.

As a result, the assertion is untrue.

Explanation:

Because a tower's top is 30° above ground,

We must determine that if the tower's height is twice, its top's elevation angle will also be doubled, or θ = 60°.

Let AC represent the tower's height.

BC = x units and AC = h units.

30-degree elevation angle

Pythagorean theorem states that tan 30° = AC/BC.

Tan 30° = 1/3 1/√3 = h/x can be calculated using the trigonometric ratio of angles. ------------ (1)

Explanation:

Provided, the tower's height and the gap between its foot and the point of view are both increased by 10%.

We must ascertain whether the top's elevation angle hasn't changed.

Let AB represent the tower's height.

Let BC represent the distance between the object under observation and its foot.

Let AB equal x units and BC equal y units.

Be the elevation's angle

Tan = AB/BC and Tan = x/y in the triangle ABC. ------------------ (1)

Now, 10% more is added to x and y.

x = x + (1/10)x = (1 + 10)/10, and y = y + (1/10)y = 11y/10

Tan = (11/10)x/(11/10)y Tan = (11/10)x/(11/10)y ------------------- (2)

From (1) and (2), stay the same.

As a result, the elevation angle doesn't alter.

Explanation:

We've got

LHS = 1+sin/cos+ 1+sin/cos

LHS is equal to sin2+(1+cos)2/sin(1+cos)/.

LHS = sin(1+cos), sin2(1+cos), +1+2cos, + cos2.

LHS = (sin2 + cos2 )+1 + 2 cos / sin(1 + cos) [sin2 + cos2 = 1]

LHS = 2+2cosec/sin(1+cos) = 2(1+cos)/sin(1+cos) = 2(1+cos)/sin(1+cos) = 2cosec=RHS.

Explanation:

tanA/1+secA-tanA/1sec A=2⋅cosecA

L.H.S]

tanA/1+sec A-tanA/1-sec A

=tanA(1−secA)−tanA(1+secA)/1-sec 2A

=tanA(1−secA−1−secA)/tan 2 A

=−2secA/-tanA

=2⋅1/cosA/sinA/cosA

=2/sinA

=2⋅cosecA

=R.H.S

∴ L.H.S=R.H.S.

Explanation:

LHS is equal to (sin α+ cosα) (tan α+ cotα).

We are aware that tan = sinα/cosα and cotα = cosα/sinα = sinα + cosα. (cos/sin plus sin/cos)

By using the formula LCM = (sinα + cosα) [(sin2 α- cos2α )/cosα - cosα]

Due to the fact that sin2 + cos2 = 1 = (sin + cos)/cos sin 

You may write it as = 1/cos + 1/sin.

In this case, 1/cosα = sec α and 1/sin α= cosec α = secα + cosec α= RHS.

It has therefore been proven.

Explanation:

LHS is equal to (√3 + 1) (cot 30°)

We are aware that 30° cot = √3 = (3 + 1) (3 - √3)

Writen as (√3 - 3), = √3 (√3 - 1) = (√3 + 1) √3 (√3 - 1)

Making use of the algebraic formula (3 + 1)(3 - 1) = ((3)2 - 1)]

= ((√3)2 - 1) √3 = (3 - 1) √3 = 2√3

RHS = tan3 60° - 2 sin 60° should be solved.

Due to the fact that tan 600 = 3 and sin 600 = 3/2, we have (3)3 - 2.(√3/2) = 3√3 - √3 = 2√3

Thus, it is established that LHS equals RHS.

Explanation:

LHS is equal to 1 + cot2 α / (1 + cosecα)

We are aware that cot 2 α= cosec2 α - 1.

LHS then equals 1 + (cosec2 α - 1)/ (1 + cosecα).

With the help of the algebraic formula a2 - b2 = (a + b) (a - b),

LHS is equal to 1 plus [(cosec α+ 1) (cosec α- 1)/ (1 + cosec α )]

RHS = 1 + cosec α- 1 = cosec α

It has therefore been proven.

Explanation:

We understand that tan (90° - θ) = cot θ

LHS = Tan θ(90° - Tan) = Tan (Tan) + Cotθ

We are aware that tan = sin/cos and cot = cos/sin.

LHS is equal to sinθ/cosθ + cosθ/sinθ.

Using the formula LCM = (sin2θ + cos2θ ) / sin θcosθ 

Since sin2θ + cos2 θ= 1 = 1/sin2θ + cos2θ,

It can be expressed mathematically as = 1/ cosθ x 1/ sinθ = sec cosec.

In this instance, cosec = sec (90° - ) = sec and sec (90° - ) = RHS.

It has therefore been proven.

Explanation:

We understand that tan (90° - θ) = cotθ

LHS = Tan (90° - Tan) = Tanθ (Tanθ) + Cotθ

We are aware that tanθ = sinθ/cosθ and cotθ = cosθ/sinθ.

LHS is equal to sinθ/cosθ + cosθ/sinθ.

Using the formula LCM = (sin2 + cos2 ) / sin cos 

Since sin2θ + cos2 θ= 1 = 1/sin2 θ+ cos2θ,

It can be expressed mathematically as = 1/ cos θx 1/ sin θ= secθ cosecθ.

In this instance, cosec θ= sec θ(90° - ) = secθ and sec (90° -θ ) = RHS.

It has therefore been proven.

Explanation:

It is assumed that 3 tanθ = 1.

You can write it as tan θ= 1/3

By using the tan inverse, we get 30° = tan-1 (1/3)

In order to calculate sin2 - cos2,

Let's replace sin2 30° - cos2 30° = (1/2) with its value.2 - (√3/2)2

By simplifying even further, 1/4 to 3/4

We thus have = -2/4 = -1/2.

Therefore, sin2 - cos2 has a value of -1/2.

As a result, the Sun is elevated at a 30° angle.

Explanation:

Think about the ladder's length, which is 15 meters (Hypotenuse).

based on the figure

Angle between ladder and ground: <CAB = 90° - 60° = 30° Angle between ladder and wall: <BCA = 60°

We are aware that BC is equal to the height of the wall sin 30°, which is BC/15 1/2.

The result is BC = 15/2 BC = 7.5 m.

Consequently, the wall is 7.5 meters tall.

Explanation:

Given are (1 + tan2θ ) (1 - sinθ ) and (1 + sinθ )

Since 1 + tan2 θ= sec2θ,

Thus, it can be expressed as = sec2θθ  (1 - sinθ) (1 + sinθ).

We obtain = sec2θ  (1 - sin2θ ) by applying the algebraic identity a2 - b2 = (a + b) (a - b).

We are aware that 1 - sin2θ = cos2θ

By changing it to it = sec2 θcos2θ 

In this case, cos2 = 1/sec2θ

Thus, we obtain = sec2 θ1/sec2 θ = 1

As a result, by simplifying, we obtain 1.

Explanation:

The formula 2 sin2 θ- cos2 θ= 2 is known.

Because sin2 θ+ cos2 θequals 1,

We obtain sin2θ - (1 - sin2θ ) = 2 sin2 θ= co2θ = 1 - sin2θ = 2

Simplifying even further, 2 sin2 θ- 1 + sin2 θ- = 2

3 sin2 θ = 3

Multiply both sides by 3 sin2 to get 1.

Sin = 1 after applying square roots to both sides.

Thus, we have = sinθ-1 (1) = 90°.

As a result, has a value of 90°.

Explanation:

LHS = [cos2 (45° +θ ) + cos2 (45° - θ)] is an example.[sunny (60° + θ) sunny (30° - θ)]

As we are aware, cos A = sin (90° - A).

(90° - A) = tan A

Tan A = Cot A

We get = [cos2 (45° + θ) + sin2 (90° - 45° +θ )] by swapping it.[Cotton (90° - 30° + θ) in tan (60° + θ)]

Consequently, we have [cos2 (45° +θ ) + sin2 (45° +θ )].[Cotton (60° + θ) in a tan shade]

In this case, cos2A + sin2A = 1, and tanA cotA = 1.

By changing it to = 1/1 = 1.

Explanation:

Given that the observer is 1.5 inches tall

AE=22−1.5=20.5

In △AED

tanθ=EDAE​=20.520.5​=1⇒θ=45^o

Explanation:

LHS equals tan4θ plus tan2θ.

removing tan2 θas common results in tan2 θ(tan2 θ+ 1).

We are aware that tan2θ = sec2 θ- 1 and therefore 1 + tan2 θ= sec2θ.

It is possible to write it as = (sec2 θ- 1) sec2θ.

Thus, we obtain sec4 θ- sec2 θ= RHS.

It has therefore been proven.

Explanation:

Cosecθ + Cotθ = p given -------- (1)

It must be demonstrated that cos = p2 - 1 / 22 + 1

Trigonometric identities show that cot2 A + 1 = cose22 A.

Therefore, cosec 2 A - cot2 A = 1.

The equation (a2 - b2) can be written as (a + b)(a - b) (cosecθ  - cotθ )(cosec θ+ cotθ ) = 1 (cosecθ  - cotθ )(P) = 1 cosec θ- cotθ  = 1/P. ---------------------- (2)

Cosecθ  - cot θ+ cosecθ  + cotθ equals P + 1/P after adding (1) and (2).

P plus 1/P + 2cosecθ

(P + 1/P)/2 cosecθ

When (1) and (2) are subtracted, cosecθ + cotθ - cosec θ+ cotθ = P - 1/P.

2cot θ = (P - 1/P)

cot θ = (P - 1/P)/2 -------------------------- (3)

Coefficient / cosec is equal to (P - 1/P) / (P + 1/P) when (3) is divided by (2).

As we are aware, cotθ A = cosθ A/sinθ A.

Additionally, cosecθ A = 1/sinθ A

Now, cot θ/ cosec θ= (cosθ / sinθ) / (1/sinθ) = (cosθ / sinθ) (sinθ / cosθ)

Thus, cosθ = (P.(P2 - 1)/(P2 + 1) = (P2 - 1)/(P2 + 1) = (P2 - 1)/(P2 + 1)

Consequently, cos = (P2 - 1)/(P2 + 1).

Explanation:

The trigonometric identities sec2θ=tan22+1 and cosec2θ=cot2θ+1 are used.

These are obtained by adding them together: sec2θ+cosec2θ=tan2θ+cot2θ+2 sec2θ+cosec2θ=tan2θ+cot2θ+2 tanθ+cotθ=(tanθ+cotθ)2 sec2θ+cosec2θ=tanθ+cotθ

Thusly Proven.

Explanation:

Because a tower's top is 30 degrees higher than a particular point,

The angle of elevation of the top increases by 15° as the spectator approaches the tower by 20 m. We must determine the height of the tower.

Let D be the observation point, AB be the tower, and BCA equal 30 degrees.

Assuming CD = 20 m

The angle of elevation rises by 15° as the viewer advances from point C to point D.

BDA is therefore 30° + 15° + 45°.

Triangle BAD has

Python's Theorem states that tan 45° = AB/AD 1 = h/AD

AD = h ----------------- (1)

triangular BAC

Pythagorean theorem states that tan 30° = AB/AC.

1/√3 = h/AC

As we are aware, AC = CD + AD AC = 20 + h

Now, 1/√3 = h/(20 + h)

20 + h = 3(h) on cross multiplication

When calculated simply, 3h - h = 20 h(3 - 1) = 20 h(1.732 - 1) = 20 h(0.732) = 20 h = 20/0.732 h = 27.3 m

Consequently, the tower has a height of 27.3 m.

Explanation:

Provided, Sin2 θ+ 1 = 3sin2 θ+ cos2θ

It must be demonstrated that tanθ equals 1 or 1/2.

When both sides are divided by sin2θ, the result is 1/sin2 θ+ sin2θ/sin2θ = 3sinθ cosθ/sin2θ.

1/sin2θ + 1 equals 3cosθ/sinθ.

As we are aware, cosecθ A = 1/sinθ A

Additionally, sin θA/cosθ A = cot θA cosecθ2 + 1 = 3cotθ ------------ (1)

Trigonometric identity reveals that cot2 θA + 1 = cosec2 θA.

3cot θ= cot2 θ+ 1 + 1

3cot2 θ- 2cot2θ = 0

If cotθ = x,

So, x2 - 3x + 2 = 0

Factoring reveals that x2 - x - 2x + 2 = 0 x(x - 1) - 2(x - 1) = 0 (x - 1)(x - 2) = 0

Now, x - 1 = 0 x = 1

Also, x = 2 and x - 2 = 0.

Cot now equals 1, 2

We are aware that tanθ = 1/cotθ.

Tan then equals 1/1 or 1/2.

Tan thus equals 1 or 1/2.

Explanation:

Provided that sinθ + cosθ  2 = 1,

It must be demonstrated that 2sinθ - cosθ = 2.

Putting a square on both sides, (sinθ + 2cosθ)² = 1

(A + B)² = A²+ 2AB + B² is an algebraic identity.

Therefore, (sinθ + 2cosθ)2 is equal to sin² θ+ 4sin² θ+ 4cos²θ

Therefore, sin² θ+ 4sin θ+ cos² θ= 1.

Trigonometric identity reveals that cos²θ A = 1 - sin² θA and sin² θA = 1 - cos² θA.

As a result, (1 - cos² θ) + 4 cosθ + 4 (1 - sin²θ ) = 1

1 + cos²θ + 4 + 4sin²θ + 4 sinθ cosθ = 1 5 + cos² θ+ 4 + 4sinθ cosθ = 1

Rearranging results in 5 - 1 = cos²θ + 4sin²θ - 4 sinθ cosθ, where cos² θ+ 4sin² θ- 4sinθ cos equals 4. --------------- (1) The algebraic identity (a - b)² = a² - 2ab + b² can be found. --------------------- (2) When comparing (1) and (2), we see that a²= 4sin²θ a = 2sin ²ab = 4sin cos ab = 2sinθ cosθ.

b² = cos²θ.

Explanation:

Because the two points s and t are spaced apart from the base of a tower, the angle of elevation at the tower's top is complimentary.

We must provide evidence that the tower is √st in height.

Make AC represent the tower's height.

units of AC = h

The places of observation are P and B.

If PC = t units and BC = s units, then

Given that ABC = APC = 90° -, the angle of elevation is complimentary.

Tan θ = AC/BC in triangle ABC

tan θ = h/s ------------------ (1)

Tan θ (90°- ) = AC/PC in the triangle APC

Tan θ(90° - A) = cot θ A is the result of the trigonometric ratio of complementary angles.

Consequently, tan (90°-) = cot θ = cot  θ= AC/PC

cot θ = h/t -------------------- (2)

Sun θ = cot  θ= (h/s)(h/t) is obtained by multiplying (1) and (2).

We are aware that tan cot 1

Thus, 1 = h2/st and h2 = st

Using square root, h equals st.

The tower's height is thus st.

Explanation:

Because the length of a tower's shadow on a level plane is determined to be 50 m longer at 30° elevation than at 60°, we must determine the tower's height.

Let SQ represent the tower's height.

SQ = h m

The initial angle of elevation: ∠SRQ = 60°

The shadow's initial length is equal to x meters.

When the angle of elevation ∠SPQ = 30°, the shadow of the tower is extended by 50 meters.

Tan 60° in the triangle ∠SRQ equals SQ/RQ 3 = h/x x = h/3 m. --------------- (1)

Tan 30° = SQ/PQ in the triangle SPQ

PQ = PR + RQ, which means that PQ = 50 + x 1/3 = h/(50+x).

When multiplying by themselves, 50 + x = 3h x = 3h - 50. ------------ (2)

According to (1) and (2), h/3 = 3h - 50

According to grouping, h/3 - 3h = -50 3h - h/3 = 50

If you simplify, (3h-h)/3 equals 50.

2h/√3 = 50

2h = 50√3 h = 50√3/2 h = 25√3 m

As a result, the tower has a height of 25 3 m.

Explanation:

Provided, a vertical tower rests on a horizontal plane and is topped by an h-foot-tall vertical flagstaff.

The flag staff's bottom and top are elevated at an angle of and, respectively.

We must demonstrate that the tower is [h tan]/(tan - tan)].

Let PO represent the vertical tower and FP represent the flagstaff.

Given the vertical staff's height, FP = h units

Let the tower's height be in H units.

The angle of depression, ∠FRO =  ꞵ

∠PRO = ⍺Angle of elevation

Let RO be the distance in units between the observation point and the tower's base.

Tan = PO/RO in triangle PRO

H/x x = H/tan = tan ⍺------------------- (1)

Tan⍺  = FO/RO for the triangle FRO

The figure shows that FO = FP + PO.

FO = h + H

Thus, x = (h + H)/tan = tan = (h + H)/x. ----------------- (2)

According to (1) and (2), H/tan⍺ = (h + H)/tan

When cross-multiplied, H(tan ) equals (h+H) tan.

H(tanꞵ ) equals H(tanꞵ ) plus h(tanꞵ ).

In order to find H, we get H(tan ꞵ) - H(tanꞵ ) = h(tan ꞵ).

Tanꞵ - Tanꞵ = H(tan ⍺- Tan⍺)

H is equal to h(tan ꞵ ) / (tanꞵ - tanꞵ ).

As a result, the tower's height is [h tan⍺]/(tan ꞵ - tan⍺)].

Explanation:

It is known that tan θ+ secθ = l ------------------- (1)

We must demonstrate that secθ = l²+ 1/2l.

Thinking about LHS: tanθ + secθ

Using the operations (sec θ- tanθ), (tanθ + secθ)(secθ - tanθ)/(secθ - tanθ), multiply and divide the numbers

Algebraic identity yields (a² - b²) = (a - b)(a + b).

As a result, (tanθ + secθ)(sec θ- tanθ) = (sec²θ - tan²θ)

LHS is now equal to (sec²θ - tan²θ) / (secθ - tanθ).

Trigonometric identity demonstrates that 1 + tan²θ A = sec²θ A.

When arranged, sec² θA - tan θA equals 1.

This means that (sec² θ- tan² θ) / (sec θ- tanθ) = 1/(sec θ- tanθ)

The result is now 1/(secθ - tanθ) = l (sec θ- tanθ) = 1/l. ---------------------- (2)

When (1) and (2) are combined, (tanθ + secθ) + (secθ - tanθ) = l + 1/l.

2secθ = l + 1/l

In a simplified formula, 2secθ = (l² + 1)/l secθ

Consequently, secθ = l² + 1/2l.

Explanation:

Assuming that sinθ + cosθ = p and secθ + cosecθ = q.

It must be demonstrated that q (p2 - 1) = 2p.

Regarding LHS: q (p² - 1)

When p and q are substituted, the result is (sec θ+ cosecθ)[(sin θ+ cosθ)2 - 1].

Algebraic identity reveals that (a + b)2 = a2 + 2ab + b2 (sinθ + cosθ)² is equal to sin² θ+ 2sinθ cosθ + cos²θ.

Using the trigonometric identity, cos² θA plus sin²θ A equals one.

Consequently, sin² θ+ 2sinθ cosθ + cos² θ= 1 + 2sinθ cosθ.

[(sin θ+ cosθ)² - 1] now equals 1 + 2 sinθ cosθ - 1 = 2 sinθ cosθ.

At this point, (sec θ+ cosecθ)[(sinθ + cosθ)² - 1] = (secθ + cosecθ)(2sin θcosθ) is the result.

We are aware that sec θ= 1/cosθ A and cosecθ = 1/sinθ A are equivalent to (1/cosθ + 1/sinθ)(2sinθ cosθ).

With regard to simplification, = [(sinθ + cosθ)/sinθ cosθ](2sinθ + cosθ) = 2(p + sinθ + cosθ) = RHS

RHS LHS

Consequently, q (p² - 1) = 2p.

Explanation:

\asinΦ+bcosΦ=c

balancing the two sides

A2sin2 + B2cos2 + 2absin2 cos = C2

a²(1-cos²Φ)+b²(1-sin²Φ)+2absinΦcosΦ=c²

a²-a²cos²Φ+b²-b²sin²Φ+2abcosΦsinΦ=c²

a²+b²-c²=(acosΦ-bsinΦ)².

Explanation:

L.H.S= 

1+secθ−tanθ/ 1+secθ+tanθ

= secθ−tanθ+1/ secθ+tanθ+1

= secθ−tanθ+(secθ+tan2 θ)/sec θ+tan θ+1   [sec2θ−tanθ=1]

= secθ−tanθ+(secθ−tanθ)(secθ+tanθ)/sec θ+tan θ+1

= (secθ−tanθ)(1+secθ+tanθ)/1+secθ+tanθ

=1/cosθ-sinθ/cosθ

=1-sinθ/cosθ=R.H.S

So it was confirmed.

Explanation:

Because a tower's top is 30 meters high and its base is 60 degrees from another structure in the same plane,

The top of the second tower is 30° higher than the bottom of the first tower.

We must determine the separation between the two towers as well as the second tower's height.

With a height of 30 m and an angle of elevation of <QAB = 60°, let BQ be the first tower.

PA should be the second tower, standing at h m.

The separation between the two towers is given by the angle of elevation, <PBA = 30° AB.

Triangular AQB

Pythagorean theorem states that tan 60° = QB/AB.

Tan 60° = 3 using the trigonometric ratio of angles

Consequently, 3 = 30/AB AB = 30/3 m AB = 3(10)/AB AB = 103 m

As a result, there are 103 meters between each tower.

Pythagorean theorem: Tan 30° = AP/AB for triangle APB

Tan 30° = 1/3 using the trigonometric ratio of angles

The result is that 1/3 = AP/(30/√3) AP = (30/√3)/3 AP = 30/√3 AP = 10 m.

Consequently, the second tower has a 10 m height.

Explanation:

The angles of depression of two objects that are parallel to the tower's foot are provided as and β (β > ), and we must determine their distances from the tower's top.

Make AB the tower.

The tower is h meters tall.

The angle of depression, <ABD = 𝛼

< ACD = β  Angle of depression.

Think about BC = x CD = y.

We can infer that a triangle's alternate angles are equal by looking at its attributes.

Therefore, ABD = <BAX = 𝛼

<  ACD = <CAY = β.

Triangle ABD results in tan = AD/BD, tan = h/BC + CD, tan = h/x+y, and tan = h/tan. - x -------------------(1)

Triangle ACD results in tan = AD/CD tan = h/y y = h/tan. -----------------------(2)

As a result of comparing (1) and (2), we arrive at the following equation: h/tan - x = h/tan - h/tan x = h(1/tan - 1/tan) x = h(cot - cot)

The necessary distance is therefore h(cot - cot).

Explanation:

Assume that a ladder is leaning horizontally against a vertical wall.

The ladder establishes α an angle to the horizontal by pulling its foot away from the wall by a distance p, allowing its upper end to slide down the wall by a distance q.

It must be demonstrated that p/q = cosβ - cosα and sinα - sinβ.

Let AB represent the ladder leaning up against the wall.

<BAO =𝛼 Angle of Inclination

The upper end of the ladder slides a distance q down the wall when the foot is moved away via a distance p.

SQ is the ladder following the slide.

SA = p, the pulled-away distance

BQ = q, distance slide down the wall

QSO = Angle of Inclination

Let's think about,

QO = x and AO = y

Now consider the wall's height: BO = q + x

In light of the triangle BAO, we obtain

cos OA/AB =

y/AB = cos

If y = ABcos,

Further, OA = AB cos𝛼 --------------- (1)

sin(OB, AB)

OB = BA sin𝛼 ----------------------- (2)

Given triangle QSO, we obtain sin(QO/SQ)

QO = Sinβ SQ

c0 = SO/SQ

SQcosβ = SO

We have AB because the ladder's length stays the same.

Since = SQ,

AB = QO sinβ ---------------------- (3)

AB = SO cos β---------------------- (4)

As far as we are aware, SA = OS - AO p = AB cosβ - AB cos.

Cosβ - Cos𝛼 = p = AB ------------------- (5)

BO - QO = BQ

q = BA AB sin𝛼 - sinβ

(sin𝛼 - sinβ) = q ----------------------- (6)

We are required to demonstrate that p/q = cos - cos𝛼/ sin𝛼 - sinβ in order to answer the question.

p/q is equal to AB (cosβ - cos)/BA (sin - sinβ).

P/Q is equal to (cosβ - cos)/ (sin𝛼 - sinβ).

Consequently, p/q = (cosβ - cos𝛼)/ (sin𝛼 - sin)β.

Explanation:

Because a vertical tower's top is 60 degrees higher than a point on the ground,

We need to determine the tower's height since the angle of elevation is 45 degrees from a second point that is vertically above the first.

AP should act as the vertical tower.

Vertical tower height, AP = 10m

<TPO's angle of elevation is 60 degrees, while <TAB's angle of elevation is 45 degrees.

According to the diagram above, OB = AP = 10m.

Let's think about,

PO = x m, which is the distance from the observation point to the tower's base

TO = h m, or the tower's overall height

The figure shows that TB = (h - 10)m.

Taking into account the <TPO, we obtain tan60 = TO/PO.

Given that tan60° = 3 3 = h/x x = h/3, ---------(1)

Taking into account the TAB, we obtain tan45° = TH/AB.

45° Tan = TO - TH/AB

tan45° = 1, so 1 = h - 10/AB.

AB = h -10 x = h -10-----------------(2)

The result of comparing (1) and (2) is h/3 = h -10 h - h/3 = 10 h(1 - 13) = 10 h(3 - 1/3) = 10 h = 103/3 - 1

Consider conjugate, h

= 10√3/√3 - 1 × √3 - 1/√3 - 1 

h = 30 + 10√3/3 - 1

 h = 30 + 10√3/2 

h = 15 + 5√3

 h = 5(3 + √3) 

h = 5√3 (√3 + 1)

The tower's height is therefore 5√3 (3 + 1)m.

Explanation:

Assume that a window is h m above the ground.

From the window, it can be seen that another house on the other side of the lane has an angle of elevation and depression of the top and bottom that is α and β

We must provide evidence that the other house is h (1 + tan α cot β) meters tall.

W represents the house's window in the figure.

BW = h m, the window's height above the ground

Make OQ a different dwelling.

Height of the neighboring building, QO = Hm

Angle of depression, WOB = Angle of elevation, <QWM𝛼

Think about BO = xm.

Tan = QM/WM in the triangle QWM𝛼

tan = H - h/x x = H - h/tan since QM = QO - MO.--------------------(1)

Taking into account triangular WOB, tan = WB/BO

Tan = h/x x, and Tan = h/tan𝛼--------------------------(2) When we compare (1) and (2), we obtain,

(H - h)/tan = (H - h)/tan Tan equals h Tan

(H - h)/tan α = h/tan β

(H - h) tan β = h tan𝛼

H tanβ - h tanβ = h tan𝛼

H tanβ = h tan𝛼 + h tanβ

H tanβ = h (tan𝛼 + tanβ )

H = (h (tan𝛼 + tanβ))/tanβ

H = h (tan𝛼 /tanβ + 1)

H = h (1 + tan𝛼 /tanβ)

H = h (1 + tan𝛼 cotβ)

Consequently, the second house's height.

Explanation:

Because a house's lower window is 2 meters above the ground

The higher window is 4 m above the lower window vertically.

From these windows, a balloon is observed to rise at angles of 60° and 30°.

We must determine the balloon's height above the ground.

W1 is the upper window in the figure.

W2 is the bottom window.

Lower window's height above the ground, OR = 2m

RQ = 4m is the distance between the top of the lower window and the upper window.

The angle at which the balloon is elevated from the lower window, <BW₂R = 60°

Height of the balloon from the ground, BO = h m Angle of elevation of the balloon from an upper window, <BW₁R = 30°

PO = xm is the distance from the point of observation to the foot.

Since tan60° = 3 3 = BO - RO/PO 3 = h - 2/x x = h - 2/3 in the triangle BW₂R ----------------(1)

In the triangle BW₁R, tan30° = BQ/W1R since W1R = PO = x x = 3(h - 6) and tan30° = 1/3 1/3 = h - 6/x------------------------(2) When we compare (1) and (2), we discover that h - 2/3 = 3(h - 6) h.- 2 = 3(h - 6)

h - 2 = 3h - 18

3h - h = 18 - 2

2h = 16

h = 8m.

Consequently, the balloon is 8 meters above the earth.