Trigonometry

Let us consider a parallelogram ABDC.
Let us draw a perpendicular on base AB from C say, CE

Let length of AE = x cm.

In Δ AEC:

∠CAE = 45° and ∠CEA = 90°
⇒ ∠ACE = 45° (∵ ∠C + ∠E + ∠A = 180° )

We know that sides opposite to equal angles are also equal. So the perpendicular side opposite to the angle 45° is also equal to x cm
∴ AE = CE = x cm.

Using the Pythagoras theorem in Δ AEC ,
we get,

AC² = AE² + EC²
⇒ (2)² = (x)² + (x)²
⇒ (2)² = 2(x)²
⇒ (2) = (x)²
⇒ x = √2 cm

OR

We know that sides of any triangle of angles 45°, 45° and 90°
are in the ratio 1:1:√2.

⇒ x:x:2 = 1:1: √2
⇒ x = √2 cm

⇒ CE = √2 cm and also line segment CE is a height of parallelogram ABDC.
Also base AB = 4 cm.

Thus area of parallelogram = base × height

∴ area (ABDC) = 4 × √2 cm² = 4√2 cm²

Area = 4√2 cm²

Consider a parallelogram ABDC.
Let us draw a perpendicular on base AB from C say, CE.

Let length of AE = x cm and CE = y cm.

In Δ AEC:

∠CAE = 60° and ∠CEA = 90°
⇒ ∠ACE = 30° (∵ ∠C + E + ∠A = 180° )

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ x:y:2 = 1:√3:2
⇒ x = 1 cm and y = √3 cm

OR
We can consider Δ AEC as a half of equilateral Δ ACL.

Since, Δ ACL is an equilateral triangle.
We get,

AL = AC
⇒ (x + x) = 2 cm
⇒ 2x = 2 cm
⇒ x = 1 cm

Using the Pythagoras theorem in Δ AEC ,
we get,

AC² = AE² + EC²
⇒ (2)² = (1)² + (y)²
⇒ 4 - 1 = (y)²
⇒ (3) = (y)²
⇒ y = √3 cm

⇒ CE = √3 cm and also line segment CE is a height of parallelogram ABDC.
Also base AB = 4 cm.

Thus area of parallelogram = base × height

∴ area (ABDC) = 4 × √3 cm² = 4√3 cm²

Area = 4√3 cm²

Let ABCD be the rectangle and BD be the diagonal and EFG be the derived equilateral triangle. Let EH be perpendicular to FG.

Given,
Δ EFG is an equilateral triangle
⇒ FG = FE = EG = 50 cm.
⇒ FH = HG
(∵ In equilateral triangle median is same as altitude)

⇒ FG = FH + GH = 50 cm
⇒ 2(FH) = 2(GH) = 50 cm
⇒ FH = GH = 25 cm

Also, ∠FEH = 2(∠FEH) = 2(∠GEH) = 60°
⇒ ∠FEH = ∠GEH = 30°
(∵ In equilateral triangle altitude bisects the angle)

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ 25 : EH : 50 = 1:√3:2
⇒ 25 : EH = 1:√3
⇒ EH = 25√3 cm

∴ IN Δ EFG and rect. ABCD
We get,

Δ DBC ≈ Δ EFG

(∵ Δ EFG is same as Δ BDC because both are part of same rectangle divided be diagonal BD.)

∴ AB = CD = FH = HG = 25 cm and BC = EH = 25√3 cm

Length and Breadth of the rectangle is 25√3 cm and 25 cm.

Let each polygon be assigned with name.

In regular hexagon AB₁F₁DF₂B₂A

AB₁ = B₁F₁ = F₁D = DF₂ = F₂B₂ = B₂A = 30 cm

∠B₂AB₁ = ∠AB₁F₁ = ∠B₁F₁D = ∠F₁DF₂ = ∠DF₂B₂ = ∠F₂B₂A = 120°

Δ AB₁G and Δ AB₂G are the two equal halves of one rectangle.
and Δ F₁DH and Δ F₂DH are two equal halves of another rectangle.

⇒ ∠B₁AG = ∠B₂AG

Now, ∠B₁AB₂ = ∠B₁AG + ∠B₂AG = 120°
⇒ ∠B₁AG = ∠B₂AG = 60°

In Δ AGB_1,

∠AGB₁ + ∠AB₁G + ∠GAB₁ = 180°
⇒ 90° + 60° + ∠GAB₁ = 180°
⇒ ∠GAB₁ = 30°

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ AG : B₁G: AB₁ = 1: √3: 2
⇒ AG: B₁G: 30 = 1: √3: 2
⇒ AG = 15 cm and B₁G = 15√3 cm

Similarly, In Δ AB₁B₂
⇒ B₁B₂ = B₁G + B₂G = 2(B₁G) = 2(15√3) cm = 30√3 cm

∴ For small rectangles :
Length = B₂G = B₁G = 15√3 cm
Breadth = AG = 15 cm

∴ For rectangle B₁B₂ H₂H₁ :
Length = B₁B₂ = 30√3 cm
Breadth = B₁F₁ = 30 cm

Length and Breadth of rectangles in cm are
(15√3 ,15),(15√3, 15) and ( 30√3, 30).

Construction, Let the centre of the circle be A and radius be r.
Let the triangle be CDE. Also, join AD and AE.

Now, AD = AE = r (radius)
We know that,
The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.
∴ ∠OAE = 2 (∠OCE)
⇒ ∠OAE = 2 (60°) = 120°

In Δ DAE :
Draw a perpendicular bisector on DE from A say AS.

AS is a perpendicular bisector of DE.
∴ ∠DAS = ∠EAS , DS = DE and ∠ASD = ∠ASE = 90°

⇒ ∠DAE = ∠DAS + ∠EAS = 120°
⇒ ∠DAE = 2(∠DAS) = 2(∠EAS) = 120°
⇒ ∠DAS = ∠EAS = 60°

In Δ ADE,
ΔDAE is an isosceles triangle (∵ AD = AE)
∴ ∠ADE = ∠AED = δ

Now, ∠ADE + ∠DEA + ∠DAE = 180°
⇒ ∠DAE + δ + δ = 180°
⇒ 120° + δ + δ = 180°
⇒ δ = ∠ADS = 30°
We get,
∠ADS = 30° , ∠DAS = 60° and ∠DAS = 90°

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1:√3:2.

⇒ AS :DS: AD = 1: √3: 2
⇒ AS :1.5 :AD = 1: √3: 2
⇒ AD = √3 cm = r

Radius of circle is √3 cm.

Let ABC be the required triangle.
Draw perpendicular from AS on BC.

Let AS = x cm.

In Δ BAS,

∠ABS + ∠BAS + ∠ASB = 180°
⇒ 45° + ∠BAS + 90° = 180°
⇒ ∠BAS = 45°
⇒ Δ BAS is an isosceles triangle.

∴ AS = BS = x cm

BC = BS + CS = 4 cm
⇒ x + CS = 4 cm
⇒ CS = (4–x) cm

In Δ CAS,

∠ACS + ∠CAS + ∠ASC = 180°
⇒ 60° + ∠CAS + 90° = 180°
⇒ ∠CAS = 30°

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1: √3: 2 .

⇒ CS :AS: AC = 1: √3: 2
⇒ (4-x) :x : AC = 1: √3 :2
⇒ (4 - x) : x = 1 : √3
⇒ √3(4 - x) = x
⇒ 4√3 - x√3 = x
⇒ x(√3 + 1) = 4√3

Multiplying and Dividing by (√3 - 1)

⇒ x = 2(√3 - 1) cm

In Δ BAC,

Height = x = 2(√3 - 1) cm and Base = 4 cm

Area (Δ) = 4(√3 - 1) cm² = 10.92 cm²

Let the equilateral triangle be Δ ABC, Circumcentre and circumradius be E and r respectively.

We know that in equilateral triangle median and internal angle bisector are same.
Here AE, BE and CE are part of medians.
⇒ EA acts as an internal angle bisector for ∠CAB.
⇒ ∠CAE = ∠EAB

In Δ ABC,
∠CAB = ∠CAE + ∠EAB = 60°
⇒ ∠CAB = 2(∠CAE ) = 2(∠EAB) = 60°
⇒ ∠CAE = ∠EAB = 30°

Extend CE to meet at AB at H.
CH is a median as well as altitude.
⇒ CH ⊥ AB and ∠CHA = ∠CHB = 90°

In Δ AEH,

∠EAH + ∠AHE + ∠HEA = 180°
⇒ 30° + 90° + ∠HEA = 180°
⇒ ∠HEA = 60°

We know that sides of any triangle of angles 30°, 60° and 90°
are in the ratio 1: √3: 2 .

⇒ EH :AH :AE = 1: √3: 2
⇒ EH :4 :r = 1: √3: 2
⇒ 4 :r = √3: 2

Let ABCD be the required rhombus.

⇒ AB = BC = CD = DA = 5 cm and ∠ ABC = 100°

In ⋄ ABCD,

Diagonals are the perpendicular bisector of each other

⇒ AO = OC and BO = OD and AC ⊥ BD

⇒ ∠ AOB = ∠ BOC = ∠ COD = ∠ DOA = 90°

Diagonals are the internal angle bisector.

⇒ ∠ ABO = ∠ CBO

⇒ ∠ ABC = ∠ ABO + ∠ CBO

⇒ 100° = 2(∠ ABO) = 2(∠ CBO)

⇒ ∠ ABO = ∠ CBO = 50°

In Δ AOB,

AO = sin 50° × AB

⇒ AO = sin50° × 5 cm

(From table: sin 50° = 0.776 )

⇒ AO = 0.776 × 5 = 3.88 cm

Again,

BO = cos 50° × AB

⇒ BO = cos 50° × 5 cm

(From table: cos 50° = 0.642 )

⇒ BO = 0.642 × 5 = 3.21 cm

Now,

AC = AO + OC

⇒ AC = 2(AO) = 2(OC)

⇒ AC = 2 × 3.88 cm

⇒ AC = 7.76 cm

BC = BO + OD

⇒ BC = 2(BO) = 2(OD)

⇒ BC = 2 × 3.21 cm

⇒ BC = 6.42 cm

Area of Rhombus = 24.9096 cm² = 25 cm²

Area of Rhombus = 25 cm².

Let us draw a perpendicular from D on AB say DH ⇒ DH ⊥ AB

In Δ ADH,

DH = sin 50° × AD

(From table: sin 50° = 0.776)

⇒ DH = 0.776 × 8 cm

⇒ DH = 6.208 cm

Area of parallelogram = Base × Height

Area = 12 × 6.208 cm²

Area = 74.496 cm²

Area of parallelogram is 74.496 cm².

Let us draw a perpendicular from D on AB say DH ⇒ DH ⊥ AB

In Δ ADH,

DH = sin 30° × AD

(From table: sin 30° = 0.5)

⇒ DH = 0.5 × 6 cm

⇒ DH = 3 cm

Area of parallelogram = Base × Height

Area = 14 × 3 cm²

Area = 42 cm²

Area of parallelogram is 42 cm².

We know that perpendicular length drawn between two lines is the smallest length.

Let AB = 8 cm and ∠ BAX = 40°

From B draw perpendicular on AX say BC.

In Δ ABC,

BC = AB × sin 40°

(From table, sin 40° = 0.64)

⇒ BC = 8 × 0.64 cm

⇒ BC = 5.12 cm

The minimum length of the side opposite to angle 40° is 5.12 cm.

Consider a pentagon PQRST with circumcircle having centre C and radius r = 15 cm.

Join CQ and CR and draw a perpendicular CH on QR.

Since, PQRST is a regular pentagon.

We get,

Each internal angle = 108°

⇒ ∠ PQR = 108°

Also CQ acts as an internal angle bisector for each internal angle of regular pentagon.

∴ ∠ PQC = ∠ RQC

⇒ ∠ PQR = ∠ PQC + ∠ RQC = 108°

⇒ ∠ PQR = 2(∠ PQC) = 2(∠ RQC) = 108°

⇒ ∠ PQC = ∠ RQC = 54°

Now in Δ QCH,

CD = sin 54° × QC

⇒ CD = sin 54° × 15 cm

(Frome table, sin 54° = 0.809)

⇒ CD = 0.809 × 15 = 12.135 cm

⇒ CD = 12.13 cm

We know that in regular pentagon all sides are equal.

∴ Length of each side = 12.13 cm

Length of each side = 12.13 cm.

Consider a triangle ABC with AB = 10 cm, AC = 8 cm and

∠ BAC = 40°

Draw perpendicular from C on AB, say CH.

From right angles triangle ABC,

CH = CA × sin 40°

⇒ CH = 8 × sin 40°

(From the table, sin 40° = 0.642 )

⇒ CH = 5.136 cm

Area (Δ ABC) = 25.68 cm²

If given angle is 140° i.e. ∠ BAC = 140°

Draw perpendicular from C on AB and extend AB to meet it at H.

∴ CH ⊥ HB

HB is a straight line.

⇒ ∠ HAB = ∠ HAC + ∠ BAC = 180°

⇒ ∠ HAC + 140° = 180°

⇒ ∠ HAC = 40°

In right triangle CHA,

∠ HAC = 40° and AC = 8 cm

CH = CA × sin 40°

⇒ CH = 8 × sin 40°

(From the table, sin 40° = 0.642 )

⇒ CH = 5.136 cm

Area (Δ ABC) = 25.68 cm²

Hence, Area is same only the position of perpendicular line was changed if the angle is considered as 140°.

Area of triangle (given angle is 40°) = 25.68 cm² and area of triangle ( angle is taken as 140° ) = 25.68 cm².

Construction, Let the centre of the circle be O and radius be r.

Let the triangle be ABC . Also, join OC and OB.

Now, OB = OC = r (radius )

We know that,

The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.

∴ ∠ BOC = 2 (∠ BAC)

⇒ ∠ BOC = 2 (70°) = 140°

In Δ BOC :

Draw a perpendicular bisector on BC from O say OH.

OH is a perpendicular bisector of BC.

∴ ∠ BOH = ∠ COH, OB = OC and ∠ OHB = ∠ OHC = 90°

⇒ ∠ BOC = ∠ BOH + ∠ COH = 140°

⇒ ∠ BOC = 2(∠ BOH) = 2(∠ COH) = 140°

⇒ ∠ BOH = ∠ COH = 70°

In Δ BOC,

Δ BOC is an isosceles triangle (∵ OB = OC)

∴ ∠ OBC = ∠ OCB = α

Now, ∠ OBC + ∠ OCB + ∠ BOC = 180°

⇒ α + α + 140° = 180°

⇒ 2(α) + 140° = 180°

⇒ α = ∠ OBC = 20°

We get,

∠ OBC = 20° , ∠ BOH = 70° and ∠ OHB = 90°

In right triangle BOH,

BH = BO × sin 70°

(From table, sin 40° = 0.939 )

BO = r = 1.86 cm

Radius of circle is 1.86 cm.

Construction, Let the centre of the circle be O and radius be r.

Let the triangle be ABC . Also, join OC and OB.

Now, OB = OC= r ( radius )

We know that,

The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.

∴ ∠BOC = 2 (∠BAC)

⇒ ∠BOC = 2 (70°) = 140°

In Δ BOC :

Draw a perpendicular bisector on BC from O say OH.

OH is a perpendicular bisector of BC.

∴ ∠BOH = ∠COH , OB = OC and ∠OHB = ∠OHC = 90°

⇒ ∠BOC = ∠BOH + ∠COH = 140°

⇒ ∠BOC = 2(∠BOH) = 2(∠COH) = 140°

⇒ ∠BOH = ∠COH = 70°

In Δ BOC,

Δ BOC is an isosceles triangle (∵ OB = OC)

∴ ∠OBC = ∠OCB = α

Now, ∠OBC + ∠OCB + ∠BOC = 180°

⇒ α + α + 140° = 180°

⇒ 2(α) + 140° = 180°

⇒ α = ∠OBC = 20°

We get,

∠OBC = 20° , ∠BOH = 70° and ∠OHB = 90°

In right triangle BOH,

BH = BO × sin 70°

(From table, sin 40° = 0.939 )

BO = r = 2.13 cm

Radius of circle is 2.13 cm.

Let AB = 5 cm and ∠BAC = 80°.

Now Let O be a midpoint of AB.

With O as centre and radius OA = OB,

Construct a circumcircle for Δ ABC.

We know that,

In circle, The angle formed by the diameter on its circumference

is always equal to 90° .

∴ ∠ABC = 90° (∵ AC is a diameter of circle O)

AC = AO + OC = r + r = 2r

In Δ ABC,

AB = AC × cos 80°

(From table, cos 80° = 0.173)

⇒ AC = 2r

⇒ r = 14.45 cm

Radius of a circle is 14.45 cm

Let us draw a complete circle.

Let the centre be O. Join OA and OB.

We know that,

The angle formed at the centre of the circle by lines originating from two points on the circle's circumference is double the angle formed on the circumference of the circle by lines originating from the same points. i.e. a = 2b.

∴ ∠AOB (External) = 280°

⇒ ∠AOB (Internal) = 360° - 280° = 80°

Let radius be r.

⇒ OA = OB = r.

Draw perpendicular bisector OH on AB

In Δ OHB

OH is a perpendicular bisector of AB.

∴ ∠AOH = ∠BOH , OA = OB and ∠OHA = ∠OHB = 90°

⇒ ∠BOA = ∠BOH + ∠AOH = 80°

⇒ ∠BOA = 2(∠BOH) = 2(∠AOH) = 80°

⇒ ∠BOH = ∠AOH = 40°

In Δ BOA,

Δ BOA is an isosceles triangle (∵ OB = OA)

∴ ∠OBA = ∠OAB = α

Now, ∠OBA + ∠OAB + ∠BOA = 180°

⇒ α + α + 80° = 180°

⇒ 2(α) + 80° = 180°

⇒ α = ∠OBA = 50°

We get,

∠OBA = 50° , ∠BOH = 40° and ∠OHB = 90°

In right triangle BHO,

BH = BO × sin 40°

(From table, sin 40° = 0.642)

⇒ BO = r = 6.23 cm

Radius of a circle is 6.23 cm

Let us draw a triangle ABC with all the given angles.

∠BAC + ∠ABC + ∠ACB = 180°

⇒ 45° + 65° + ∠ACB = 180°

⇒ ∠ACB = 180° - ( 45° + 65° ) = 70°

Since, We do not know the length of any side of triangle , So we cannot use the concept of perpendicular bisector to construct a circumcircle.

But we know that every triangle formed by a circumcentre with the two consecutive points is an isosceles triangle.

Let each internal angle of a triangle be divided into two angles by a point S’ as shown in fig.

Here,

α + β = 45° …(i)

α + γ = 65° …(ii)

β + γ = 70° …(iii)

Subtracting eq. (i) from (ii),

We get,

( α + γ) – (α + β) = 65° - 45°

⇒ γ - β = 20° …(iv)

Adding eq. (iii) and (iv),

We get,

(β + γ) + (γ – β) = 70° = 20°

⇒ 2γ = 90°

⇒ γ = 45°

Substituting the value of γ in eq. (ii) and eq. (iii)

we get,

α + 45° = 65° and β + 45° = 70°

⇒ α = 20° and β = 25°

Steps of construction :

1.Mark the angle α on AX, such that ∠RAX = α = 20°.

2. Cut the arc of length 2.5 cm on AR say BO. (r = 2.5 cm)

3. With O as a centre and radius equal to 2.5 cm cut the arc

on AX say OB.

4. With base as AB construct ∠BAQ = 45° and ∠ABR = 65°.

Let them meet at C.

5. Join BC and AC.

6. With O as a centre and radius = 2.5 cm draw a circle.

Thus we find that the circle with centre O is the circumcircle of Δ ABC.

We know that perpendicular from the circumcentre act as a perpendicular bisector.

Construct OF ⊥ AB, OG ⊥ AC and OH ⊥ BC.

In right Δ OFA,

∠OAF = 20° and OA = r = 2.5 cm

AF = OA × cos 20°

⇒ AF = 2.5 × 0.939 cm (From table, cos 20° = 0.939)

⇒ AF = 2.347 cm = 2.35 cm

In Δ AOB

AF = BF

⇒ AB = 2(AF) = 2(2.35) = 4.7 cm

In right Δ OGC,

∠OCG = 25° and OC = r = 2.5 cm

GC = OC × cos 25°

⇒ GC = 2.5 × 0.906 cm (From table, cos 25° = 0.906)

⇒ GC = 2.265 cm

In Δ AOC

AG = GC

⇒ AC = 2(GC) = 2(2.265) = 4.53 cm

In right Δ OHB,

∠OBH = 45° and OB = r = 2.5 cm

BH = OB × cos 45°

⇒ BH = 2.5 × 0.707 cm (From table, cos 45° = 0.707)

⇒ BH = 1.767 cm

In Δ BOC

BH = HC

⇒ BC = 2(BH) = 2(1.767) = 3.53 cm

Thus AB = 4.7 cm, BC = 3.53 cm and CA = 4.53 cm

Length of all sides of triangle are 4.7 cm, 3.53 cm and 4.53 cm.

Let ABC be the required triangle.

Draw perpendicular from AS on BC.

Let BS = x cm

⇒ SC = (5 – x) cm

In Δ BAS,

AS = BS × tan 50°

⇒ AS = x × tan 50° cm …(i)

In Δ CAS,

AS = CS × tan 65°

⇒ AS = (5 - x) × tan 65° cm …(ii)

From eq. (i) and (ii), We get :

x × tan 50° = (5 - x) × tan 65° cm

⇒ x tan 50° = 5 tan 65° - x tan 65°

⇒ x tan 50° + x tan 65° = 5 tan 65°

⇒ x (tan 50° + tan 65°) = 5 tan 65°

(From table, tan 65° = 2.144 and tan 50° = 1.191)

In Δ BAC,

Height = x = 3.21 cm and Base = 5 cm

Area (Δ) = 8.025 cm²

The rhombus ABCD is as shown in the figure shown below,

where,

DB = 5 cm

∠ADC = 50°

We know that,

Diagonals of a rhombus are perpendicular bisector to each other.

Also, diagonals bisect the corner angles of rhombus

So,

∠ODC = 25° [∵ DB bisects ∠D]

OD = 2.5 cm [∵ AC bisects BD]

Now OC is given by,

OC = OD × tan25°

OC = 2.5 × 0.466

OC = 1.166 cm

Also, AC = 2 × OC [∵ BD bisects AC]

AC = 2 × 1.166

AC = 2.332 cm

We know, Area of rhombus =

Area

Area

Area

The question can be figured out as:

Here ladder is forming a right angled triangle with wall and floor,

By using trigonometry,

⇒ h = b × tan40°

⇒ h = 2 × 0.839

⇒ h = 1.678 m

So, top end of ladder is 1.678 m high from ground.

For the figures,

Given, WX = XY = YZ = ZY = YW = 30 cm

So, It can be easily seen that,

SQ = PN = WX = 30 cm [∵ Side of pentagon given]

CD = AB =

CD = AB =

Let’s say in rectangle PQRS,

∠PQS = ∠RSQ

[∵ Alternate interior angles for PQ ∥ SR on rectangle PQRS]

Similarly,

∠MNP = ∠OPN

∠RQS = ∠PSQ

[∵ Alternate interior angles for SP ∥ RQ on rectangle PQRS]

∠ONP = ∠MPN

So ∠WYJ = ∠ZYJ = ∠WXI = ∠YXI

And,

∠WYJ =

∠WYJ = [∵ Interior angle of regular pentagon = 108° ]

∠WYJ = 54°

So,

WJ = WY × sin54°

WJ = 30 × 0.81

WJ = 24.3 cm (i)

So, WZ = 2× WJ = 2× 24.3

WZ = 48.6

And, YJ = WY× cos54°

YJ = 30 × 0.59

YJ = 17.7 cm (ii)

Also, YZ = 2 × KZ

⇒ 30 = 2× KZ

⇒ KZ = 15 cm (iii)

By applying Pythagoras Theorem to ΔWKZ, we have,

WK² = WZ² - KZ²

⇒ WK² = 48.6² - 15²

⇒ WK² = 2361.96 – 225

⇒ WK² = 2136.96

⇒ WK = 46.2 cm (iv)

Thus dimensions of rectangles are as follows:

In ABCD,

Length AD = BC = 46.2 cm [From (iv) WK = 46.2 cm]

Breadth AB = CD = 15 cm [From (iii) KZ = 15 cm]

In PQRS and MNOP,

Length PS = QR = MP = NO = 24.3 cm [From (i) WJ = 24,3 cm]

Breadth PQ = SR = MN = PO = 17.7 cm [From (ii) YJ = 17.7 cm]

Let us label the diagram

Let the distance between two consecutive vertical lines be 'x' and height of vertical lines be h₁, h₂, h₃, …, and so on.

Let AB₁ = x

Now, we know

⇒ perpendicular = base × tanθ

Therefore, in consecutive triangles, we have

B₁C₁ = AB₁ × tan 40°

⇒ h₁ = atan40°

B₂C₂ = AB₂ × tan 40°

⇒ h₂ = (a + x)tan40°

B₃C₃ = AB₃ × tan 40°

⇒ h₃ = (a + 2x)tan40°

.

.

.

and so on.

Now, we have

h₁, h₂, h₃,… = atan40°, (a + x)tan40°, (a + x)tan40°, …

Let us calculate the common difference.

h₂ - h₁ = (a + x)tan40° - atan40° = xtan40°

h₃ - h₂ = (a + 2x)tan40° - (a + x)tan40° = xtan40°

As, the common difference between the terms is same, terms are in AP and hence,

h₁, h₂, h₃, …. are in AP.

Therefore, heights of vertical lines are in AP with common difference xtan40°, where 'x' is the space between two consecutive lines. [tan 40° = 0.8391]

Let ABC be a triangle, with BC = 6 cm and angles at its ends are ∠ABC = 40° and ∠ACB = 65° respectively.

Draw AP ⊥ BC, such that APB and APC are right-angled triangles,

Let AP = 'h' cm

BP = 'x' cm and PC = (6 - x) cm

In ΔAPB,

…[1]

In ΔAPC

⇒ 2.1445(6 - x) = h

= h [From 1]

⇒ 12.867 - 2.556h = h

⇒ 3.556h = 12.867

⇒ h = 3.618 cm

Also, area of a triangle

⇒ area(ΔABC) = 3 × 3.618

= 10.854 cm² [appx]

Given :
Angle of elevation = 40° and Length of shadow = 18 m.
Let the required triangle be Δ ABC.
∴ Angle of elevation = ∠ ABC = 40° and Length of shadow = AB = 18 m and Height of tree be AC.

In right Δ BAC,
AC = AB × tan 40°
⇒ AC = 18 × tan 40° (From table, tan 40° = 0.839)

⇒ AC = 18 × 0.839 = 15.102 m = 15.1 m

Height of tree is 15.1 metre.

Let the height of the hill be x m and height of tower be y m.

Let BD be height of boy 1.75 m.

The below diagram shows the relation for the given condition.

In right Δ DFC,
FC = AC – AF = (x – 1.75) m and FD = AB = 40 m

FC = tan 60° × FD
⇒ (x – 1.75) = tan 60° × 40
(From table, tan 60° = 1.732 )
⇒ (x – 1.75) = 1.732× 40 = 69.28
⇒ x = 69.28 + 1.75 = 71.03 m

In right Δ EGC,
GC = AC – AG = (71.03 – y) m and EG = AB = 40 m

GC = tan 50° × EG
⇒ (71.03 – y) = tan 50° × 40
(From table, tan 50° = 1.19)
⇒ (71.03 – y) = 1.19× 40 = 47.6
⇒ y = 71.03 – 47.6 = 23.43 m

The height of the hill is 71.03 m and height of tower is 23.43 m.

Let x m be actual height of building and BD be the height of

boy i.e. BD = 1.5 m.

Let the distance between the boy and the building be y m.

In right Δ DFC,
FC = AC – AF = (x – 1.5) m and FD = AB = y m

FC = tan 60° × FD
⇒ (x – 1.5) = tan 60° × y
(From table, tan 60° = 1.732 )
⇒ (x – 1.75) = 1.732× y …(i)

In right Δ DFG,
FG = AC – CG – AF = (x – 10 – 1.5) m = (x – 11.5) m
and FD = AB = y m

FG = tan 30° × FD

⇒ (x – 11.5) = tan 30° × y
(From table, tan 30° = 0.57)

⇒ (x – 11.5) = 0.57 × y …(ii)

Dividing eq. (i) from eq. (ii)

⇒ 1.732(x – 11.5) = 0.57(x – 1.75)

⇒ 1.732(x) – 1.732(11.5) = 0.57(x) – 0.57(1.75)

⇒ 1.732(x) – 0.57(x) = 1.732(11.5) – 0.57(1.75)

⇒ (1.732 – 0.57)x = 19.918 – 0.9975

⇒ 1.162(x) = 18.92

Height of the building is 16.28 m.

Let the height of tower be x metre and distance between tower and building be y metre.

Let CD be the height of man i.e. is 1.8 m.

In right Δ ABD,
BD = (x + 1.8) m and AB = y m

BD = tan 60° × AB
⇒ (x + 1.8) = tan 60° × y
(From table, tan 60° = 1.732 )
⇒ (x + 1.8) = 1.732× y …(i)

In right Δ DEJ,
DJ = (1.8 + x – 10) m = (x – 8.2) m and EJ = AB = y m

DJ = tan 40° × EJ
⇒ (x – 8.2) = tan 40° × y
(From table, tan 40° = 0.839 = 0.84)
⇒ (x – 8.2) = 0.84× y …(ii)

Dividing eq. (i) from eq. (ii),

⇒ 1.732(x – 8.2) = 0.84(x + 1.82)

⇒ 1.732(x) – 1.732(8.2) = 0.84(x) + 0.84(1.82)

⇒ 1.732(x) – 0.84(x) = 1.732(8.2) + 0.84(1.82)

⇒ x(1.732 – 0.84) = 14.2024 + 1.5288

⇒ x(0.892) = 15.7312

⇒ x = 17.63 m
Substituting the value of x in eq. (i)

⇒ (x + 1.8) = 1.732× y

⇒ (17.63 + 1.8) = 1.732(y)

⇒ 19.43 = 1.732(y)

⇒ y = 11.21 m

Height of tower is 17.63 m and distance between tower and building is 11.21 m.

Let the height of the post be x metre and distance between post and one of the wire fixed point be y metre.

In right Δ ADC,

AD = y m and CD = x m
CD = tan 55° × y

⇒ x = tan 55° × y

(From table, tan 55° = 1.428 = 1.43)

⇒ x = 1.43 × y …(i)

In right Δ CDB,

BD = AB – AD = (25 –y) m and CD = x

CD = tan 40° × BD

⇒ CD = tan 40° × (25 – y)

(From table, tan 40° = 0.839 = 0.84)

⇒ x = tan 40° × (25 – y)

⇒ x = 0.84 × (25 – y) …(ii)

Dividing eq. (i) from eq. (ii)

⇒ 0.84(25) – 0.84(y) = 1.43(y)

⇒ 1.43(y) + 0.84(y) = 0.84(25)

⇒ 2.27 (y) = 21

⇒ y = 9.25 m
Substituting value of y in eq. (i)

⇒ x = (1.43 × 9.25) m = 13.2275 m

Height of post is 9.25 m

Let the height of the tree be x metre.

In right Δ BAC,

AC = x and AB = AD + DB = (10 + y) m

AC = tan 25° × AB

⇒ x = tan 25° × (10 + y)

(From table, tan 25° = 0.466)

⇒ x = 0.466 × (10 + y) …(i)

In right Δ DAC,

AC = x and AD = y

AC = tan 35° × AD

⇒ x = tan 35° × (y)

(From table, tan 35° = 0.7)

⇒ x = 0.7 × y …(ii)

Dividing eq. (i) from eq. (ii),

⇒ 0.7(y) = 0.466(10) + 0.466(y)

⇒ 0.7(y) – 0.466(y) = 0.466(10)

⇒ 0.234 (y) = 4.66

⇒ y = 19.91 m

AB = 10 + y = 10 + 19.91 = 29.91 m

Length of shadow at 25° = AB = 29.91 m

Length of shadow = 29.91 m