Solution of Right Triangles

Given that Δ ABC is isosceles, ∠ BAC = ∠ BCA.

Let ∠BAC = ∠BCA = x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + x + 90° = 180°

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\frac{90°}{2}$

⇒ x = 45°

Hence, ∠BAC = 45°.

Hence, option 2 is the correct option.

Let ∠ABC be x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠CAB + ∠ ACB + ∠ ABC = 180°

⇒ 60° + 90° + x = 180°

⇒ 150° + x = 180°

⇒ x = 180° - 150°

⇒ x = 30°

Now, cot ∠ABC = cot 30°

= $\sqrt3$

Hence, option 3 is the correct option.

Let ∠BAC be x.

Using the angle sum property of a triangle, which states that the sum of all angles in a triangle is 180°:

⇒ ∠BAC + ∠BCA + ∠ABC = 180°

⇒ x + 45° + 90° = 180°

⇒ x + 135° = 180°

⇒ x = 180° - 135°

⇒ x = 45°

Thus, ∠BAC = ∠BCA = 45°. Therefore, Δ ABC is an isosceles triangle, meaning AB = BC = 8 m.

Hence, option 3 is the correct option.

In Δ ABD,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AB}{BD}\\[1em] ⇒ 1 = \frac{10}{BD}\\[1em] ⇒ BD = 10 m\end{aligned}$$

In Δ ABC,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{BC}\\[1em] ⇒ \frac{1}{1.732} = \frac{10}{BC}\\[1em] ⇒ BC = 10 \times 1.732 m\\[1em] ⇒ BC = 17.32 m\end{aligned}$$

As, BC = BD + CD

⇒ 17.32 = 10 + CD

⇒ CD = 17.32 - 10

⇒ CD = 7.32

Hence, option 1 is the correct option.

Let AC = x. Therefore, DC = x.

In Δ ABC,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{BC}{x}\\[1em] ⇒ BC = \frac{x}{2}\end{aligned}$$

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AB}{x}\\[1em] ⇒ AB = \frac{x\sqrt3}{2}\end{aligned}$$

$$\begin{aligned}BD = BC + CD\\[1em] ⇒ BD = \frac{x}{2} + x\\[1em] ⇒ BD = \frac{x + 2x}{2}\\[1em] ⇒ BD = \frac{3x}{2}\end{aligned}$$

In Δ ABD, according to Pythagoras theorem,

⇒ AD² = BD² + AB² (∵ AD is hypotenuse)

$$\begin{aligned}⇒ 80^2 = \Big(\frac{3x}{2}\Big)^2 + \Big(\frac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 6400 = \frac{9x^2}{4} + \frac{3x^2}{4} \\[1em] ⇒ 6400 = \frac{9x^2 + 3x^2}{4}\\[1em] ⇒ 6400 = \frac{12x^2}{4}\\[1em] ⇒ 6400 = 3x^2\\[1em] ⇒ x^2 = \frac{6400}{3}\\[1em] ⇒ x = \sqrt\frac{6400}{3}\\[1em] ⇒ x = \frac{80}{\sqrt3}\end{aligned}$$

Substituting the value of x in AB,

$$\begin{aligned}AB = \frac{x\sqrt3}{2}\\[1em] ⇒ AB = \frac{\frac{80}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AB = \frac{\frac{80}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AB = 40 m\end{aligned}$$

Hence, option 4 is the correct option.

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{20}{x}\\[1em] ⇒ x = \frac{20 \times 2}{\sqrt3}\\[1em] ⇒ x = \frac{40}{\sqrt3} \\[1em] ⇒ x = 23.1\end{aligned}$$

Hence, x = 23.1.

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{20}{x}\\[1em] ⇒ x = 20 \times \sqrt3\\[1em] ⇒ x = 34.64\end{aligned}$$

Hence, x = 34.64.

$$\begin{aligned}\text{sin 45°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{1}{\sqrt2} = \frac{20}{x}\\[1em] ⇒ x = 20 \times \sqrt2\\[1em] ⇒ x = 28.28\end{aligned}$$

Hence, x = 28.28.

$$\begin{aligned}\text{cos A} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \text{cos A} = \frac{10}{20}\\[1em] ⇒ \text{cos A} = \frac{1}{2}\\[1em] ⇒ \text{cos A} = \text{cos 60°}\end{aligned}$$

Hence, A = 60°.

$$\begin{aligned}\text{sin A} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \text{sin A} = \frac{\frac{10}{\sqrt2}}{10}\\[1em] ⇒ \text{sin A} = \frac{1}{\sqrt2}\\[1em] ⇒ \text{sin A} = \text{sin 45°}\end{aligned}$$

Hence, A = 45°.

$$\begin{aligned}\text{tan A} = \frac{Perpendicular}{Base}\\[1em] ⇒ \text{tan A} = \frac{10\sqrt3}{10}\\[1em] ⇒ \text{tan A} = \sqrt3\\[1em] ⇒ \text{tan A} = \text{tan 60°}\end{aligned}$$

Hence, A = 60°.

In Δ ABC,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{CB}{AC}\\[1em] ⇒ \sqrt3 = \frac{30}{AC}\\[1em] ⇒ AC = \frac{30}{\sqrt3}\\[1em] ⇒ AC = \frac{30 \times \sqrt3}{3}\\[1em] ⇒ AC = 10\sqrt3\end{aligned}$$

In Δ ADC,

$$\begin{aligned}\text{sin x} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \text{sin x} = \frac{AC}{AD}\\[1em] ⇒ \text{sin x} = \frac{10\sqrt3}{20}\\[1em] ⇒ \text{sin x} = \frac{\sqrt3}{2}\\[1em] ⇒ \text{sin x} = \text {sin 60°}\end{aligned}$$

Hence, x = 60°.

BE = CD = 50 m

ED = BC = 10 m

In Δ ABE,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AE}{BE}\\[1em] ⇒ 1 = \frac{AE}{50}\\[1em] ⇒ AE = 50\end{aligned}$$

AD = AE + ED

⇒ AD = 50 + 10 m

⇒ AD = 60 m

Hence, AD = 60 m.

Let AC = x. Therefore, BC = x.

In Δ ADC,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{CD}{x}\\[1em] ⇒ CD = \frac{x}{2}\\[1em]\end{aligned}$$

And,

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AD}{x}\\[1em] ⇒ AD = \frac{x\sqrt3}{2}\\[1em]\end{aligned}$$

$$\begin{aligned}\text{BD = BC + CD}\\[1em] ⇒ BD = x + \frac{x}{2}\\[1em] ⇒ BD = \frac{2x + x}{2}\\[1em] ⇒ BD = \frac{3x}{2}\end{aligned}$$

In Δ ABD, according to Pythagoras theorem,

⇒ AB² = BD² + AD² (∵ AB is hypotenuse)

$$\begin{aligned}⇒ 100^2 = \Big(\frac{3x}{2}\Big)^2 + \Big(\frac{\sqrt3x}{2}\Big)^2 \\[1em] ⇒ 10000 = \frac{9x^2}{4} + \frac{3x^2}{4} \\[1em] ⇒ 10000 = \frac{9x^2 + 3x^2}{4}\\[1em] ⇒ 10000 = \frac{12x^2}{4}\\[1em] ⇒ 10000 = 3x^2\\[1em] ⇒ x^2 = \frac{10000}{3}\\[1em] ⇒ x = \sqrt\frac{10000}{3}\\[1em] ⇒ x = \frac{100}{\sqrt3}\end{aligned}$$

Substituting the value of x in AD,

$$\begin{aligned}AD = \frac{x\sqrt3}{2}\\[1em] ⇒ AD = \frac{\frac{100}{\sqrt3} \times \sqrt3}{2}\\[1em] ⇒ AD = \frac{\frac{100}{\cancel {\sqrt3}} \times \cancel {\sqrt3}}{2}\\[1em] ⇒ AD = 50 m\end{aligned}$$

Hence, AD = 50 m.

In Δ BDC,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{DC}{BC}\\[1em] ⇒ 1 = \frac{DC}{40}\\[1em] ⇒ DC = 40\end{aligned}$$

In Δ ABC,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{AC}{BC}\\[1em] ⇒ \sqrt3 = \frac{AC}{40}\\[1em] ⇒ AC = 40\sqrt3 = 69.28\end{aligned}$$

AD = AC - DC

⇒ AD = 69.28 - 40 = 29.28 cm

Hence, AD = 29.28 cm.

ABCD is a rhombus as AB = BC = CD = DA and ∠ BAD = 60°.

Now in Δ ABD,

AB = AD (Sides of rhombus)

⇒ ∠ ABD = ∠ ADB (Angles opposite to equal side of Δ)

Let ∠ ABD = x.

According to the angle sum property,

∠ ABD + ∠ ADB + ∠ BAD = 180°

⇒ x + x + 60° = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 180° - 60°

⇒ 2x = 120°

⇒ x = $\frac{120°}{2}$

⇒ x = 60°

ABCD is a rhombus. So, diagonals AC and BD bisect each other at 90°.

AO = OC and BO = OD

In Δ AOD,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{OD}{60}\\[1em] ⇒ OD = \frac{60}{2} = 30\end{aligned}$$

And,

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AO}{60}\\[1em] ⇒ AO = \frac{60\sqrt3}{2} = 51.96\end{aligned}$$

AC = 2 x AO = 2 x 51.96 = 103.92 cm

BD = 2 x BO = 2 x 30 = 60 cm

Hence, AC = 103.92 cm and BD = 60 cm.

Draw a line FP parallel to CD such that FP = CD and FC = PD.

In Δ ACF,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{20}{AC}\\[1em] ⇒ AC = 20\end{aligned}$$

In Δ DEB,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{30}{BD}\\[1em] ⇒ BD = \frac{30}{\sqrt3}\\[1em] ⇒ BD = \frac{30 \times \sqrt3}{\sqrt3 \times \sqrt3}\\[1em] ⇒ BD = \frac{30\sqrt3}{3}\\[1em] ⇒ BD = 10\sqrt3 = 17.32\end{aligned}$$

As, FC = 20 and ED = 30

EP = ED - PD = ED - FC = 30 - 20 = 10 cm

In Δ EFP,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{FP}{10}\\[1em] ⇒ FP = 10 \sqrt3 = 17.32\end{aligned}$$

Thus AB = AC + CD + BD

= 20 + 17.32 + 17.32

= 54.64

Hence, AB = 54.64.

(i) Draw two perpendicular lines, DP and CM, from points D and C to AB, respectively. Since AB is parallel to CD, PMCD will form a rectangle.

In Δ APD,

cos 60° = $\frac{Base}{Hypotenuse}$

⇒ $\frac{1}{2} = \frac{AP}{AD}$

⇒ $\frac{1}{2} = \frac{AP}{20}$

⇒ AP = 10

Now, Δ APD and Δ BMC,

DP = CM (Height of same quadrilateral)

AD = CB (Both are 20 cm)

∠ DPA = ∠ CMB

So, Δ APD ≅ Δ BMC (∵ RHS congruency)

And, by using corresponding sides of congruent triangle,

AP = MB = 10 cm

PM = DC = 20 cm

AB = AP + PM + MB

= 10 + 20 + 10 cm

= 40 cm

Hence, AB = 40 cm.

(ii) In Δ APD,

sin 60° = $\frac{Perpendicular}{Hypotenuse}$

⇒ $\frac{\sqrt3}{2} = \frac{PD}{AD}$

⇒ $\frac{\sqrt3}{2} = \frac{PD}{20}$

⇒ PD = 10 $\sqrt3$ = 17.32 cm

Hence, PD = 17.32 cm.

In Δ APQ,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AQ}{AP}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{10}{AP}\\[1em] ⇒ AP = 10 \sqrt3 = 17.32 cm\end{aligned}$$

In Δ PBR,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{PB}{BR}\\[1em] ⇒ 1 = \frac{PB}{8}\\[1em] ⇒ PB = 8 cm\end{aligned}$$

AB = AP + PB

= 17.32 + 8 cm

= 25.32 cm

Hence, AB = 25.32 cm.

From figure,

AB = BC ....................(1)

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

From figure,

⇒ tan A = $\frac{BC}{AB}$

⇒ tan A = $\frac{AB}{AB}$ [from equation (1)]

⇒ tan A = 1

⇒ tan A = tan 45°

⇒ A = 45°.

Hence, option 1 is the correct option.

Since CD is perpendicular on AB.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

From figure,

$$\begin{aligned}\Rightarrow \text{tan B} = \frac{CD}{DB}\\[1em] \Rightarrow \text{tan 30°} = \frac{10}{DB}\\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{DB}\\[1em] \Rightarrow DB = 10\sqrt{3}.\end{aligned}$$

As ΔACD is a right angled triangle,

$$\begin{aligned}\Rightarrow \text{tan A} = \frac{CD}{AD} \\[1em] \Rightarrow \text{tan 45°} = \frac{10}{AD} \\[1em] \Rightarrow 1 = \frac{10}{AD} \\[1em] \Rightarrow AD = 10.\end{aligned}$$

From figure,

AB = AD + DB = 10 + $10\sqrt{3} = 10(1 + \sqrt{3})$ cm.

Hence, option 4 is the correct option.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

In triangle BCD,

$$\begin{aligned}\Rightarrow \text{tan B} = \frac{CD}{BC}\\[1em] \Rightarrow \text{tan 45°} = \frac{10}{BC}\\[1em] \Rightarrow 1 = \frac{10}{BC}\\[1em] \Rightarrow BC = 10.\end{aligned}$$

In triangle ACD,

$$\begin{aligned}\Rightarrow \text{tan A} = \frac{CD}{AC}\\[1em] \Rightarrow \text{tan 30°} = \frac{10}{AC}\\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{10}{AC}\\[1em] \Rightarrow AC = 10\sqrt{3}.\end{aligned}$$

From figure,

⇒ AB = AC - BC

⇒ AB = $10\sqrt{3}$ - 10 cm

⇒ AB = 10($\sqrt{3}$ - 1) cm.

Hence, option 2 is the correct option.

From figure,

DC = CA = x (let)

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

In triangle ABD,

$$\begin{aligned}\Rightarrow \text{tan D} = \frac{AB}{DB} \\[1em] \Rightarrow \text{tan 30°} = \frac{AB}{DC + CB} \\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{x + 20} \\[1em] \Rightarrow AB = \frac{x + 20}{\sqrt{3}}.\end{aligned}$$

Since ΔABC is a right angled triangle, using pythagoras theorem,

⇒ Hypotenuse² = Base² + Height²

⇒ AC² = CB² + AB²

$$\begin{aligned}\Rightarrow x^2 = 20^2 + \Big(\frac{x + 20}{\sqrt{3}}\Big)^2\\[1em] \Rightarrow x^2 = 400 + \Big(\frac{x^2 + 400 + 40x}{3}\Big)\\[1em] \Rightarrow x^2 = \Big(\frac{1200 + x^2 + 400 + 40x}{3}\Big)\\[1em] \Rightarrow 3x^2 = 1200 + x^2 + 400 + 40x\\[1em] \Rightarrow 3x^2 = 1600 + x^2 + 40x\\[1em] \Rightarrow 3x^2 - 1600 - x^2 - 40x = 0\\[1em] \Rightarrow 2x^2 - 40x - 1600 = 0\\[1em] \Rightarrow 2(x^2 - 20x - 800) = 0 \\[1em] \Rightarrow x^2 - 20x - 800 = 0\\[1em] \Rightarrow x^2 - 40x + 20x - 800 = 0\\[1em] \Rightarrow x(x - 40) + 20(x - 40) = 0\\[1em] \Rightarrow (x - 40)(x + 20) = 0\\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 20) = 0\\[1em] \Rightarrow x = 40 \text{ or } x = -20.\end{aligned}$$

As length of side of triangles cannot be negative. So, x = 40 cm.

DC = AC = x = 40 cm.

$AB = \frac{x + 20}{\sqrt{3}} = \frac{40 + 20}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3}$ cm.

Hence, option 1 is the correct option.

In figure,

Let AB be the tower and BC be the shadow.

Join AC.

Given, AB = 50 m and BC = 50 $\sqrt{3}$ m.

As tower is perpendicular to its shadow, thus ΔABC is a right angled triangle,

By formula,

$$\begin{aligned}\Rightarrow \text{tan θ} = \frac{\text{Perpendicular}}{\text{Base}} \\[1em] \Rightarrow \text{tan θ} = \frac{50}{50\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \frac{1}{\sqrt{3}} \\[1em] \Rightarrow \text{tan θ} = \text{tan 30°} \\[1em] \Rightarrow \text{θ} = 30°.\end{aligned}$$

∴ Both the statements are true.

Hence, option 1 is the correct option.

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

In ΔABP,

$$\begin{aligned}\Rightarrow \text{tan P} = \frac{AB}{BP}\\[1em] \Rightarrow \text{tan 60°} = \frac{3}{BP}\\[1em] \Rightarrow \sqrt{3} = \frac{3}{BP}\\[1em] \Rightarrow BP = \frac{3}{\sqrt{3}}\\[1em] \Rightarrow BP = \sqrt{3}\text{ cm}.\end{aligned}$$

In ΔPCD,

$$\begin{aligned}\Rightarrow \text{tan P} = \frac{DC}{PC}\\[1em] \Rightarrow \text{tan 30°} = \frac{2}{PC}\\[1em] \Rightarrow \frac{1}{\sqrt{3}} = \frac{2}{PC}\\[1em] \Rightarrow PC = 2 \sqrt{3}\text{ cm}.\end{aligned}$$

From figure,

BC = BP + PC $= \sqrt{3} + 2\sqrt{3} = 3\sqrt{3}$.

∴ Statement 1 is false, and statement 2 is true.

Hence, option 4 is the correct option.

Given, angle ABC = 120° and length of its each side = 20 cm.

As we know that, each diagonal divides the angles at its endpoints into two equal parts.

The diagonal BD bisects ∠ABC and ∠ADC.

⇒ ∠ABD = ∠CBD = $\frac{120°}{2}$ = 60°

By formula,

cos θ = $\frac{\text{Base}}{\text{Hypotenuse}}$

We know that,

Diagonals of rhombus are perpendicular to each other.

Thus, triangle AOB is a right angle triangle.

In ΔAOB,

$$\begin{aligned}\Rightarrow \text{cos B} = \frac{\text{OB}}{\text{AB}}\\[1em] \Rightarrow \text{cos 60°} = \frac{\text{OB}}{20}\\[1em] \Rightarrow \frac{1}{2} = \frac{\text{OB}}{20}\\[1em] \Rightarrow \text{OB} = \frac{20}{2}\\[1em] \Rightarrow \text{OB} = 10\end{aligned}$$

Since, diagonals of a rhombus bisect each other.

⇒ BD = 2 x OB

So, reason (R) is true.

⇒ BD = 2 x 10 cm = 20 cm

So, assertion (A) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

From figure,

⇒ BC = AP

⇒ AP = 50 m

By formula,

tan θ = $\frac{\text{Perpendicular}}{\text{Base}}$

In ΔPBC,

$$\begin{aligned}\Rightarrow \text{tan 45°} = \frac{BC}{PB}\\[1em] \Rightarrow 1 = \frac{50}{PB}\\[1em] \Rightarrow PB = 50 \text{ m}.\end{aligned}$$

From figure,

AB = AP + PB = 50 m + 50 m = 100 m.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

In Δ ABC, according to Pythagoras theorem,

⇒ AC² = BC² + AB² (∵ AC is hypotenuse)

⇒ 2² = BC² + ($\sqrt3$)²

⇒ 4 = BC² + 3

⇒ BC² = 4 - 3

⇒ BC² = 1

⇒ BC = $\sqrt1$

⇒ BC = 1

$$\begin{aligned}\text{cot x} = \frac{Base}{Perpendicular}\\[1em] \text{cot x} = \frac{CB}{AB}\\[1em] \text{cot x} = \frac{1}{\sqrt3}\end{aligned}$$

Hence, cot x = $\frac{1}{\sqrt3}$.

DE = CB = 30 cm

In Δ ADE,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AE}{DE}\\[1em] ⇒ 1 = \frac{AE}{30}\\[1em] ⇒ AE = 30 cm\end{aligned}$$

In Δ DEB,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{EB}{DE}\\[1em] ⇒ \sqrt3 = \frac{EB}{30}\\[1em] ⇒ EB = 30\sqrt3 = 51.96\end{aligned}$$

AB = AE + EB

= 30 + 51.96 cm

= 81.96 cm

Hence, AB = 81.96 cm.

(i) In Δ ADC,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AD}{DC}\\[1em] ⇒ 1 = \frac{2}{DC}\\[1em] ⇒ DC = 2\end{aligned}$$

DC = AB = 2 cm

Hence, AB = 2 cm.

(ii) In Δ ADC,

$$\begin{aligned}\text{sin 45°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{1}{\sqrt2} = \frac{AD}{AC}\\[1em] ⇒ \frac{1}{\sqrt2} = \frac{2}{AC}\\[1em] ⇒ AC = 2\sqrt2 = 2.83 cm\end{aligned}$$

Hence, AC = 2.83 cm.

(iii) In Δ ADE,

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AD}{AE}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{2}{AE}\\[1em] ⇒ AE = \frac{4}{\sqrt3} = 2.31 cm\end{aligned}$$

Hence, AC = 2.31 cm.

(i) In Δ ABE,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{BE}{AB}\\[1em] ⇒ \frac{1}{2} = \frac{BE}{16}\\[1em] ⇒ BE = \frac{16}{2} = 8 cm\end{aligned}$$

Hence, BE = 8 cm.

(ii) In Δ ABE, according to Pythagoras theorem,

⇒ AB² = BE² + EA² (∵ AB is hypotenuse)

⇒ 16² = 8² + EA²

⇒ 256 = 64 + EA²

⇒ EA² = 256 - 64

⇒ EA² = 192

⇒ EA = $\sqrt{192}$

⇒ EA = 8 $\sqrt{3}$

EC = BC - BE

= 23 - 8 cm

= 15 cm

In Δ AEC, according to Pythagoras theorem,

⇒ AC² = AE² + EC² (∵ AC is hypotenuse)

⇒ AC² = (8 $\sqrt{3}$)² + 15²

⇒ AC² = 192 + 225

⇒ AC² = 417

⇒ AC = $\sqrt{417}$ = 20.42 cm

Hence, AC = 20.42 cm.

(i) In Δ ABC,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{BC}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{12}{BC}\\[1em] ⇒ BC = 12 \sqrt3\\[1em] ⇒ BC = 20.78\end{aligned}$$

Hence, BC = 20.78 cm.

(ii) In Δ ABC, according to angle sum property,

∠ ABC + ∠ BAC + ∠ ACB = 180°

⇒ 90° + ∠ BAC + 30° = 180°

⇒ 120° + ∠ BAC = 180°

⇒ ∠ BAC = 180° - 120°

⇒ ∠ BAC = 60° = ∠ BAD

In Δ ABD,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{AD}{AB}\\[1em] ⇒ \frac{1}{2} = \frac{AD}{12}\\[1em] ⇒ AD = \frac{12}{2}\\[1em] ⇒ AD = 6\end{aligned}$$

Hence, AD = 6 cm.

(iii) In Δ ABC,

$$\begin{aligned}\text{sin 30°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{AB}{AC}\\[1em] ⇒ \frac{1}{2} = \frac{12}{AC}\\[1em] ⇒ AC = 12 \times 2\\[1em] ⇒ AC = 24\end{aligned}$$

Hence, AC = 24 cm.

(i) AB = ${\sqrt3}$ BC

$\frac{AB}{BC} = \sqrt3$

$$\begin{aligned}\text{cot θ} = \frac{AB}{BC}\\[1em] \text{cot θ} = \sqrt3\\[1em] \text{cot θ} = \text{cot 30°}\\[1em]\end{aligned}$$

So, θ = 30°

Hence, magnitude of angle A = 30°.

(ii) BC = ${\sqrt3}$ AB

$\frac{BC}{AB} = \sqrt3$

$$\begin{aligned}\text{tan θ} = \frac{BC}{AB}\\[1em] \text{tan θ} = \sqrt3\\[1em] \text{tan θ} = \text{tan 60°}\\[1em]\end{aligned}$$

So, θ = 60°

Hence, magnitude of angle A = 60°.

AC is the ladder and BC is the tower.

In Δ ABC,

$$\begin{aligned}\text{sin 30°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{BC}{AC}\\[1em] ⇒ \frac{1}{2} = \frac{15}{AC}\\[1em] ⇒ AC = 15 \times 2\\[1em] ⇒ AC = 30\end{aligned}$$

Hence, length of the ladder = 30 m.

AC is the string and BC is the height of kite.

In Δ ABC,

$$\begin{aligned}\text{sin 60°} = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{BC}{AC}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{BC}{100}\\[1em] ⇒ BC = \frac{\sqrt3 \times 100}{2}\\[1em] ⇒ BC = 50\sqrt3 = 86.60\end{aligned}$$

Hence, the greatest height reached by the kite = 86.60 m.

Let BC be x cm.

BD = BC + CD = (x + 20) cm

In Δ ABD,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{BD}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{x + 20}\\[1em] ⇒ x + 20 = AB\sqrt3 ..............(1)\end{aligned}$$

In Δ ABC,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AB}{BC}\\[1em] ⇒ 1 = \frac{AB}{x}\\[1em] ⇒ x = AB ..............(2)\end{aligned}$$

From equation (1),

AB + 20 = AB $\sqrt3$

⇒ AB $\sqrt3$ - AB = 20

⇒ AB($\sqrt3$ - 1) = 20

⇒ AB(1.73 - 1) = 20

⇒ AB x (0.73) = 20

⇒ AB = $\frac{20}{0.73}$

⇒ AB = 27.32 cm

And BC = x = AB = 27.32 cm

Hence, AB = BC = 27.32 cm.

Let BC be x cm

BD = BC + CD = x + 20 cm

In Δ ABD,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{BD}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{x + 20}\\[1em] ⇒ x + 20 = AB\sqrt3 ..............(1)\end{aligned}$$

In Δ ABC,

$$\begin{aligned}\text{tan 60°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \sqrt3 = \frac{AB}{BC}\\[1em] ⇒ \sqrt3 = \frac{AB}{x}\\[1em] ⇒ x = \frac{AB}{\sqrt3} ..............(2)\end{aligned}$$

From equation (1),

$$\begin{aligned}\frac{AB}{\sqrt3} + 20 = AB\sqrt3\\[1em] ⇒ AB + 20\sqrt3 = 3AB\\[1em] ⇒ 34.64 = 3AB - AB\\[1em] ⇒ AB = \frac{34.64}{2}\\[1em] ⇒ AB = 17.32\end{aligned}$$

From equation (2),

x = $\frac{AB}{\sqrt3}$

x = $\frac{17.32}{\sqrt3}$

x = 10 cm

Hence, AB = 17.32 cm and BC = 10 cm.

Let BC = $x$ cm.

Given,

CD = 20 cm.

∴ BD = BC + CD = $(x + 20)$ cm.

In △ABD,

$$\begin{aligned}\tan 45^\circ = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow 1 = \frac{AB}{BD}\\[1em] \Rightarrow 1 = \frac{AB}{x + 20}\\[1em] \Rightarrow AB = x + 20 \qquad .....(1)\end{aligned}$$

In △ABC,

$$\begin{aligned}\tan 60^\circ = \frac{\text{Perpendicular}}{\text{Base}}\\[1em] \Rightarrow \sqrt{3} = \frac{AB}{BC}\\[1em] \Rightarrow \sqrt{3} = \frac{AB}{x}\\[1em] \Rightarrow x = \frac{AB}{\sqrt{3}} \qquad .....(2)\end{aligned}$$

Substituting the value of $x$ from equation (2) in equation (1), we get:

$$\begin{aligned}AB = \frac{AB}{\sqrt{3}} + 20\\[1em] \Rightarrow AB\sqrt{3} = AB + 20\sqrt{3}\\[1em] \Rightarrow AB\sqrt{3} - AB = 20\sqrt{3}\\[1em] \Rightarrow AB(\sqrt{3} - 1) = 20\sqrt{3}\\[1em] \Rightarrow AB = \frac{20\sqrt{3}}{\sqrt{3} - 1}\\[1em] \Rightarrow AB = \frac{20\sqrt{3}(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\\[1em] \Rightarrow AB = \frac{20\sqrt{3}(\sqrt{3} + 1)}{3 - 1}\\[1em] \Rightarrow AB = 10\sqrt{3}(\sqrt{3} + 1)\\[1em] \Rightarrow AB = 30 + 10\sqrt{3}\\[1em] \Rightarrow AB = 30 + 10 \times 1.73205\\[1em] \Rightarrow AB = 47.3205\\[1em] \Rightarrow AB \approx 47.32 \text{ cm}\end{aligned}$$

Substituting the exact value of AB in equation (2), we get:

$$\begin{aligned}x = \frac{AB}{\sqrt{3}}\\[1em] \Rightarrow x = \frac{30 + 10\sqrt{3}}{\sqrt{3}}\\[1em] \Rightarrow x = \frac{30}{\sqrt{3}} + 10\\[1em] \Rightarrow x = \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} + 10\\[1em] \Rightarrow x = 10\sqrt{3} + 10\\[1em] \Rightarrow x = 10 \times 1.73205 + 10\\[1em] \Rightarrow x = 27.3205\\[1em] \Rightarrow x \approx 27.32 \text{ cm}\end{aligned}$$

∴ BC = 27.32 cm.

Hence, AB = 47.32 cm and BC = 27.32 cm.

In Δ APB,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{PB}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{150}{PB}\\[1em] ⇒ PB = 150\sqrt3 = 259.80\end{aligned}$$

And, in Δ ABQ,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AB}{BQ}\\[1em] ⇒ 1 = \frac{150}{BQ}\\[1em] ⇒ BQ = 150\end{aligned}$$

As, PQ = PB + BQ

= 259.80 + 150 m

= 409.80 m

Hence, PQ = 409.80 m.

In Δ APB,

$$\begin{aligned}\text{tan 30°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{AB}{PB}\\[1em] ⇒ \frac{1}{\sqrt3} = \frac{150}{PB}\\[1em] ⇒ PB = 150\sqrt3 = 259.80\end{aligned}$$

And, in Δ ABQ,

$$\begin{aligned}\text{tan 45°} = \frac{Perpendicular}{Base}\\[1em] ⇒ 1 = \frac{AB}{BQ}\\[1em] ⇒ 1 = \frac{150}{BQ}\\[1em] ⇒ BQ = 150\end{aligned}$$

As, PQ = PB - BQ

= 259.80 - 150 m

= 109.80 m

Hence, PQ = 109.80 m.

Given:

$\text{tan x°} = \frac{5}{12}$, $\text{tan y°} = \frac{3}{4}$ and AB = 48 m

Let BC be x m.

AC = AB + BC = (48 + x) m.

In Δ ADC,

$$\begin{aligned}\text{tan x°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{5}{12} = \frac{DC}{AC}\\[1em] ⇒ \frac{5}{12} = \frac{DC}{48 + x}\\[1em] ⇒ \text{12 CD = 240 + 5x} ................(1)\end{aligned}$$

And, in Δ BDC,

$$\begin{aligned}\text{tan y°} = \frac{Perpendicular}{Base}\\[1em] ⇒ \frac{3}{4} = \frac{DC}{BC}\\[1em] ⇒ \frac{3}{4} = \frac{DC}{x}\\[1em] ⇒ 4 CD = 3x \\[1em] ⇒ x = \frac{4 CD}{3}................(2)\end{aligned}$$

From equation (1),

240 + 5 $\times \frac{4CD}{3}$ = 12 CD

⇒ 240 + $\frac{20CD}{3}$ = 12 CD

⇒ 720 + 20 CD = 36 CD

⇒ 720 = 36 CD - 20 CD

⇒ 16 CD = 720

⇒ CD = $\frac{720}{16}$

⇒ CD = 45

Hence, CD = 45 m.

ABCD is a rhombus.

Perimeter of rhombus = 4 x side

Hence, side AB = BC = CD = DA = $\frac{96}{4}$ = 24 cm

∠ ABC = 120°

As we know that diagonal of a rhombus bisect each other at 90°.

In Δ ABO,

∠ ABO = $\frac{120°}{2}$ = 60°

$$\begin{aligned}sin 60° = \frac{Perpendicular}{Hypotenuse}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AO}{AB}\\[1em] ⇒ \frac{\sqrt3}{2} = \frac{AO}{24}\\[1em] ⇒ AO = \frac{24 \sqrt3}{2}\\[1em] ⇒ AO = 20.78\end{aligned}$$

∴ AC = 2 x AO = 2 x 20.78 = 41.56 cm

Similarly,

$$\begin{aligned}\text{cos 60°} = \frac{Base}{Hypotenuse}\\[1em] ⇒ \frac{1}{2} = \frac{BO}{AB}\\[1em] ⇒ \frac{1}{2} = \frac{BO}{24}\\[1em] ⇒ BO = \frac{24}{2}\\[1em] ⇒ BO = 12\end{aligned}$$

∴ BD = 2 x BO = 2 x 12 = 24 cm

Hence, the lengths of the diagonals are: AC = 41.56 cm and BD = 24 cm.