SELINA Solution Class 9 Chapter 24 Solution of Right Triangles [Simple 2-D Problems Involving One Right -angled Triangle] Exercise 24

Question 1.1

Find 'x', if :

Sol:

From the figure we have

sin 60° = ${\displaystyle \frac{20}{x}}$

${\displaystyle \frac{\sqrt{3}}{2}=\frac{20}{x}}$

x = ${\displaystyle \frac{40}{\sqrt{3}}}$

Question 1.2

Find 'x', if :

Sol:

From the figure we have

tan 30° = ${\displaystyle \frac{20}{x}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{20}{x}}$

x = 20${\displaystyle \sqrt{3}}$

Question 1.3

Find 'x', if :

Sol:

From the figure we have

sin 45° = ${\displaystyle \frac{20}{x}}$

${\displaystyle \frac{1}{\sqrt{2}}=\frac{20}{x}}$

x = 20${\displaystyle \sqrt{2}}$

Question 2.1

Find angle 'A' if :

Sol:

From the figure we have

cos A = ${\displaystyle \frac{10}{20}}$

cos A = ${\displaystyle \frac{1}{2}}$

cos A = cos 60°

A = 60°

Question 2.2

Find angle 'A' if :

Sol:

From the figure we have

sin A = ${\displaystyle \frac{\frac{10}{\sqrt{2}}}{10}}$

sin A =${\displaystyle \frac{1}{\sqrt{2}}}$

sin A = sin 45°

A = 45°

Question 2.3

Find angle 'A' if :

Sol:

From the figure we have

tan A = ${\displaystyle \frac{10\sqrt{3}}{10}}$

tan A = ${\displaystyle \sqrt{3}}$

tan A = sin 60°

A = 60°

Question 3

Find angle 'x' if :

Sol:

The above figure we have

tan 60° = ${\displaystyle \frac{30}{\text{AD}}}$

${\displaystyle \sqrt{3}=\frac{30}{\text{AD}}}$

AD = ${\displaystyle \frac{30}{\sqrt{3}}}$

Again

sin x = ${\displaystyle \frac{\text{AD}}{20}}$
AD = 20 sin x

Now

20 sin x = ${\displaystyle \frac{30}{\sqrt{3}}}$

sin x = ${\displaystyle \frac{30}{20\sqrt{3}}}$

sin x = ${\displaystyle \frac{\sqrt{3}}{2}}$

sin x = sin 60°
x = 60°

Question 4.1

Find AD, if :

Sol:

From the right triangle ABE

tan 45° = ${\displaystyle \frac{\text{AE}}{\text{BE}}}$

1 = ${\displaystyle \frac{\text{AE}}{\text{BE}}}$

AE = BE

Therefore  AE = BE = 50 m.

Now from the rectangle BCDE we have

DE = BC = 10 m.

Therefore the length of AD will be:

AD = AE + DE = 50 + 10 = 60 m.

Question 4.2

Find AD, if :

Sol:

From the triangle ABD we have

sin B = ${\displaystyle \frac{\text{AD}}{\text{AB}}}$

sin 30 = ${\displaystyle \frac{\text{AD}}{100}}$  ...[Since ∠ACD is the exterior angle of the triangle ABC]

${\displaystyle \frac{1}{2}=\frac{\text{AD}}{100}}$
AD = 50 m

Question 5

Find the length of AD.
Given: ∠ABC = 60^o.
∠DBC = 45^o
and BC = 40 cm.

 

Sol:

From right triangle ABC,

tan 60° = ${\displaystyle \frac{\text{AC}}{\text{BC}}}$

⇒ ${\displaystyle \sqrt{3}=\frac{\text{AC}}{40}}$

⇒ AC = ${\displaystyle 40\sqrt{3}\text{cm}}$

From right triangle BDC,

tan 45° = ${\displaystyle \frac{\text{DC}}{\text{BC}}}$

⇒ 1 = ${\displaystyle \frac{\text{DC}}{40}}$

⇒ DC = 40 cm

From the figure, it is clear that AD = AC - DC

⇒ AD = ${\displaystyle 40\sqrt{3}-40}$

⇒ AD = ${\displaystyle 40(\sqrt{3}-1)}$

⇒ AD = 29.28 cm

Question 6

Find the lengths of diagonals AC and BD. Given AB = 60 cm and ∠ BAD = 60°.

Sol:

We know, diagonals of a rhombus bisect each other at right angles and also bisect the angle of the vertex.
The figure is shown below:

Now
OA = OC =${\displaystyle \frac{1}{2}}$AC, OB = OD =${\displaystyle \frac{1}{2}}$BD; ∠AOB = 90°

And ∠OAB = ${\displaystyle \frac{60°}{2}}$ = 30°

Also given AB = 60 cm

In right triangle AOB

sin 30° = ${\displaystyle \frac{\text{OB}}{\text{AB}}}$

${\displaystyle \frac{1}{2}=\frac{\text{OB}}{60}}$
OB = 30 cm

Also

cos 30° = ${\displaystyle \frac{\text{OA}}{\text{AB}}}$

${\displaystyle \frac{\sqrt{3}}{2}=\frac{\text{OA}}{60}}$

OA = 51.96 cm

Therefore,

Length of diagonal AC = 2 x OA = 2 x 51.96 = 103.92 cm.

Length of diagonal BD = 2 x OB = 2 x 30 = 60 cm.

Question 7

Find AB.

Sol:

Consider the figure

From right triangle ACF

tan 45° = ${\displaystyle \frac{20}{\text{AC}}}$

1 = ${\displaystyle \frac{20}{\text{AC}}}$

AC = 20 cm

From triangle DEB

tan 60° = ${\displaystyle \frac{30}{\text{BD}}}$

${\displaystyle \sqrt{3}=\frac{30}{\text{BD}}}$

BD = ${\displaystyle \frac{30}{\sqrt{3}}}$ = 17.32 cm

Given FC = 20, ED = 30, So EP = 10 cm
Therefore

tan 60° = ${\displaystyle \frac{\text{FP}}{\text{EP}}}$

${\displaystyle \sqrt{3}=\frac{\text{FP}}{10}}$

FP = ${\displaystyle 10\sqrt{3}}$ = 17.32 cm

Thus AB = AC + CD + BD = 54.64 cm.

Question 8.1

In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and ∠ A = 60°. Find: length of AB

Sol:

First, draw two perpendiculars to AB from point D and C respectively. Since AB || CD therefore PMCD will be a rectangle.
Consider the figure,


From right triangle ADP we have

cos 60° = ${\displaystyle \frac{\text{AP}}{\text{AD}}}$

${\displaystyle \frac{1}{2}=\frac{\text{AP}}{20}}$

AP = 10

Similarly from the right triangle BMC we have BM = 10 cm.

Now from the rectangle PMCD we have CD = PM = 20 cm.

Therefore,
AB = AP + PM + MB = 10 + 20 + 10 = 40 cm.

Question 8.2

In trapezium ABCD, as shown, AB // DC, AD = DC = BC = 20 cm and A = 60^°. Find: distance between AB and DC.

Sol:

First, draw two perpendiculars to AB from point D and C respectively. Since AB || CD therefore PMCD will be a rectangle.
Consider the figure,

Again from the right triangle APD we have

sin 60° = ${\displaystyle \frac{\text{PD}}{20}}$

${\displaystyle \frac{\sqrt{3}}{2}=\frac{\text{PD}}{20}}$

PD = ${\displaystyle 10\sqrt{3}}$

Therefore the distance between AB and CD is ${\displaystyle 10\sqrt{3}}$.

Question 9

Use the information given to find the length of AB.

Sol:

From right triangle AQP

tan 30° = ${\displaystyle \frac{\text{AQ}}{\text{AP}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{10}{\text{AP}}}$

AP = ${\displaystyle 10\sqrt{3}}$

Also from triangle PBR

tan 45° = ${\displaystyle \frac{\text{PB}}{\text{BR}}}$

${\displaystyle 1=\frac{\text{PB}}{8}}$

Therefore,

AB = AP + PB = ${\displaystyle 10\sqrt{3}+8}$.

Question 10

Find the length of AB.

Sol:

From right triangle = ADE

tan 45° = ${\displaystyle \frac{\text{AE}}{\text{DE}}}$

1 = ${\displaystyle \frac{\text{AE}}{30}}$

AE = 30 cm

Also, from triangle DBE

tan 60° = ${\displaystyle \frac{\text{BE}}{\text{DE}}}$

${\displaystyle \sqrt{3}=\frac{\text{BE}}{30}}$

BE = ${\displaystyle 30\sqrt{3} \text{cm}}$

Therefore AB = AE + BE = 30 + ${\displaystyle 30\sqrt{3}=30(1+\sqrt{3})\text{cm}}$

Question 11.1

In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.

Given that ∠ AED = 60° and ∠ ACD = 45°. Calculate: AB.

Sol:

From the triangle ADC we have

tan 45° = ${\displaystyle \frac{\text{AD}}{\text{DC}}}$

1 = ${\displaystyle \frac{2}{\text{DC}}}$

DC = 2

Since AD || DC and AD⊥EC, ABCD is a parallelogram and hence opposite sides are equal.

Therefore AB = DC = 2 cm.

Question 11.2

In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.

Given that ∠AED = 60° and ∠ACD = 45°. Calculate: AC

Sol:

Again

sin 45° = ${\displaystyle \frac{\text{AD}}{\text{AC}}}$

${\displaystyle \frac{1}{\sqrt{2}}=\frac{2}{\text{AC}}}$

AC = ${\displaystyle 2\sqrt{2}}$.

Question 11.3

In the given figure, AB and EC are parallel to each other. Sides AD and BC are 2 cm each and are perpendicular to AB.

Given that ∠ AED = 60° and ∠ ACD = 45°. Calculate: AE.

Sol:

From the right triangle ADE we have

sin 60° = ${\displaystyle \frac{\text{AD}}{\text{AE}}}$

${\displaystyle \frac{\sqrt{3}}{2}=\frac{2}{\text{AE}}}$

AE = ${\displaystyle \frac{4}{\sqrt{3}}}$.

Question 12

In the given figure, ∠B = 60° , AB = 16 cm and BC = 23 cm,
Calculate: (i) BE, (ii) AC

Sol:

In ΔABE,

sin 60° = ${\displaystyle \frac{\text{AE}}{\text{AB}}}$

⇒ ${\displaystyle \frac{\sqrt{3}}{2}=\frac{\text{AE}}{16}}$

⇒  AE = ${\displaystyle \frac{\sqrt{3}}{2}\times 16}$

= ${\displaystyle 8\sqrt{3} \text{cm}}$

(i) In ∆ABE,

m∠AEB = 90°

∴ By Pythagoras Theorem, we get
BE² = AB² - AE²
⇒ BE² = (16)² - (8√3)²
⇒ BE² = 256 - 192
⇒ BE² = 64
⇒ BE = 8 cm

(ii) EC = BC - BE = 23 - 8 = 15
In ∆AEC, 
m∠AEC = 90°

∴ By Pythagoras Theorem, we get
AC² = AE²+ EC²
⇒ AC² = (8√3)²  + (15)²
⇒ AC² = 192 + 225
⇒ AC² = 417
⇒ AC = 20. 42 cm

Question 13.1

Find: BC

Sol:

From right triangle ABC

tan 30° = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{12}{\text{BC}}}$

BC = ${\displaystyle 12\sqrt{3}\text{cm}}$.

Question 13.2

Find: AD

Sol:

From the right triangle ABD

cos A = ${\displaystyle \frac{\text{AD}}{\text{AB}}}$

cos 60° = ${\displaystyle \frac{\text{AD}}{\text{AB}}}$

${\displaystyle \frac{1}{2}=\frac{\text{AD}}{12}}$

AD = ${\displaystyle \frac{12}{2}}$

 = 6 cm

Question 13.3

Find: AC

Sol:

From right triangle ABC

sin B = ${\displaystyle \frac{\text{AB}}{\text{AC}}}$

sin 30° = ${\displaystyle \frac{\text{AB}}{\text{AC}}}$

${\displaystyle \frac{1}{2}=\frac{12}{\text{AC}}}$

AC = 24 cm

Question 14.1

In right-angled triangle ABC; ∠ B = 90°. Find the magnitude of angle A, if: AB is √3 times of BC.

Sol:

Consider the figure

 Here AB is ${\displaystyle \sqrt{3}}$ times of BC means

${\displaystyle \frac{\text{AB}}{\text{BC}}=\sqrt{3}}$

cot θ = cot 30°

θ = 30°

Question 14.2

In right-angled triangle ABC; ∠ B = 90°. Find the magnitude of angle A, if:
BC is ${\displaystyle \sqrt{3}}$ times of AB.

Sol:

Consider the figure

Again from the figure

${\displaystyle \frac{\text{BC}}{\text{AB}}=\sqrt{3}}$

tan θ = ${\displaystyle \sqrt{3}}$

tan θ = tan 60°

θ = 60°

Therefore, magnitude of angle A is 30°

Question 15

A ladder is placed against a vertical tower. If the ladder makes an angle of 30° with the ground and reaches upto a height of 15 m of the tower; find length of the ladder.

Sol:

Given that the ladder makes an angle of 30° with the ground and reaches up to a height of 15 m of the tower which is shown in the figure below:

Suppose the length of the ladder is x m

From the figure

${\displaystyle \frac{15}{x}}$ = sin 30°    ...${\displaystyle [\because \frac{\text{Perp.}}{\text{Hypot.}}=\text{sin}]}$

${\displaystyle \frac{15}{x}=\frac{1}{2}}$

x = 30 m

Therefore the length of the ladder is 30 m.

Question 16

A kite is attached to a 100 m long string. Find the greatest height reached by the kite when its string makes an angles of 60° with the level ground.

Sol:

Given that the kite is attached to a 100 m long string and it makes an angle of  60° with the ground level which is shown in the figure below:

Suppose that the greatest height is x m.

From the figure

${\displaystyle \frac{x}{100}}$ = sin 60°     ...${\displaystyle [\because \frac{\text{Perp.}}{\text{Hypot.}}=\text{sin}]}$

${\displaystyle \frac{x}{100}=\frac{\sqrt{3}}{2}}$

x = 86.6 m

Therefore the greatest height reached by the kite is 86.6 m.

Question 17.1

Find AB and BC, if:

Sol:

Let BC = x m

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

In ΔABD,

tan 30° = ${\displaystyle \frac{\text{AB}}{\text{BD}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{\text{AB}}{x+20}}$

x + 20 = ${\displaystyle \sqrt{3}\text{AB}}$           .....(1)

In ΔABC

tan 45° = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

1 = ${\displaystyle \frac{\text{AB}}{x}}$

AB = x         ...(2)

From (1)

AB + 20 = ${\displaystyle \sqrt{3}\text{AB}}$

AB${\displaystyle (\sqrt{3}-1)=20}$

AB = ${\displaystyle \frac{20}{(\sqrt{3}-1)}}$

= ${\displaystyle \frac{20}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}}$

= ${\displaystyle \frac{20(\sqrt{3}+1)}{3-1}}$

= 27.32cm

From (2)

AB = x = 27.32 cm

Therefore BC = x = AB = 27.32 cm

Therefore, AB = 27.32 cm, BC = 27.32 cm

Question 17.2

Find AB and BC, if:

Sol:

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

In ΔABD,
tan 30° = ${\displaystyle \frac{\text{AB}}{\text{BD}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{\text{AB}}{x+20}}$

x + 20 = ${\displaystyle \sqrt{3}\text{AB}}$      ...(1)

In ΔABC
tan 60° = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

${\displaystyle \sqrt{3}=\frac{\text{AB}}{x}}$
 x = ${\displaystyle \frac{\text{AB}}{\sqrt{3}}}$         ...(2)
From (1)

${\displaystyle \frac{\text{AB}}{\sqrt{3}}+20=\sqrt{3}\text{AB}}$

AB + ${\displaystyle 20\sqrt{3}}$ = 3AB

2AB = ${\displaystyle 20\sqrt{3}}$

2AB = ${\displaystyle \frac{20\sqrt{3}}{2}}$

AB = ${\displaystyle 10\sqrt{3}}$

AB = 17.32 cm

From (2)

x = ${\displaystyle \frac{\text{AB}}{\sqrt{3}}}$

x = ${\displaystyle \frac{17.32}{\sqrt{3}}}$

x = 10 cm

Therefore BC = x = 10 cm
Therefore, AB = 17.32 cm, BC = 10 cm.

Question 17.3

Find AB and BC, if:

Sol:

Let  BC = x m

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

In ΔABD,
tan 45° = ${\displaystyle \frac{\text{AB}}{\text{BD}}}$

1 = ${\displaystyle \frac{\text{AB}}{x+20}}$

x + 20 = AB      ...(1)

In ΔABC

tan 60° = ${\displaystyle \frac{\text{AB}}{\text{BC}}}$

${\displaystyle \sqrt{3}=\frac{\text{AB}}{x}}$

x = ${\displaystyle \frac{\text{AB}}{\sqrt{3}}}$      ...(2)

From (1)

${\displaystyle \frac{\text{AB}}{\sqrt{3}}+20=\text{AB}}$

AB + ${\displaystyle 20\sqrt{3}=\sqrt{3}\text{AB}}$

AB${\displaystyle (\sqrt{3}-1)=20\sqrt{3}}$

AB = ${\displaystyle \frac{20\sqrt{3}}{(\sqrt{3}-1)}}$

AB = ${\displaystyle \frac{20\sqrt{3}}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}}$

AB = ${\displaystyle \frac{20\sqrt{3}(\sqrt{3}+1)}{3-1}}$

AB = 47.32 cm

From (2)

x = ${\displaystyle \frac{\text{AB}}{\sqrt{3}}}$

x = ${\displaystyle \frac{47.32}{\sqrt{3}}}$

x = 27.32 cm

∴ BC = x = 27.32 cm

Therefore, AB = 47.32 cm, BC = 27.32 cm.

Question 18.1

Find PQ, if AB = 150 m, ∠ P = 30° and ∠ Q = 45°.

Sol:

From ΔAPB

tan 30° = ${\displaystyle \frac{\text{AB}}{\text{PB}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{150}{\text{PB}}}$

PB = ${\displaystyle 150\sqrt{3}}$

PB = 259.80 m

Also, from ΔABQ

tan 45° = ${\displaystyle \frac{\text{AB}}{\text{BQ}}}$

1 = ${\displaystyle \frac{150}{\text{BQ}}}$

BQ = 150 m

Therefore,

PQ = PB + BQ

PQ = 259.80 + 150

PQ = 409.80 m

Question 18.2

Find PQ, if AB = 150 m, ∠P = 30° and ∠Q = 45°.

.

Sol:

From ΔAPB

tan 30° = ${\displaystyle \frac{\text{AB}}{\text{PB}}}$

${\displaystyle \frac{1}{\sqrt{3}}=\frac{150}{\text{PB}}}$

PB = ${\displaystyle 150\sqrt{3}}$

PB = 259.80 m

Also, from ΔABQ

tan 45° = ${\displaystyle \frac{\text{AB}}{\text{BQ}}}$

1 = ${\displaystyle \frac{150}{\text{BQ}}}$

BQ = 150 m

Therefore,

PQ = PB - BQ

PQ = 259.80 - 150

PQ = 109.80 m

Question 19

If tan x° = ${\displaystyle \frac{5}{12}}$,

tan y° = ${\displaystyle \frac{3}{4}}$ and AB = 48 m; find the length of CD.

SoL:

Given tan x° = ${\displaystyle \frac{5}{12}}$ tan t° = ${\displaystyle \frac{3}{4}}$ and AB = 48 m;

Let length of  BC = x m

From ΔADC

tan x°  = ${\displaystyle \frac{\text{DC}}{\text{AC}}}$

${\displaystyle \frac{5}{12}=\frac{\text{DC}}{48+x}}$

240 + 5x = 12 CD     ...(1)

Also, from ΔBDC

tan y° = ${\displaystyle \frac{\text{CD}}{\text{BC}}}$

${\displaystyle \frac{3}{4}=\frac{\text{CD}}{x}}$

x = ${\displaystyle \frac{4\text{CD}}{3}}$       ...(2)

From (1)

240 + 5 ${\displaystyle \left(\frac{4\text{CD}}{3}\right)}$ = 12CD

240 + ${\displaystyle \frac{20\text{CD}}{3}}$ = 12CD

720 + 20CD = 36CD

16CD = 720

CD = 45

Therefore, length of CD is 45 m.

Question 20

The perimeter of a rhombus is 96 cm and obtuse angle of it is 120°. Find the lengths of its diagonals.

Sol:

Since in a rhombus all sides are equal.

The diagram is shown below:

Therefore PQ = ${\displaystyle \frac{96}{4}}$ = 24 cm, Let ∠ PQR = 120°.

We also know that in rhombus diagonals bisect each other perpendicularly and diagonals bisect the angle at vertex.

Hence POR is a right angle triangle and

POR = ${\displaystyle \frac{1}{2}\left(\text{PQR}\right)}$ = 60°

sin 60° = ${\displaystyle \frac{\text{Perp.}}{\text{Hypot.}}=\frac{\text{PO}}{\text{PQ}}=\frac{\text{PO}}{24}}$
But
sin 60° = ${\displaystyle \frac{\sqrt{3}}{2}}$

${\displaystyle \frac{\text{PO}}{24}=\frac{\sqrt{3}}{2}}$

PO = ${\displaystyle 12\sqrt{3}}$ = 20.784

Therefore,

PR = 2PO
= 2 x 20.784
= 41.568 cm

Also,
cos 60° = ${\displaystyle \frac{\text{Base}}{\text{Hypot}}=\frac{\text{OQ}}{24}}$
But
cos 60° = ${\displaystyle \frac{1}{2}}$

${\displaystyle \frac{\text{OQ}}{24}=\frac{1}{2}}$

OQ = 12

Therefore, SQ = 2 x OQ

= 2 x 12

= 24 cm
So, the length of the diagonal PR = 41.568 cm and SQ = 24 cm.