Heights and Distances
One of the main uses of trigonometry is to find the height of an object or the distance between two points. The instruments called theodolite and sextant are used to measure certain angles and then method of solution of triangles is used to find the required height or distance.
Angle of elevation
In the adjoining figure, O is the position of the observer, P is the position of the object and OP is the line of vision. OA is the horizontally through the observation point O. the angle AOP formed when the observer observes the object at P above horizontal line is called an angle of elevation.
The angle of depression
In the adjoining figure, O is the position of the observer, is the position of the object and P is the line of vision.
OP is the horizontal line through the observation point. The angle AOP formed when the observer observes the object at P below the horizontal line is called angle of depression.
Here,
Height of the tower (PQ) = ?
Distance between R and S (RS) = 30m
Angle of elevation (\(\angle\) PRQ) = 45°
Angle of elevation (\(\angle\) PSQ) = 30°
We have,
In right angled triangle \(\triangle\) PQR:
tan 45° = \(\frac {PQ}{PR}\)
1 = \(\frac {PQ}{PR}\)
∴ PR = PQ
In right angled triangle \(\triangle\) PQS:
tan 30° = \(\frac {PQ}{PS}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {PQ}{30 + PR}\) [\(\because\) PQ = PR]
or, 30 + PQ = \(\sqrt 3\) PQ
or, \(\sqrt 3\) PQ  PQ = 30
or, PQ (\(\sqrt 3\)  1) = 30
or, PQ = \(\frac {30}{\sqrt 3  1}\)
or, PQ = \(\frac {30}{1.732  1}\)
or, PQ = \(\frac {30}{0.732}\)
∴ PQ = 40.98 m
∴ The height of tower (PQ) = 40.98 m _{Ans}
Let: AB be the tower and C be initial point of observation.
∴ \(\angle\) ACB = 60°
D is the point 200 m away from C.∴ \(\angle\) ADB = 30°
Height of the tower (AB) = ?
In right angled triangle ABC,
tan 60° = \(\frac {AB}{BC}\)
or, \(\sqrt 3\) = \(\frac {AB}{BC}\)
∴ BC = \(\frac {AB}{\sqrt 3}\) .............................(i)
In right angled triangle ABD,
tan 30° = \(\frac {AB}{BC}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{BC + CD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{\frac {AB}{\sqrt 3} + 200}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{\frac {AB + 200\sqrt 3}{\sqrt 3}}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {\sqrt 3 AB}{AB + 200\sqrt 3}\)
or, AB + 200 \(\sqrt 3\) = 3 AB
or, 3 AB  AB = 200\(\sqrt 3\)
or, 2 AB = 200\(\sqrt 3\)
or, AB = \(\frac {200\sqrt 3}{2}\)
or, AB = 100× 1.732
∴ AB = 173.2 m
Hence, the height of tower = 173.2 m _{Ans}
Let:Height of tower (PS) = 30 m
Position of two men Q and R.
Angle of elevation \(\angle\)PQS = 60° and \(\angle\)PRS = 30°
Distance between two men (QR) = ?
In right angled triangle PQS,
tan 60° = \(\frac {PS}{QS}\)
or, \(\sqrt 3\) = \(\frac {30}{QS}\)
or, QS = \(\frac {30}{\sqrt 3}\)
or, QS = \(\frac {10.\sqrt 3.\sqrt 3}{\sqrt 3}\)
∴ QS = 10\(\sqrt 3\) m
In right angled triangle PRS,
tan 30° = \(\frac {PS}{RS}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{RS}\)
∴ RS = 30\(\sqrt 3\)
QR
= QS + SR
= 10\(\sqrt 3\) + 30\(\sqrt 3\)
= 40\(\sqrt 3\)
= 40× 1.732
= 69. 28 m _{Ans}
Let: AB be the height of the tower. The angle of elevation observed from a point C to the top of the tower B is \(\angle\)ACB = 60° and from the D is \(\angle\)ADB = 45°.
Distance of CD = 60 m
In the right angled triangle ABC,
tan 60° = \(\frac {AB}{AC}\)
or, \(\sqrt 3\) = \(\frac {AB}{AC}\)
∴ AB = \(\sqrt 3\) AC.........................(i)
In right angled triangle ABD,
tan 45° = \(\frac {AB}{AD}\)
or, 1 = \(\frac {AB}{AD}\)
∴ AB = AD...............................................(ii)
From equation (i) and (ii)
AD = \(\sqrt 3\) AC
or, AC + CD = \(\sqrt 3\) AC
or, \(\sqrt 3\)AC  AC = CD
or, AC (\(\sqrt 3\)  1) = 60
or, AC (1.732  1) = 60
or, AC = \(\frac {60}{0.732}\)
∴ AC = 81.97 m
AD = AC + CD = 81.97 + 60 = 141.97 m
AB = AD = 141.97 m
∴ The height of tower = 141.97 m _{Ans}
Let:
The height of the pole = AB
The height of the flagstaff = BC = ?
The angle of elevation bottom and top of the flagstaff are 30° and 45°.
The distance between the pole and observation point = 30 m
In right angled triangle ACD,
tan 45° = \(\frac {AC}{AD}\)
or, 1 = \(\frac {AC}{30}\)
∴ AC = 30 m
In right angled triangle ABD,
tan 30° = \(\frac {AB}{AD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {AB}{30}\)
or, AB \(\sqrt 3\) = 30
or, AB = \(\frac {30}{\sqrt 3}\)
or, AB = \(\frac {30}{1.732}\)
∴ AB = 17.32 m
Hence, the height of flagstaff BC = AC  AB = (30  17.32) = 12.68 m _{Ans}
Let: AB and CD be two lamp posts of equal heights. E be the point of observation.
BD = 80 m
\(\angle\)AEB = 60°
\(\angle\)CED = 30°
BE = ?
AB = CD = ?
In right angled triangle ABE,
tan 60° = \(\frac {AB}{BE}\)
or, \(\sqrt 3\) = \(\frac {AB}y\)
or, AB = y\(\sqrt 3\) .........................(i)
In right angled triangle CDE,
tan 30° = \(\frac {CD}{DE}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {CD}{80  y}\)
or, CD = \(\frac {80  y}{\sqrt 3}\)..........................(ii)
We have,
AB = CD
y\(\sqrt 3\) = \(\frac {80  y}{\sqrt 3}\)
or, 3y = 80  y
or, 3y + y = 80
or, 4y = 80
or, y = \(\frac {80}4\)
∴ y = 20
Putting the value of y in eq^{n} (i)
AB = CD = \(\sqrt 3\)y = 20\(\sqrt 3\) = 20× 1.732 = 34.65 m
The point is 20 m from the second post and height of posts each is 34.65 m. _{Ans}
Let:
Height of the building (AB) = 30 m
C and D be the positions of two men standing due to east from the building. The angles depression are \(\angle\)MBC = 45° and \(\angle\)MBD = 30°.
The distance between the two persons (CD) = ?
\(\angle\)MBD = \(\angle\)ADB = 30°
\(\angle\)MBC = \(\angle\)ACB = 45°
In the right angled triangle ABC,
tan 45° = \(\frac {AB}{AC}\)
or, 1 = \(\frac {30}{AC}\)
∴ AC = 30 m
In right angled triangle ABD,
tan 30° = \(\frac {AB}{AD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{AC + CD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {30}{30 + CD}\)
or, 30 + CD = 30\(\sqrt 3\)
or, CD = 30× 1.732  30
or, CD = 51.96  30
∴ CD = 21.96 m
Hence, the distance between two men = 21.96 m _{Ans}
Let: AB be the tower and C and D be two points on the ground such that BC = a meters BD = b meters and \(\angle\)ACB = \(\theta\), then:
\(\angle\) ADB = 90  \(\theta\)
In right angled triangle ABC,
tan \(\theta\) = \(\frac {AB}{BC}\)
or, tan\(\theta\) = \(\frac {AB}a\)............(i)
In right angled triangle ABD,
tan(90  \(\theta\)) = \(\frac {AB}{BD}\)
or, cot\(\theta\) = \(\frac {AB}{BD}\)
or, \(\frac 1{tan\theta}\) = \(\frac {AB}b\)
or, \(\frac 1{\frac {AB}a}\) = \(\frac {AB}b\) [\(\because\) From equation (i)]
or, \(\frac a{AB}\) = \(\frac {AB}b\)
or, (AB)^{2} = ab
∴ AB = \(\sqrt {ab}\)
Hence, the height of the tower is \(\sqrt {ab}\) m. _{Proved}
Suppose: AB be tree which is broken at the point C and the upper part makes an angle 30° with ground 9 m apart from the foot of the tree.
where,
Ac = CD
BD = 9 m
AB = ?
In right angled triangle CBD,
tan 30° = \(\frac {CB}{BD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {CB}9\)
or, \(\sqrt 3\) CB = 9
or, BC = \(\frac 9{\sqrt 3}\)
or, BC = \(\frac 9{\sqrt 3}\)× \(\frac {\sqrt 3}{\sqrt 3}\)
or, BC = 3\(\sqrt 3\) m
Again,
In right angled triangle,
sin 30° = \(\frac {BC}{CD}\)
or, \(\frac 12\) = \(\frac {3\sqrt 3}{CD}\)
or, CD = 6\(\sqrt 3\) m
Now,
The height of the tree:
= AB
= AC + BC
= CD + BC
= 6\(\sqrt 3\) + 3\(\sqrt 3\)
= 9\(\sqrt 3\) m _{Ans}
Here,
Height of the cliff (AB) = 21 m
Height of the tower (CD) = ?
Angle of depression (\(\angle\)MBD) = 45° = \(\angle\)BDE
Angle of depression (\(\angle\)MBC) = 60° = \(\angle\)BCA
In right angled triangle BDE,
tan 45° = \(\frac {BE}{DE}\)
or, 1 = \(\frac {BE}{DE}\)
or, BE = DE............................(i)
In right angled triangle ABC
tan 60° = \(\frac {AB}{AC}\)
or, \(\sqrt 3\) = \(\frac {21}{AC}\)
or, AC = \(\frac {21}{\sqrt 3}\)
∴ AC = 12.12 m
BE = AC = DE = 12.12
AE = CD = AB  BE = 21  12.12 = 8.88 m
∴ The height of tower = 8.9 m _{Ans}
Let:
Height of pole (AB) = 25 m
Height of tower = CE
Angle of elevation \(\angle\)EAD = 45°
Angle of depression \(\angle\)MEB = \(\angle\)CBE = 60°
AB = CD
AD = BC
In right angled \(\triangle\)ADE,
tan 45° = \(\frac {DE}{AD}\)
or, 1 = \(\frac {DE}{AD}\)
or, DE = AD = BC.........................................(i)
In right angled \(\triangle\)BCE,
taan 60° = \(\frac {CE}{BC}\)
or, \(\sqrt 3\) = \(\frac {DE + CD}{BC}\)
or, \(\sqrt 3\)DE = DE + 25
or, \(\sqrt 3\)DE  DE = 25
or, DE (\(\sqrt 3\)  1) = 25
or, DE (1.732  1) = 25
or, DE = \(\frac {25}{0.732}\)
∴ DE = 31.15 m
CE = CD + DE = 25 + 31.15 = 59.15 m
∴ The height of the tower = 59.15 m _{Ans}
Let:
EB be the height of man = 2 m
\(\angle\)ABC and \(\angle\)DBC are the angle of elevation.
\(\angle\)ABC = 45°
\(\angle\)DBC = 30°
DF is the height of window above the ground.
DF = 12 m
CF = BE = 2 m
CD = DF  CF = 12  2 = 10 m
In the right angled triangle BCD,
tan 30° = \(\frac {CD}{BC}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {10}{BC}\)
or, BC = 10\(\sqrt 3\) m
Again,
In right angled triangle ABC,
tan 45° = \(\frac {AC}{BC}\)
or, 1 = \(\frac {AD + CD}{10\sqrt 3}\)
or, 10\(\sqrt 3\) = AD + 10
or, AD = 10\(\sqrt 3\)  10
or, AD = 10× 1.732  10
or, AD = 17.32  10
∴ AD = 7.32 m
Hence, the height of the house (AF) = AD + DF = 7.32 + 12 = 19.32 m _{Ans}
Let:
Height of man (MN) = 1.68 m
Height of the house (PR) = ?
Height of the window (QR) = 11.68 m
Angle of elevation: \(\angle\)QMS = 30° and \(\angle\)PMS = 45°
From figure,
MN = SR = 1.68 m
QS = 11.68  1.68 = 10 m
In the right angled \(\triangle\)QMS
tan 30° = \(\frac {QS}{MS}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {10}{MS}\)
or, MS = 10\(\sqrt 3\) m
Again,
In right angled triangle PMS,
tan 45° = \(\frac {PS}{MS}\)
or, 1 = \(\frac {PS}{10\sqrt 3}\)
or, PS = 10\(\sqrt 3\)
or, PS = 10× 1.732
∴ PS = 17.32 m
Hence, the height of house (PR) = PS + SR = 17.32 + 1.68 = 19 m _{Ans}
From the figure,
Height of CD = ?
AF = BD
AB = DF = 10 m
CE = 30 m
In the right angled triangle AFE,
tan 30° = \(\frac {EF}{AF}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {EF}{AF}\)
or, AF = \(\sqrt 3\)EF....................................(i)
In the right angledtriangle ACF,
\(\angle\)CAF = 30° + 15° = 45°
tan 45° = \(\frac {CE}{AF}\)
or, 1 = \(\frac {CE + EF}{AF}\)
or, AF = CE + EF..........................(ii)
From equation (i) and (ii)
\(\sqrt 3\)FE = CE + EF
or, \(\sqrt 3\)EF  EF = 30
or, EF (\(\sqrt 3\)  1) = 30
or, EF (1.732  1) = 30
or, EF = \(\frac {30}{0.732}\)
∴ EF = 40.98 m
CD = CE + EF + DF = 30 + 40.98 + 10 = 80.98 m
Hence, the height of CD is 80.98 m. _{Ans}
Let:
Height of the cliff (AB) = 42 m
Height of the tower (CD) = ?
\(\angle\)OAC = \(\angle\)ACE = 30°
\(\angle\)OAD = \(\angle\)ADB = 45°
In the right angled triangle ABD,
tan 45° = \(\frac {AB}{BD}\)
or, 1 = \(\frac {AB}{BD}\)
∴ BD = AB...................................(1)
In the right angled triangle AEC,
tan 30° = \(\frac {AE}{CE}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {AB  BE}{BD}\) = \(\frac {AB  CD}{AB}\)
or, \(\frac {42  CCD}{42}\) = \(\frac 1{\sqrt 3}\)
or, 42  CD = \(\frac {42}{\sqrt 3}\)
or, CD = 42  \(\frac {42}{\sqrt 3}\)
or, CD = 42  24.25
∴ CD = 17.75 m
Hence, the height of the tower is 17.75 m. _{Ans}
Let:
Height of pole (AB) = 25 m
Height of tower (CE) = ?
Angle of depression (\(\angle\)MEA) = 60°
\(\angle\)DAE = \(\angle\)MEA = 60°
Angle of elevation (\(\angle\)ACB) = 30°
In right angled \(\triangle\)ABC,
tan 30° = \(\frac {AB}{BC}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {25}{BC}\)
∴ BC = 25\(\sqrt 3\)
Again,
AD = BC = 25\(\sqrt 3\)
In right angled \(\triangle\)ADE,
tan 60° = \(\frac {DE}{AD}\)
or, \(\sqrt 3\) = \(\frac {DE}{25\sqrt 3}\)
or, DE = 25\(\sqrt 3\)× \(\sqrt 3\)
∴ DE = 75 m
∴ The height of the tower = 100 m _{Ans}
Let Ab and CD be the height of the pillar and tower respectively.
Let AC be the distance between them.
From B, draw BE//FD//AC.
Ab = 60 m
\(\angle\)DBE = 60°
\(\angle\)BCA = 22°
In the right angled\(\triangle\)ABC,
tan 22° = \(\frac {AB}{BC}\)
or, 0.4 = \(\frac {60}{AC}\)
or, AC = \(\frac {60}{0.4}\)
∴ AC = 150 m
Also,
BE = AC = 150 m
In right angled \(\triangle\)BDE,
tan 60° = \(\frac {DE}{BE}\)
or, \(\sqrt 3\) = \(\frac {DE}{150}\)
∴ DE = 150\(\sqrt 3\) m
∴ Height of tower = CD = CE + DE = AB + DE = 60 + 150\(\sqrt 3\) = 319.81 m _{Ans}
Let:
AB be the tower 20 m height and CD be another tower.
\(\angle\)ADB = 30°
\(\angle\)XCA = \(\angle\)CAE = 60°
AB = ED = 20 m
CD =?
In right angled triangle ABD,
tan 30° = \(\frac {AB}{BD}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {20}{BD}\)
or, BD = 20\(\sqrt 3\) m
∴ BD = AE = 20\(\sqrt 3\) m
In right angled triangle CEA,
tan 60° = \(\frac {CE}{AE}\)
or, \(\sqrt 3\) = \(\frac {CE}{20\sqrt 3}\)
or, 20\(\sqrt 3\)× \(\sqrt 3\) = CE
∴ CE = 60 m
∴ The height of the second tower = CD = CE + ED = 60 + 20 = 80 m _{Ans}
Let:
AB be the tower and C and D are two points on the road where vehicles are stand. \(\angle\)ACB = x° and \(\angle\)ADB = y°.
Now,
In right angled triangle ABC,
tan x° = \(\frac {AB}{BC}\)
or, \(\frac 34\) = \(\frac {96}{BC}\)
or, 3 BC = 96× 4
or, BC = \(\frac {96 × 4}3\)
∴ BC = 128 m
In right angled triangle ABD,
tan y° = \(\frac {AB}{BD}\)
or, \(\frac 13\) = \(\frac {96}{BD}\)
or, BD = 3× 96
∴ BD = 288 m
∴ The distance between two vehicles = CD = BD  BC = 288  128 = 160 m _{Ans}
Let:
Height of the pillar (AB) = ?
The foot of the pillar and observation point (AC) = 200 m
The length between two observation point (CD) = 125 m
The angle of elevation from point C = \(\theta\) and the angle of elevation from point D = 2\(\theta\), \(\angle\)ACB = \(\theta\) and \(\angle\)ADB = 2\(\theta\), AD = 200  125 = 75 m
In right angled \(\triangle\)ABC,
tan\(\theta\) = \(\frac {AB}{AC}\)
or, tan \(\theta\) = \(\frac {AB}{200}\).....................................(i)
In right angled\(\triangle\)ABD,
tan 2\(\theta\) = \(\frac {AB}{AD}\)
or, \(\frac {2 tan\theta}{1  tan^2\theta}\) = \(\frac {AB}{75}\)..................................(ii)
Putting the value of tan\(\theta\) from equation (i)
or, \(\frac {2 × \frac {AB}{200}}{1  (\frac {AB}{200})^2}\) = \(\frac {AB}{75}\)
or, \(\frac {AB}{100}\)× 75 = AB (1  \(\frac {{AB}^2}{40000}\))
or, \(\frac {3AB}4\) = AB (\(\frac {40000  {AB}^2}{40000}\))
or, \(\frac {3 × 40000}4\) = 40000  AB^{2}
or, AB^{2} = 40000  30000
or, AB^{2} = 10000
∴ AB = 100 m
∴ The height of the pillar = 100 m _{Ans}
Let:Height of the pole ACbe derived into two parts BC and AB from the base. If D be the point at which they subtend equal angles.
Let:
AB = 9x
BC = x
∴ AC = 10x
\(\angle\)ADB = \(\angle\)BDC = \(\theta\)
CD = 20 m
Height of the pole (AC) = ?
In right angled\(\triangle\)BCD,
tan\(\theta\) = \(\frac {BC}{CD}\)
or, tan\(\theta\) = \(\frac x{20}\).......................(i)
In right angled \(\triangle\)ACD,
tan (\(\theta\) + \(\theta\)) = \(\frac {AC}{CD}\)
or, tan 2\(\theta\) = \(\frac {10x}{20}\)
or, \(\frac {2 tan\theta}{1  tan^2\theta}\) = \(\frac x2\)..........................(ii)
Putting the value of tan\(\theta\) on eq^{n} (ii),
\(\frac {2 (\frac x{20})}{1  (\frac x{20})^2}\) = \(\frac x2\)
or, \(\frac {\frac x{10}}{1  \frac {x^2}{400}}\) = \(\frac x2\)
or, \(\frac {\frac x{10}}{\frac {400  x^2}{400}}\) = \(\frac x2\)
or, \(\frac x{10}\)× \(\frac {400}{400  x^2}\) = \(\frac x2\)
or, \(\frac {40x × 2}x\) = 400  x^{2}
or, 80 = 400  x^{2}
or, x^{2} = 400  80
or, x^{2} = 320
∴ x = 17.89 m
AC = 10x = 10× 17.89 = 178.9 m
∴ The height of the pole = 17.89 m _{Ans}
Let:
Height of the building = BC
The height of the antenna = CD = h. The distance between observer and foot of the building = AB. The angle of elevation of the antenna = \(\theta\) and the angle of elevation bottom of the antenna = \(\alpha\).
In right angled \(\triangle\)ABD,
tan\(\theta\) = \(\frac {BD}{AB}\) = \(\frac {BC + CD}{AB}\)
∴ AB = \(\frac {BC + CD}{tan\theta}\).................................(i)
In right angled \(\triangle\)ABC,
tan\(\alpha\) = \(\frac {BC}{AB}\)
or, AB = \(\frac {BC}{tan\alpha}\).........................................(ii)
From equation (i) and (ii)
\(\frac {BC + CD}{tan\theta}\) = \(\frac {BC}{tan\alpha}\)
or, BC tan\(\alpha\) + CD tan\(\alpha\) = BC tan\(\theta\)
or, BC tan\(\theta\)  BC tan\(\alpha\) = CD tan\(\alpha\)
or, BC (tan\(\theta\)  tan\(\alpha\)) = h tan\(\alpha\) [\(\because\) CD = h]
∴ BC =\(\frac {h tan\alpha}{tan\theta  tan\alpha}\)
Hence, Proved.
Let:
Height of the pillar=BD = ?
angle of elevation top of the pillar = \(\angle\)BAD = 60°
Angle of elevation 20 m below the top of the pillar = \(\angle\)BAC = 30°
∴ The distance between pillar and observation point = AB
In the right angled \(\triangle\)ABC,
tan 30° = \(\frac {BC}{AB}\)
or, \(\frac 1{\sqrt 3}\) = \(\frac {BC}{AB}\)
∴ AB = \(\sqrt 3\)BC
In right angled \(\triangle\)ABD,
tan 60° = \(\frac {BD}{AB}\)
or, \(\sqrt 3\) = \(\frac {BC + CD}{\sqrt 3 BC}\)
or, 3 BC = BC + 20
or, 3 BC  BC = 20
or, 2 BC = 20
or, BC = \(\frac {20}2\)
∴ Bc = 10 m
BD = BC + CD = 10 + 20 = 30 m
∴ The height of the pillar = 30 m _{Ans}
Let:
Height of building (AC) = ?
Length of ladder (BD) = 9 m
Angle of elevation on the top of building \(\angle\)ABC = 60°
Angle of elevation makes by ladder with ground \(\angle\)CBD = 45°
In the right angled \(\triangle\)BCD,
cos 45° = \(\frac {BC}{BD}\)
or, \(\frac 1{\sqrt 2}\) = \(\frac {BC}9\)
or, BC = \(\frac 9{\sqrt 2}\)
∴ BC = 6.36 m
In the right angled \(\triangle\)BCD,
tan 45° = \(\frac {CD}{BC}\)
or, 1 = \(\frac {CD}{6.34}\)
∴ CD = 6.34 m
In right angled \(\triangle\)ABC,
tan 60° = \(\frac {AC}{BC}\)
or, \(\sqrt 3\) = \(\frac {AD + CD}{BC}\)
or, \(\sqrt 3\)× 6.36 = AD + 6.36
∴ AD = 4.66 m
∴ Height of building = AD + CD = 6.36 + 4.66 = 11.02 m _{Ans}
Let: The height of posts AB and DE distance between the two post is 120 m. Angle of elevation \(\angle\)BCA = \(\alpha\) and \(\angle\)DCE = 90  \(\alpha\), CD = 60 m = BC
DE = 2 AB
In the right angled \(\triangle\)ABC,
tan\(\alpha\) = \(\frac {AB}{BC}\) = \(\frac {DE}{2× 60}\)
∴ DE = 120 tan\(\alpha\)............................(i)
In the right angled \(\triangle\)CDE,
tan (90  \(\alpha\)) = \(\frac {DE}{Cd}\) = \(\frac {120 tan\alpha}{60}\) = 2 tan\(\alpha\)
cot\(\alpha\) = 2 tan\(\alpha\)
or, \(\frac 1{tan\alpha}\) = 2 tan\(\alpha\)
or, tan^{2}\(\alpha\) = \(\frac 12\)
∴ tan\(\alpha\) = \(\frac 1{\sqrt 2}\)
Putting the value of tan\(\alpha\) in equation (i),
DE = 120×\(\frac 1{\sqrt 2}\) = 84.85 m
AB = \(\frac {DE}2\) = \(\frac {84.85}2\) = 42.43 m
∴ Height of the posts are 42.43 m and 84.85 m. _{Ans}
Let: AB and DE are the height of the poles where, DE = 2 AB.
Let: angle of elevation \(\angle\)BCA = \(\theta\), \(\angle\)DCE = 90  \(\theta\)
Distance between the two poles BD = 40 m
BC = CD = 20 m
In the right angled \(\triangle\)ABC,
tan\(\theta\) = \(\frac {AB}{BC}\) = \(\frac {DE}{2× 20}\) = \(\frac {DE}{40}\)
or, DE = 40 tan\(\theta\)....................................(i)
In the right angled \(\triangle\)CDE,
tan (90  \(\theta\)) = \(\frac {DE}{CD}\)
or, cot\(\theta\) = \(\frac {40 tan\theta}{20}\)
or, \(\frac 1{tan\theta}\) = tan\(\theta\)× 2
or, \(\frac 12\) = tan^{2}\(\theta\)
or, tan\(\theta\) = \(\frac 1{\sqrt 2}\)
POutting the value of tan\(\theta\) in equation (i)
DE = \(\frac 1{\sqrt 2}\)× 40 = 28.28 m
AB = \(\frac {DE}2\) = \(\frac {28.28}2\) = 14.14 m
∴ Height of two poles are 28.28 m and 14.14 m. _{Ans}
Let:
The tower's height (AB) = ?
Distance of AD = 16 m
Distance of Ac = 36 m
\(\angle\)ACB = \(\theta\)
\(\angle\)ADB = 90  \(\theta\)
CD = 36  16 = 20 m
In right angled \(\triangle\)ABD,
tan (90  \(\theta\)) =\(\frac {AB}{AD}\)
or, cot\(\theta\) = \(\frac {AB}{16}\)
or, AB = 16 cos\(\theta\)...........................(i)
In right angled \(\triangle\)ABC,
tan\(\theta\) = \(\frac {AB}{AC}\) = \(\frac {AB}{36}\)
or, AB = 36 tan\(\theta\)............................(ii)
From equation (i) and (ii),
16 cot\(\theta\) = 36 tan\(\theta\)
or, 16 \(\frac 1{tan\theta}\) = 36 tan\(\theta\)
or, \(\frac {16}{36}\) = tan^{2}\(\theta\)
or, tan^{2}\(\theta\) = \(\frac 49\)
or, tan^{2}\(\theta\) = \(\frac 23\)^{2}
∴ tan\(\theta\) = \(\frac 23\)
Putting the value of tan\(\theta\) in equation (ii)
AB = 36 tan\(\theta\) = 36× \(\frac 23\) = 12× 2 = 24 m
∴ The height of tower = 24 m _{Ans}
Let: Height of the tower (PQ) = ?
The height of flagstaff (QR) = 7 m
Angle of elevation:
\(\angle\)PSQ = 45° and \(\angle\)QSR = 15°
∴ \(\angle\)PSR = 45° + 15° = 60°
In right angled \(\triangle\)PRS,
tan 60° = \(\frac {PR}{PS}\)
or, 1 = \(\frac {PR}{PS}\)
∴ PQ = PS...............................(i)
In right angled \(\triangle\)PRS,
tan 60° = \(\frac {PR}{PS}\)
or, \(\sqrt 3\) = \(\frac {PQ + QR}{PQ}\) [From eq^{n} (i)]
or, PQ \(\sqrt 3\) = PQ + 7
or, PQ \(\sqrt 3\)  PQ = 7
or, PQ (\(\sqrt 3\)  1) = 7
or, PQ = \(\frac 1{\sqrt 3  1}\)
or, PQ = \(\frac 1{1.732  1}\)
or, PQ = \(\frac 1{0.732}\)
∴ PQ = 9.56 m
∴ Height of the tower = 9.56 m _{Ans}
Suppose, AD be the flagstaff and DB be the tower. C be the point of observation. \(\angle\)DCB = 45°, \(\angle\)ACD = 15° so that \(\angle\)ABC = 60°, \(\angle\)ACD = 60° and BC = 100 m, AD = ?
In the right angled\(\triangle\)DBC,
tan 45° = \(\frac {DB}{BC}\)
or, 1 = \(\frac {DB}{100}\)
∴ DB = 100 m
In right angled \(\triangle\)ABC,
tan 60° = \(\frac {AB}{BC}\)
or, \(\sqrt 3\) = \(\frac {AD + DB}{100}\)
or, \(\sqrt 3\) = \(\frac {AD + 100}{100}\)
or, 100\(\sqrt 3\) = AD + 100
or, AD = 100\(\sqrt 3\)  100
or, AD = 173.2  100
∴ AD = 73.2 m
∴ The length of the flagstaff is 73.2 m. _{Ans}

The angle of elevation of the top of a tower from a point was observed to be 45(^0) and on walking 60 metre away from that point it was found to be 30(^o) .Find the height of the tower .
81m
81. 97m
71.67m
81.90m

The angle of elevation of the top of the tower from a point was observed to be 45(^o).On walking 30 m away from that point was found to be 30(^o) . Find the height of the tower .
59.88m
40 m
40.98 m
0m

The angle of elevation of the top of a point on the ground was observed to be 60(^o) . On walking 60 m away from that point it was found to be 30(^o) . If the house and these points are in the same line of the same plain , find the height of the house .
20(sqrt{2})
90(sqrt{9})
100(sqrt{10})
30(sqrt{3})

A pole is surrounded on its top by a flagstaff . The angles of elevation of the top and the bottom of the flagstaff as observed from a point 30 metres away from the top of the pole are found to be 45(^o)and 30(^o) respectively . Find the height of the flagstaff .
12.69 m
12.68 Km
11.667 m
12.68 m

A flag of height 7 metres stands on the top of a tower . The angles subtended by the tower and the flagstaff to a point on the gorund are 45^{o} and 15^{o} respectively . Find the height of the tower.
9.56 m
12.68 m
10.11 m
11.11 m

The angle of elevation observed from a place to the roof of house standing infront of the place and to a window 6 m below its roof are found to be 45^{o} and 30^{o} respectively . Find the height of the house .
15.116 m
16.197 m
14.116 m
14.196 m

A pillar is 20 m high . The angle of elevation of the top of a house just infront of it , from the top pillar is 30^{o} and from the foot is 45^{o. }
Find the distance between the pillar & the house .48. 76 m
47.32 m
43. 44 m
46.77 m

From the foot of a tree 50 m high , the angle of elevation of a tower is 60^{o }and from the top the angle of elevation is 30^{o} . Find the height of the tower .
75 m
78.66 m
77 m
70 m

A water tank of height 12ft. is fixed on the roof of a building . The angles subtended by the building and the tank to a point on the ground are 45(^o) and 15(^o respectively . Find the height of the building .
12.56ft
15.99ft
16.93 ft
13.56ft

The angle of depression of the top of 70 m high building from the top of a tower is 45(^o) and the angle of elevation of the top of the tower from foot of building is 60(^o. Find the height of the tower .
98.76m
98.34 m
45.45m
95.62 m

The angle of depression of the top of 60 m high building from the tower is 30(^o) If the angle of elevation of the tower from foot of building is 60(^o) , find the height of tower .
20m
45m
90m
70m

From the top of the tower , 80 m high , the measures of the angles of depression of two objects due east of the tower are found to be 45(^o) and 60(^o).Find the distance between the objects.
33.81m
76.65m
58.57m
33.65m

From of the top of tower 100m high the measures of depression of two objects due east of the tower are found to be 45(^o) and 60(^o). Find the distance between the objects.
46.67m
42.26m
78.45m
23.45m

The angles of elevation of the top of tower from two points 242 m and 288 m away from it on the same side a straight line , are found to be complementary. Find the height of tower.
264 m
278 m
267 m
462 m

The angle of elevation of the top of tower from two points 169 m and 196 m away from it on the same side a straight line , are found to be complementary. Find the height of the tower.
182 m
187 m
298m
128 m

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krishnait is the best and simple way to teach height and distance everyone have to try it 
Dec 29, 2016 
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