Height and Distance

Let CB be the pole and point D divides it such that BD : DC = 1 : 9
Given that AB = 15 m
Let the the two parts subtend equal angles at point A such that ∠ CAD = ∠ BAD = $$\theta $$
From "Angle Bisector Theorem", we have
$$\frac{{BD}}{{DC}} = \frac{{AB}}{{AC}}$$
$$ ⇒ \frac{1}{9} = \frac{{15}}{{AC}}$$     [∵ BD : DC = 1 : 9 and AB = 15(given)]
$$ ⇒ AC = 15 \times 9\,{\text{m}}\,......\left( {eq:1} \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$CB = \sqrt {A{C^2} - A{B^2}} $$     (∵ Pythagorean theorem)
$$ = \sqrt {{{\left( {15 \times 9} \right)}^2} - {{15}^2}} $$     [∵AC=15 × 9(eq : 1) and AB=15 m(given)]
$$ = \sqrt {{{15}^2} \times {9^2} - {{15}^2}} $$
$$ = \sqrt {{{15}^2}\left( {{9^2} - 1} \right)} $$
$$ = \sqrt {{{15}^2} \times 80} $$
$$ = \sqrt {{{15}^2} \times 16 \times 5} $$
$$ = 15 \times 4 \times \sqrt 5 $$
$$ = 60\sqrt 5 \,{\text{m}}$$

Consider the diagram shown above.
Let AB be the tower. Let C and D be the positions of the boat
Then, ∠ ACB = 45°, ∠ ADC = 30°, BC = 100 m
$$\tan {45^ \circ } = \frac{{AB}}{{BC}}$$
$$ ⇒ 1 = \frac{{AB}}{{100}}$$
$$ ⇒ AB = 100\,......\left( {eq:1} \right)$$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{{100}}{{BD}}$$     (∵ Substituted the value of AB from equation 1)
$$ ⇒ BD = 100\sqrt 3 $$
$$CD = \left( {BD - BC} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\, = \left( {100\sqrt 3 - 100} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\, = 100\left( {\sqrt 3 - 1} \right)$$
It is given that the distance CD is covered in 10 seconds.
i.e., the distance $$100\left( {\sqrt 3 - 1} \right)$$     is covered in 10 seconds.
$${\text{Required}}\,{\text{speed}}$$
$$ = \frac{{{\text{Distance}}}}{{{\text{Time}}}}$$
$$ = \frac{{100\left( {\sqrt 3 - 1} \right)}}{{10}}$$
$$ = 10\left( {1.73 - 1} \right)$$
$$ = 7.3\,{\text{meter/seconds}}$$
$$ = 7.3 \times \frac{{18}}{5}\,{\text{km/hr}}$$
$$ = 26.28\,{\text{km/hr}}$$

Let C and D be the position of the aeroplanes.
Given that CB = 900 m, ∠ CAB = 60°, ∠ DAB = 45°
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$\tan {60^ \circ } = \frac{{CB}}{{AB}}$$
$$\sqrt 3 = \frac{{900}}{{AB}}$$
$$AB = \frac{{900}}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\, = \frac{{900 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\, = \frac{{900\sqrt 3 }}{3}$$
$$\,\,\,\,\,\,\,\,\,\, = 300\sqrt 3 $$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABD,$$
$$\tan {45^ \circ } = \frac{{DB}}{{AB}}$$
$$1 = \frac{{DB}}{{AB}}$$
$$DB = AB = 300\sqrt 3 $$
$$$$
$${\text{Required}}\,{\text{height}}$$
$$ = CD$$
$$ = \left( {CB - DB} \right)$$
$$ = \left( {900 - 300\sqrt 3 } \right)$$
$$ = \left( {900 - 300 \times 1.73} \right)$$
$$ = \left( {900 - 519} \right)$$
$$ = 381\,{\text{m}}$$

Consider the diagram shown above where QR represents the tree and PQ represents it's shadow
$${\text{We}}\,{\text{have}},\,QR = PQ$$
$${\text{Let}}\,\angle QPR = \theta $$
$$\tan \theta = \frac{{QR}}{{PQ}} = 1$$     (since QR = PQ)
$$ ⇒ \theta = {45^ \circ }$$
i.e., required angle of elevation = 45°

Let BD be the tower and A and C be the positions of the persons.
Given that BD = 50 m, ∠ BAD = 30°, ∠ BCD = 60°
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta \,ABD,$$
$$\tan {30^ \circ } = \frac{{BD}}{{BA}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{{50}}{{BA}}$$
$$ ⇒ BA = 50\sqrt 3 $$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CBD,$$
$$\tan {60^ \circ } = \frac{{BD}}{{BC}}$$
$$ ⇒ \sqrt 3 = \frac{{50}}{{BC}}$$
$$ ⇒ BC = \frac{{50}}{{\sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50 \times \sqrt 3 }}{{\sqrt 3 \times \sqrt 3 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{50\sqrt 3 }}{3}$$
Distance between the two persons
$$ = AC = BA + BC$$
$$ = 50\sqrt 3 + \frac{{50\sqrt 3 }}{3}$$
$$ = \sqrt 3 \left( {50 + \frac{{50}}{3}} \right)$$
$$ = \frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
i.e., the distance travelled by the car in 10 seconds = $$\frac{{200\sqrt 3 }}{3}\,{\text{m}}$$
$${\text{Speed}}\,{\text{of the car}} = \frac{{{\text{Distance}}}}{{{\text{Time}}}}$$
$$ = \frac{{\left( {\frac{{200\sqrt 3 }}{3}} \right)}}{{10}} = \frac{{200\sqrt 3 }}{3}\,{\text{m/s}}$$
$$ = \frac{{200\sqrt 3 }}{3} \times \frac{{18}}{5}\,{\text{km/hr}}$$
$$ = {\text{24}}\sqrt 3 \,{\text{km/hr}}$$

Consider the diagram shown above. Let AB be the tower. Let D and C be the positions of the car.
Then, ∠ ADC = 30° , ∠ ACB = 45°
Let AB = h, BC = x, CD = y
$$\tan {45^ \circ } = \frac{{AB}}{{BC}} = \frac{h}{x}$$
$$ ⇒ 1 = \frac{h}{x}$$
$$ ⇒ h = x\,......\left( 1 \right)$$
$$\tan {30^ \circ } = \frac{{AB}}{{BD}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{AB}}{{\left( {BC + CD} \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{h}{{x + y}}$$
$$ ⇒ \frac{1}{{\sqrt 3 }} = \frac{h}{{x + y}}$$
$$ ⇒ x + y = \sqrt 3 \,h$$
$$ ⇒ y = \sqrt 3 \,h -×\cr} $$
$$ ⇒ y = \sqrt 3 \,h - h$$     (∵ Substituted the value of x from equation 1 )
$$ ⇒ y = h\left( {\sqrt 3 - 1} \right)$$
Given that distance y is covered in 8 minutes.
i.e, distance $$h\left( {\sqrt 3 - 1} \right)$$   is covered in 8 minutes.
Time to travel distance x
= Time to travel distance h (∵ Since×= h as per equation 1).
Let distance h is covered in t minutes.
since distance is proportional to the time when the speed is constant, we have
$$h\left( {\sqrt 3 - 1} \right) \propto 8\,......\left( {\text{A}} \right)$$
$$h \propto t\,..............\left( {\text{B}} \right)$$
$$\frac{{\left( {\text{A}} \right)}}{{\left( {\text{B}} \right)}} ⇒ \frac{{h\left( {\sqrt 3 - 1} \right)}}{h} = \frac{8}{t}$$
$$ ⇒ \left( {\sqrt 3 - 1} \right) = \frac{8}{t}$$
$$ ⇒ t = \frac{8}{{\left( {\sqrt 3 - 1} \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{\left( {1.73 - 1} \right)}}$$
$$\,\,\,\,\,\,\,\,\,\,\, = \frac{8}{{.73}}$$
$$\,\,\,\,\,\,\,\,\,\,\, = \frac{{800}}{{73}}\,{\text{minutes}}$$
$$\,\,\,\,\,\,\,\,\,\,\, = 10\frac{{70}}{{73}}\,{\text{minutes}}$$
$$ \approx {\text{10}}\,{\text{minutes}}\,{\text{57}}\,{\text{seconds}}$$

Let AB and CD be the poles with heights h and 2h respectively.
Given that distance between the poles, BD = 200 m
Let E be the middle point of BD,
∠ AEB = $$\theta $$
∠ CED = (90 - $$\theta $$)   (∵ given that angular elevations are complementary)
Since E is the middle point of BD, we have BE = ED = 100 m
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE,$$
$$\tan \theta = \frac{{AB}}{{BE}}$$
$$\tan \theta = \frac{h}{{100}}$$
$$h = 100\tan \theta \,......\left( 1 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta EDC,$$
$$\tan\left( {90 - \theta } \right) = \frac{{CD}}{{ED}}$$
$$\cot \theta = \frac{{2h}}{{100}}$$     $$\left[ {\because \tan \left( {90 - \theta } \right) = \cot \theta } \right]$$
$$2h = 100\cot \theta \,......\left( 2 \right)$$
$$\left( 1 \right) \times \left( 2 \right)$$
$$ ⇒ 2{h^2} = {100^2}$$     $$\left[ {\because \tan \theta \times \cot\theta = \tan\theta \times \frac{1}{{\tan \theta }} = 1} \right]$$
$$ ⇒ \sqrt 2 \,h = 100$$
$$ ⇒ h = \frac{{100}}{{\sqrt 2 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{100 \times \sqrt 2 }}{{\sqrt 2 \times \sqrt 2 }}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 50\sqrt 2 $$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 50 \times 1.41$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 70.5$$
$$ ⇒ 2h = 2 \times 70.5 = 141$$
i.e. the height of the poles are 70.5 m and 141 m:

Let ED be the building and AC be the tower.
Given that ED = 230 m, ∠ADC = b, ∠AEB = a
Also given that tan a = $$\frac{5}{{12}}$$ and tan b = $$\frac{4}{{5}}$$
Let AC = h
Required Distance = Distance between the top of these buildings = AE
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABE,$$
$$\tan \left( a \right) = \frac{{AB}}{{BE}}$$
$$ ⇒ \frac{5}{{12}} = \frac{{\left( {h - 230} \right)}}{{BE}}$$     [∵ tan(a) = $$\frac{5}{{12}}$$ (given), AB = (AC - BC) = (AC - ED) = (h - 230)]
$$ ⇒ BE = \frac{{12\left( {h - 230} \right)}}{5}\,.....\left( 1 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ACD,$$
$$\tan \left( b \right) = \frac{{AC}}{{CD}}$$
$$ ⇒ \frac{4}{5} = \frac{h}{{CD}}$$    [∵ tan(b) = $$\frac{4}{5}$$ (given), AC = h]
$$ ⇒ CD = \frac{{5h}}{4}\,.....\left( 2 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{diagram,}}\,BE = CD$$
$$ ⇒ \frac{{12\left( {h - 230} \right)}}{5} = \frac{{5h}}{4}$$     (from equation 1 & 2)
$$ ⇒ 48h - \left( {4 \times 12 \times 230} \right) = 25h$$
$$ ⇒ 23h = \left( {4 \times 12 \times 230} \right)$$
$$ ⇒ h = \frac{{\left( {4 \times 12 \times 230} \right)}}{{23}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\, = 480\,{\text{m}}\,......\left( 3 \right)$$
$$AB = \left( {AC - BC} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\, = \left( {480 - 230} \right)$$    [∵ Since AC = h = 480 (from 3) and BC = ED = 230 m(given)]
$$\,\,\,\,\,\,\,\,\,\,\,\, = 250\,{\text{m}}$$
In the triangle ABE, tan(a) =$$\frac{5}{{12}}$$ . Let's figure out the value of sin(a) now.
Consider a triangle with opposite side = 5 and adjacent side = 12 such that tan(a) = $$\frac{5}{{12}}$$
$${\text{hypotenuse}} = \sqrt {{5^2} + {{12}^2}} = 13$$
$${\text{i}}{\text{.e}}{\text{.,}}$$
$$\sin \left( a \right) = \frac{{{\text{opposite}}\,{\text{side}}}}{{{\text{hypotenuse}}}} = \frac{5}{{13}}$$
We have seen that $$\sin \left( a \right) = \frac{5}{{13}}$$
$$ ⇒ \frac{{AB}}{{AE}} = \frac{5}{{13}}$$
$$ ⇒ AE = AB \times \frac{{13}}{5}$$
$$ = 250 \times \frac{{13}}{5}$$
$$ = 650\,{\text{m}}$$
i.e., Distance between the top of the buildings = 650 m

Let AC be the tower and B be the position of the bus.
Then BC = the distance of the bus from the foot of the tower.
Given that height of the tower, AC = 80 m and the angle of depression, ∠DAB = 30°
∠ABC = ∠DAB = 30° (because DA || BC)
$$\tan {30^ \circ } = \frac{{AC}}{{BC}}$$
$$ ⇒ \tan {30^ \circ } = \frac{{80}}{{BC}}$$
$$ ⇒ BC = \frac{{80}}{{\tan {{30}^ \circ }}}$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{80}}{{\left( {\frac{1}{{\sqrt 3 }}} \right)}}$$
$$ = 80 \times 1.73 = 138.4\,{\text{m}}$$
i.e., Distance of the bus from the foot of the tower = 138.4 m

Let A be the foot and C be the summit of a mountain.
Given that ∠CAB = 45°
From the diagram, CB is the height of the mountain. Let CB = x
Let D be the point after ascending 2 km towards the mountain such that AD = 2 km and given that ∠DAY = 30°
It is also given that from the point D, the elevation is 60°
i.e., ∠CDE = 60°
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta ABC,$$
$$\tan {45^ \circ } = \frac{{CB}}{{AB}}$$
$$ ⇒ 1 = \frac{x}{{AB}}$$    [∵ CB = x(the height of the mountain)]
$$ ⇒ AB = x\,......\left( 1 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta AYD,$$
$$\sin {30^ \circ } = \frac{{DY}}{{AD}}$$
$$ ⇒ \frac{1}{2} = \frac{{DY}}{2}$$    (∵ Given that AD = 2)
$$ ⇒ DY = 1\,.......\left( 2 \right)$$
$$\cos {30^ \circ } = \frac{{AY}}{{AD}}$$
$$ ⇒ \frac{{\sqrt 3 }}{2} = \frac{{AY}}{2}$$    (∵ Given that AD = 2)
$$ ⇒ AY = \sqrt 3 \,......\left( 3 \right)$$
$${\text{From}}\,{\text{the}}\,{\text{right}}\,\Delta CED,$$
$$\tan {60^ \circ } = \frac{{CE}}{{DE}}$$
$$ ⇒ \tan {60^ \circ } = \frac{{\left( {CB - EB} \right)}}{{YB}}$$     [∵ CE = (CB - EB) and DE = YB)]
$$ ⇒ \tan {60^ \circ } = \frac{{\left( {CB - DY} \right)}}{{AB - AY}}$$     [∵ EB = DY and YB = (AB - AY)]
$$ ⇒ \tan {60^ \circ } = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}}$$
[∵ CB = x, DY = 1(eq:2), AB = x(eq:1) and AY = $${\sqrt 3 }$$ (eq:3)]
$$ ⇒ \sqrt 3 = \frac{{\left( {x - 1} \right)}}{{\left( {x - \sqrt 3 } \right)}}$$
$$ ⇒ x\sqrt 3 - 3 =×- 1$$
$$ ⇒ x\left( {\sqrt 3 - 1} \right) = 2$$
$$ ⇒ 0.73x = 2$$
$$ ⇒ x = \frac{2}{{0.73}} = 2.7$$
i.e., the height of the mountain = 2.7 km