Some Applications of Trigonometry test - 25

 $$ Given-\quad \\ ABCD\quad is\quad a\quad square\quad whose\quad central\quad point\quad is\quad O.\quad \\ OP\bot BO\quad \& \quad \angle APB={ 60 }^{ o }.\quad AB=b\quad \& \quad OP=h.\\ To\quad find\quad out-\\ The\quad relation\quad between\quad b\quad \& \quad h.\\ Solution-\\ O\quad is\quad the\quad central\quad point\quad of\quad the\quad square\quad ABCD.\\ \therefore \quad OB=\frac { 1 }{ 2 } \times diagonal=\frac { \sqrt { 2 } b }{ 2 } =\frac { b }{ \sqrt { 2 }  } .........(i)\\ Now\quad O\quad is\quad the\quad central\quad point\quad of\quad the\quad square\quad ABCD.\\ \therefore \quad PA=PB\quad i.e\quad \angle ABP=\angle BAP.........(ii)\\ Again\quad \angle APB={ 60 }^{ o }.\\ \therefore \quad \angle ABP+\angle BAP={ 180 }^{ o }-{ 60 }^{ o }\quad (by\quad angle\quad sum\quad property\quad of\quad triangles)\quad \\ \therefore \quad \Longrightarrow \angle ABP+\angle BAP={ 120 }^{ o }\quad \quad \\ \therefore \quad \angle ABP=\angle BAP={ 60 }^{ o }.(from\quad ii)\\ So\quad \Delta PAB\quad is\quad an\quad equilateral\quad one.\\ \therefore \quad PB=AB=b.....(iii)\\ Considering\quad \Delta POB\quad we\quad have\quad \angle O={ 90 }^{ o },OB=\frac { b }{ \sqrt { 2 }  } \quad (from\quad i)\\ and\quad PB=b\quad (from\quad iii)\\ \therefore \quad Applying\quad Pythagoras\quad theorem\\ { OP }^{ 2 }+{ OB }^{ 2 }={ PB }^{ 2 }\\ \Longrightarrow { h }^{ 2 }+{ \left( \frac { b }{ \sqrt { 2 }  }  \right)  }^{ 2 }={ b }^{ 2 }\\ \Longrightarrow 2{ h }^{ 2 }={ b }^{ 2 }\\ Ans-\quad Option\quad B $$ 

Let PQ $$=$$ h m be a plane
From right angled $$\Delta PAQ,$$

$$tan45^0=\frac{PQ}{AQ}$$

$$\therefore 1=\frac{h}{x}$$

or $$x = h$$..............(i)

Again from right angled $$\Delta$$ PBQ,

$$tan 60^0=\frac{PQ}{QB}$$

 or $$\sqrt{3}=\frac{h}{2-x}$$

From equation $$(i),\sqrt{3}=\frac{h}{2-h}$$

or $$2\sqrt{3}- \sqrt{3}h=h$$

or $$h+\sqrt{3}h=2\sqrt{3}$$

or $$h(1+\sqrt{3})=2\sqrt{3}$$

or $$h=\frac{2\sqrt{3}}{1+\sqrt{3}}=\frac{2\sqrt{3}}{1+\sqrt{3}}\times\frac{1-\sqrt{3}}{1-\sqrt{3}}$$

$$-\frac{2\sqrt{3}-6}{1-3}-\frac{-2(3-\sqrt{3})}{-2}$$

$$=(3-\sqrt{3})$$km

Consider the following diagram with the cloud at the point A
Let the perpendicular distance of the cloud from the the elevation be $$x=AC$$
Angle ABC$$=30^0$$
angle DBC$$=60^0$$
Therefore
Consider triangle ABC
$$tan30^0=\frac{x}{BC}=\frac{1}{\sqrt{3}}$$
$$BC=\sqrt{3}x$$ ...(i)
Similarly consider triangle BDC
$$tan60^0=\frac{h+h+x}{BC}=\sqrt{3}$$
$$2h+x=\sqrt{3}BC$$ ...(ii)
Substituting the value of BC from equation i we get
$$2h+x=\sqrt{3}\sqrt{3}x$$
$$3x=2h+x$$
$$h=x$$
Therefore the height of the cloud above the surface of the water
$$=h+x$$
$$=2h$$
Hence answer is D

$$\tan60^0=\dfrac{AB}{BC}$$

Let AB $$=$$ 1 m be the stick.
Let AC be the shadow of length $$x.$$
From right angled $$\Delta ACB,$$

$$\tan 60^0=\dfrac{1}{x}$$

or $$\sqrt{3}=\dfrac{1}{x}$$

or $$x=\dfrac{1}{\sqrt{3}}$$m

Consider the above diagram
In triangle POM
Let $$OM$$ be the height of the tower
Let $$OP$$ be the horizontal distance of the marked point  from the foot of the tower
It is given that angle of depression$$=P=30^0$$
Therefore
$$\frac{MO}{PO}$$
$$=\frac{30m}{PO}$$
$$=tan30^0=\frac{1}{\sqrt{3}}$$
Therefore $$PO=30\sqrt{3}$$ metres
Hence answer is D

Consider the above given triangle
Let AB be the height of the tower and $$\angle ACB=x$$ and $$\angle ADB=y$$ ...(given)
Consider triangle ADB
$$tany=\frac{AB}{BD}$$
$$BD={AB}coty$$ ...(i)
Consider triangle ACD
$$tanx=\frac{AB}{BD+CD}$$
$$BD+CD=ABcotx$$
$${AB}coty+CD=ABcotx$$ ...from(i)
$$d=AB(cotx-coty)$$
$$AB=\frac{d}{cotx-coty}$$
Upon simplifying we get
$$AB=\frac{d(tanx tany)}{tany-tanx}$$
Hence answer is A

From the $$\triangle ABC$$
$$\tan 60^{\circ}=\dfrac{AB}{BC}$$