Some Applications of Trigonometry test

Result

Some Applications of Trigonometry test - 19

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

From the top of a tree a man observes the angle of depression of a moving car is $$30^{o}$$ and after $$3$$ minutes he finds the angle of depression is $$60^{o}$$. How much time will the car take to reach the tree?

A
$$4$$ minutes

B
$$3$$ minutes

C
$$1.5$$ minutes

D
$$2$$ minutes
Question 2
1 / -0

Assertion ($$A$$): ladder rests against a wall at an angle $$30^{0}$$ to the horizontal. Its foot is pulled away through a distance $$x$$' so that it slides a distance $$y$$' down the wall finally making an angle $$60^{0}$$ with the horizontal then $$x=y$$.
Reason ($$R$$): $$A$$ ladder rests against a wall at angle $$\alpha $$ to the horizontal. Its foot is pulled a way through a distence $$a$$' so that it slides a distence $$b$$' down the wall, finally making an angle $$\beta$$ with the horizonal then $$\displaystyle \tan(\frac{\alpha+\beta}{2})=b/a$$

A
Both $$A$$ and $$R$$ are ture and ` $$R$$' is the correct

explanation of $$A$$

B
Both $$A$$ and $$R$$ are true and ` $$R$$' is not correct

explanation of $$A$$

C
$$A$$ is true but ` $$R$$' is false

D
$$A$$' is false but $$R$$' is true.

Solution

$$AB=A^{\prime}B^{\prime}=l$$

$$\sin \alpha =\dfrac {AC}l$$

$$\sin \beta =\dfrac {A^{\prime}C}l$$

$$\sin \alpha -\sin \beta =\dfrac bl$$

$$\cos \beta -\cos \alpha =\dfrac al$$

$$\dfrac {2\sin \left(\dfrac {\alpha -\beta}2\right)\cos \left( \dfrac {\alpha +\beta}2\right)}{2\sin \left( \dfrac {\alpha +\beta}2\sin \left(\dfrac {\alpha -\beta}2\right)\right)}=\dfrac ba$$

$$\cot \left(\dfrac {\alpha +\beta}2\right)=\dfrac ba$$
Question 3
1 / -0

A flag staff of $$5$$ mts high stands on a building of $$25 $$ mt high. At an observer at a height of $$30$$ mt the flag staff and the building subtend equal angles. The distance of the observer from the top of the flag staff is

A
$$5\sqrt{3}$$

B
$$5\sqrt{2}$$

C
$$5\sqrt{6}$$

D
$$\displaystyle \frac{5\sqrt{3}}{\sqrt{2}}$$

Solution

As per the data we have drawn the diagram.
Consider OB is the bisector of angle COA of triangle COA.

$$\dfrac{OC}{OA}=\dfrac{BC}{AB}$$

$$\dfrac{x}{\sqrt{x^2+30^2}}=\dfrac{5}{25}$$

$$x = \dfrac{\sqrt{x^2+30^2}}{5}$$

Squaring on both the sides we get,

$$x^2=\dfrac{x^2+900}{25}$$

$$25x^2-x^2=900$$

$$24x^2= 900$$

$$x^2=\dfrac{900}{24}$$

$$x = \dfrac{30}{\sqrt{24}}$$

$$x = \dfrac{30}{2\sqrt{6}}$$

$$x = \dfrac{15}{\sqrt{2 \times 3}}$$

$$x = \dfrac{\sqrt{3}\sqrt{3}\times5}{\sqrt{2\times 3}}$$

$$x =\dfrac{5\sqrt{3}}{\sqrt{2}}$$ m
Question 4
1 / -0

Flag-staff of length $$d$$ stands on a tower of height $$h$$. lf at a point on the ground the angles of elevation of the tower and the top of the flag-staff be $$\alpha,\ \beta$$ respectively, then $$h=$$

A
$$\displaystyle \dfrac{d\cot\beta}{\cot\beta-\cot\alpha}$$

B
$$\displaystyle \dfrac{d\tan\beta}{\tan\alpha-\tan\beta}$$

C
$$\displaystyle d[\frac{\tan\alpha+\tan\beta}{\cot\alpha-\cot\beta}]$$

D
$$\displaystyle \frac{d\tan\alpha}{\tan\beta}$$

Solution

$$\displaystyle \tan \alpha= \frac{h}{x}$$
$$\displaystyle x = h \cot \alpha$$
where $$x$$ is the horizontal distance from base of the tower
$$\displaystyle \tan \beta= \dfrac{d+h}{x}$$
Substituting for $$x$$
$$\displaystyle \tan \beta= \dfrac{d+h}{ h \cot \alpha}$$
$$\displaystyle h(\cot \alpha \tan \beta-1)=d$$
$$\displaystyle \therefore h=\dfrac{d \cot \beta}{\cot \beta- \cot\alpha}$$
Question 5
1 / -0

A ladder $$20\ m$$ long reaches a point $$20\ m$$ below the top of a flag and makes an angle $$60^\circ$$ with the horizontal. Find the length of the flagstaff, in meters

A
$$40$$

B
$$20$$

C
$$30$$

D
$$10(2+\sqrt{3})$$

Solution

In the above diagram, $$AC$$ and point $$D$$ is the top of the flag.
In $$\triangle ABC,$$
$$\begin{aligned}{}\sin 60^\circ& = \frac{{BC}}{{AC}}\\\frac{{\sqrt 3 }}{2}& = \frac{h}{{20}}\\h &= 10\sqrt 3\ m\end{aligned}$$

So,
$$BD=BC+CD$$
$$=h+20$$
$$=10\sqrt3 + 20$$
$$=10(2+\sqrt3)\ m$$
Question 6
1 / -0

The angle of elevation of a cloud from a point $$h$$ metres above a lake is $$\Theta$$. The angle of depression of its reflection in lake is $$45^{ }$$ The height of the cloud is

A
$$h\tan(45^{0}+\Theta)$$

B
$$hcot (45^{0}-\Theta)$$

C
$$h\tan(45^{0}-\Theta)$$

D
$$h\cot(45^{0}+\Theta)$$

Solution

$$tan\theta = \ \ \frac{b}{d}$$
tan 45 = $$\frac{a}{d}$$
$$tan\theta_2\ \ \ \frac{b}{a}$$
b + h = a - h
$$tan\theta_2\frac{b}{b+2h}$$
$$\frac{1}{tan\theta}=1+\frac{2h}{b}$$
$$\frac{b}{24}=\frac{tan\theta}{1-tan\theta}$$
$$b=\frac{24\ tan\theta}{1-tan\theta}$$
bfh = $$\frac{h-h\ tan\theta+2\ h\ tan\theta}{1-h\ tan\theta}=\ \ h\ tan(\theta+45)$$
Question 7
1 / -0

$$PQ$$ is a vertical tower. $$A,\ B,\ C$$ are three points in a horizontal line through $$Q$$, the foot of the tower. If the angles of elevation of the top of the tower from $$A,\ B,\ C$$ are $$\alpha,\ \beta,\ \gamma$$ respectively, then $$BC\cot\alpha-CA\cot\beta+AB \cot$$ $$\gamma=$$

A
$$3$$

B
$$1$$

C
$$2$$

D
$$0$$

Solution

$$BC\cot { \alpha } -CA\cot { \beta } +AB\cot { \gamma } $$
$$=BC\times \cfrac { QA }{ PQ } -\cfrac { CA\times QB }{ PQ } +\cfrac { AB\times QC }{ PQ } $$
$$=\left( QC-QB \right) \times \cfrac { QA }{ PQ } -\cfrac { \left( QC-QA \right) QB }{ PQ } +\cfrac { \left( QB-QA \right) QC }{ PQ } $$
$$=\cfrac { 1 }{ PQ } \left[ QCQA-QBQA-QCQB+QAQB+QBQC-QAQC \right] $$
$$=\cfrac { 1 }{ PQ } \times 0=0$$
Question 8
1 / -0

A man on a cliff observes a boat at an angle of depression $$30^{o}$$ which is sailing towards the shore to the point immediately beneath him. $$3$$ minutes later the angle of depression of the boat is found to be $$60^{o}$$. Assuming that the boat sails at a uniform speed, the time taken by the boat to reach the shore is:

A
$$1.5 $$ min

B
$$4.5$$ min

C
$$6 $$ min

D
$$6.5$$ min

Solution

Let $$AB$$ be the cliff, speed be $$s$$, total time taken be $$T$$, total distance be $$X$$

The distance covered in $$3$$ min is $$x=3s$$

Height of cliff is $$h$$

From $$\Delta ABD$$

$$ \implies \tan 30^o =\dfrac {AB}{BD}$$

$$ \implies \dfrac 1 {\sqrt 3} =\dfrac {AB}{X}$$

$$ \implies AB= \dfrac {X}{\sqrt 3} \cdots(1)$$

From $$\Delta ABC$$

$$ \implies \tan 60^o =\dfrac {AB}{BC}$$

$$ \implies {\sqrt 3} =\dfrac {AB}{X-x}$$

$$ \implies AB= {X-x}\times {\sqrt 3} \cdots(2)$$

From $$(1),(2)$$

$$ \implies \dfrac X{\sqrt 3}=(X-3s) \times \sqrt 3$$

$$ \implies X=3X-9s$$

$$ \implies 2X=9s$$

$$ \implies 2(sT)=9s$$

$$ \implies T=\dfrac 92=4.5$$ min
Question 9
1 / -0

$$A$$ tower standing at point $$A$$ leans towards west making an angle $$\alpha$$ with the vertical. The angular elevation of $$B$$, the top most point of the tower is $$\beta$$ as observed from a point $$C$$ due east of $$A$$ at a distance $$d$$ from $$A$$. lf the angular elevation of $$B$$ from a point due east of $$C$$ at a distance $$2d$$ from $$C$$ is $$\gamma$$, then $$ 2\tan\alpha$$ can be written as

A
3 $$\cot\beta-2\cot\gamma$$

B
$$\cot\gamma -\cot\beta$$

C
$$3 \cot\beta-\cot\gamma$$

D
$$\cot\beta-3\cot\gamma$$

Solution

Let $$AB,$$ be the tower leaning towards west making an angle $$\alpha $$ with vertical at $$C$$, angle of elevation of $$B$$ is $$\beta $$ $$abd $$at $$D$$ us $$\gamma $$
$$CA=AD=d$$ in $$\triangle ABH$$
$$\tan { \alpha } =\cfrac { AH }{ h }$$
$$ \Rightarrow AH=h\tan { \alpha } ......... 1$$
In $$\triangle BCH,\tan { \beta } =\cfrac { h }{ CH }$$
$$ \Rightarrow CH=h\cot { \beta } $$
$$d-AH=h\cot { \beta } ,$$ put the value of AH from $$1$$
$$d-h\tan { \alpha } =h\cot { \beta } \Rightarrow d=h\left( \cot { \beta } +\tan { \alpha } \right) ....... 2$$
In $$\triangle BDH,\tan { \gamma } =\cfrac { BH }{ HD } =\cfrac { h }{ AH+d } =AH+d=h\cot { \gamma } $$
$$\Rightarrow h\tan { \alpha } +d=h\cot { \gamma } \Rightarrow d=h\left( \cot { \gamma } -\tan { \alpha } \right) \rightarrow 3$$
From $$2$$ and $$3$$ we get
$$h\left( \cot { \beta } +\tan { \alpha } \right) =h\left( \cot { \gamma } -\tan { \alpha } \right) $$
$$2\tan { \alpha } =\cot { \gamma } -\cot { \beta } $$
Question 10
1 / -0

Assertion (A): The shadow of a tower on a level plane is found to be 60 meters longer when sun's altitude is $$30^{0}$$ than that when it is $$45^{0}$$ Then the height of the tower is $$30$$ $$(\sqrt{3}+1)$$ {meters}
Reason (R): The angle of elevation of a top of a tower standing an horizontal plane from a Point $$A$$ is $$\alpha$$. After wlaking a distance $$d$$ meters towards the foot of the tower, the angle elveation is found to be $$\beta$$ then the height of the tower is $$h=\displaystyle \frac{d}{\cot\alpha-\cot\beta}$$

A
Both $$A$$ and $$R$$ are ture and $$R$$ is the correct explanation of $$A$$

B
Both $$A$$ and $$R$$ are true and $$R$$ is not correct explanation of $$A$$

C
$$A$$ is true but $$R$$ is false

D
$$A$$ is false but $$R$$ is true.

Solution

Let the height of the tower be $$h$$.
When the angle of elevation of the sun is $$45^{0}$$ the height of the tower and the length of its shadow is the same.
Or the length of the shadow is $$h\cdot \cot45^{0}$$
Let us denoted that by $$h$$.
When the elevation of the sun is $$30^{0}$$, the length of the shadow is $$h\cot45^{0}+60$$ m while the height of the tower is $$h$$.
Then
$$\tan30^{0}=\dfrac{h}{h\cot45^{0}+60}$$

$$h\cot45^{0}+60=h\cot30^{0}$$
$$h=\dfrac{60}{\cot30^{0}-\cot45^{0}}$$

$$=\dfrac{60}{\sqrt{3}-1}$$

$$=\dfrac{60}{2}(\sqrt{3}+1)$$

$$=30(\sqrt{3}+1)$$ m