Some Applications of Trigonometry test

Result

Some Applications of Trigonometry test - 20

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

Statement $$I$$: $$A$$ kite is flying with the string inclined at $$30^{0}$$ to the horizontal. The height of the kite above the ground when the string is $$15$$ mt long is $$7.5$$ mts.
Statement $$II$$: The altitude of the sun is $$45^{0}$$, When the length of the shadow of a pole is same as height of the pole.
Which of the above statement is/are true?

A
Only $$I$$

B
Only $$II$$

C
Both $$I$$ and $$II$$

D
Neither $$I$$ nor $$ll$$

Solution

Both the statements are true.
Statement 1: here, $$\sin 30^{\circ}= \dfrac{h}{15}$$
$$h=7.5$$
i.e $$\sin 30^{\circ}= \dfrac{1}{2}$$
we know,
$$\sin 30^{\circ}= \dfrac{1}{2}$$
Therefore, Statement 1 is true.

Statement 2 : Here, $$\tan 45^{\circ}= \dfrac{h}{h}$$
As height of pole= length of shadow=$$h$$
i.e $$\tan 45^{\circ}= 1$$
Thus, altitude of sun is $$45^{\circ}$$
Therefore, Statement 2 is true.
Question 2
1 / -0

The angle of elevation of the top of the tower at any point of the ground is $${30}^{0}$$ and if moving $$20m$$ towards the tower it becomes $${60}^{0},$$ then the height of the tower is

A
$$10$$

B
$$10\sqrt{3}$$

C
$$\displaystyle\frac{10}{\sqrt{3}}$$

D
$$30$$

Solution

Angle of elevation of the top of a tower at $$\displaystyle D={ 30 }^{ 0 };$$
distance between $$C$$ and $$D(CD)=20m$$
and angle of elevation of the top of the tower at $$C={ 60 }^{ 0 }$$

Let $$AB$$ be the tower of height $$h$$ and $$x$$ be the distance between $$A$$ and $$C.$$
We know that in $$\triangle ACB,$$
$$\displaystyle \tan { { 60 }^{ 0 } } =\frac { h }{ x } $$$$\Rightarrow\displaystyle \sqrt { 3 } =\frac { h }{ x } $$$$\Rightarrow\displaystyle h=\sqrt { 3 } x$$ ...(1)

We also know that in $$\triangle ABD,$$
$$\displaystyle \tan { { 30 }^{ 0 } } =\frac { h }{ x+20 } $$ or $$\displaystyle \frac { 1 }{ \sqrt { 3 } } =\frac { h }{ x+20 } $$
$$\Rightarrow\displaystyle \sqrt { 3 } h=x+20$$$$\Rightarrow\displaystyle \sqrt { 3 } h-20=x$$ ...(2)

Substituting the value of $$x$$ in Eq. (1), we get
$$\displaystyle h=\sqrt { 3 } \left( \sqrt { 3 } h-20 \right) $$ $$\Rightarrow\displaystyle h=3h-20\sqrt { 3 } $$ $$\Rightarrow\displaystyle-2h=-20\sqrt { 3 } $$ $$\Rightarrow\displaystyle h=10\sqrt { 3 } $$
Question 3
1 / -0

A balloon moving in a straight line passes vertically above two points $$A$$ and $$B$$ on horizontal plane $$1000\ ft$$ apart. When above $$A$$ it has an altitude of $$60^{0}$$ as seen from $$B$$. When above $$B$$ it has an altitude of $$45^{0}$$ as seen from $$A$$. The distance of $$B$$ from the point at which it will touch the plane is

A
$$500(\sqrt{3}+1)\ ft$$

B
$$1500\ ft$$

C
$$500(3+\sqrt{3})\ ft$$

D
$$500\ ft$$

Solution

tan 60=$$=\frac{h}{1000}$$
$$h=1000\sqrt{3}$$
$$\frac{1000\sqrt{3}}{1000+x}=\frac{1000}{4}$$
$$x^{2}=\ \frac{1000}{(\sqrt{3}-1)}$$
$$x^{2}=500(\sqrt{3}+1)$$
Question 4
1 / -0

The top of hill observed from the top and bottom of a building of height $$h$$ is at the angles of elevation $$p$$ and $$q$$ respectively. The height of the hill is

A
$$\displaystyle \dfrac{h\cot p}{\cot p-\cot q}$$

B
$$\displaystyle \dfrac{h\cot q}{\cot p-\cot q}$$

C
$$\displaystyle \dfrac{h\cot p}{\cot q-\cot p}$$

D
$$\displaystyle \dfrac{h\cot q}{\cot P+\cot q}$$

Solution

$$\tan p=\dfrac{h_{1}-h}{d}$$
$$\tan q= \dfrac{h_{1}}{d}$$

$$\dfrac{\tan p}{\tan q}=1-\dfrac{h}{h_{1}}$$

$$\dfrac{h}{h_{1}}=1-\dfrac{\tan p}{\tan q}$$

$$\dfrac{h_{1}}{h}=\dfrac{\cot p}{\cot p-\cot q}$$

$$h_{1}= \dfrac{h\cot p}{\cot p-\cot q}$$
Question 5
1 / -0

Three points$$A,\ B,\ C$$ are on the straight bank of a river such that $$=BC=2$$ Km. $$A$$ boat is moving towards $$B$$ on a line at right angles to the bank. At some point $$AC$$ subtends an angle of $$60^{0}$$ After moving for 10 minutes in the same direction $$AC$$ subtends an angle of $$120^{0}$$. The speed of boat is

A
$$4\sqrt{3}km/$$ hr

B
$$16\sqrt{3}km/$$ hr

C
$$8\sqrt{3}km/$$ hr

D
$$24\sqrt{3}km/$$ hr

Solution

$$\tan 60= \dfrac 2a $$ -----(1)
$$\tan 30^{\circ} = \dfrac 2 {a+v \times 10}$$
$$\dfrac {2}{ \sqrt 3} = \dfrac {2} {\dfrac {2} {\sqrt 3} +v \times 10}$$
$$2+10\sqrt 3v= 6$$
$$10\sqrt 3y= \dfrac {4\times 60}{10\sqrt 3}$$
$$v_2= 8\sqrt 3 km/hr. $$
Question 6
1 / -0

An observer finds that the angular elevation of a tower is $$\theta$$. On advancing $$a$$ metres towards the tower, the elevation is $$45^{0}$$ and on advancing $$b$$ metres nearer the elevation is $$ 90^{0}-\theta$$, then the height of the tower (in metres) is

A
$$\displaystyle \frac{ab}{a+b}$$

B
$$\displaystyle \frac{ab}{a-b}$$

C
$$\displaystyle \frac{2ab}{a+b}$$

D
$$\displaystyle \frac{2ab}{a-b}$$

Solution

Let $$b$$ be the initial point and $$AE$$ be the height of the tower.
In, $$\triangle ADE,$$
$$\Rightarrow \tan { \left( 90°-\theta \right) } =\dfrac { h }{ x } $$
$$\Rightarrow \cot { \theta } =\dfrac { h }{ x } $$
$$\Rightarrow x=h\tan { \theta } \longrightarrow (1)$$
$$\Rightarrow \tan { \theta } =\dfrac { x }{ h } $$
In, $$\triangle ACE,$$
$$\Rightarrow \tan { 45° } =\dfrac { h }{ b+x } $$
$$\Rightarrow b+x=h$$
$$\Rightarrow b+h\tan { \theta } =h$$ [From $$(1)$$ ]
$$\Rightarrow b=h\left( 1-\tan { \theta } \right) \longrightarrow (2)$$
In, $$\triangle ABE,$$
$$\Rightarrow \tan { \theta } =\dfrac { h }{ a+b+x } $$
$$\Rightarrow \tan { \theta } =\dfrac { h }{ a+h } $$
$$\Rightarrow \dfrac { x }{ h } =\dfrac { h }{ a+h } $$
$$\Rightarrow \dfrac { h-b }{ h } =\dfrac { h }{ a+h } $$
$$\Rightarrow \left( h-b \right) \left( a+h \right) ={ h }^{ 2 }$$
$$\Rightarrow ah-ab+{ h }^{ 2 }-hb={ h }^{ 2 }$$
$$\Rightarrow h\left( a-b \right) =ab$$
$$\Rightarrow h=\dfrac { ab }{ a-b } $$
Hence, the answer is $$\dfrac { ab }{ a-b }.$$
Question 7
1 / -0

Tower is observed from two stations $$A$$and $$B$$ where $$B$$ is East of $$A$$ at a distance $$100m$$. The tower is to north of $$A$$ and to North-West of $$B$$. The angles of elevation of the tower from $$A$$and $$B$$ are complementary. The height of the tower (in meters) is:

A
$$100$$ $$\sqrt{2}$$

B
$$100(2)^{1/4}$$

C
$$100(2)^{-1/4}$$

D
$$100(2)^{1/3}$$

Solution

$$h=AP\tan \theta,h=BP\tan (90^{\circ}-\theta)$$
So $$h^2=AP.BP\cdots(1)$$
Clearly $$AP=AB=100$$
$$\therefore BP=\sqrt{100^2+100^2}=100\sqrt{2}$$
Putting the values of $$AP$$ and $$BP$$ in $$(1),$$ we get
$$h^2=100\times 100\sqrt{2}$$
$$\implies h=100(2)^{1/4}$$
Question 8
1 / -0

An aeroplane flying horizontally $$1$$ km above the ground is observed at an elevation of $$60^{0}$$. If after $$10$$ secs the elevation is observed to be $$30^{0}$$. Then the uniform speed per hour the aeroplane is

A
$$20\sqrt{3}\ km$$

B
$$240\sqrt{3}\ km$$

C
$$256\sqrt{3}\ km$$

D
$$250\sqrt{3}\ km$$

Solution

in $$\triangle AED$$
$$tan60^o=\dfrac{1}{ED}\Rightarrow ED=\dfrac{1}{\sqrt3}$$

Similarly, in $$\triangle ABC$$
$$tan30^o=\dfrac{1}{EC}\Rightarrow EC=\sqrt3$$

Distance travelled by aeroplane in 10 seconds $$=EC-ED=\sqrt3-\dfrac{1}{\sqrt3}=\dfrac{2}{\sqrt3}$$

$$Speed =\dfrac{Distance}{Time}=\dfrac{\dfrac{2}{\sqrt3}}{\dfrac{10}{3600}}=240\sqrt3$$

Therefore, Answer is $$240\sqrt3$$
Question 9
1 / -0

A balloon moving in a straight line passes vertically above two points $$A$$ and $$B$$, on a horizontal plane, $$1000\ ft$$ apart. When above $$A$$, it has an altitude of $$60^{0}$$ as seen from $$B$$ and when above $$B$$ it has an altitude of $$45^{0}$$ as seen from $$A$$. The distance, from $$A$$, of the point at which it will touch the plane is

A
$$2266$$ ft

B
$$2466$$ ft

C
$$2566$$ ft

D
$$2366$$ ft

Solution

It is given that, $$AB=1000$$
Now, in $$\triangle ABE$$
$$tan60^o=\dfrac{AE}{AB}\Rightarrow AE=AB\tan60^o=1000\sqrt3$$

Similarly, in $$\triangle BAD$$
$$tan45^o=\dfrac{DB}{AB}\Rightarrow DB=AB\tan45^o=1000$$

As $$\triangle ACE$$ and $$\triangle DBC$$ are similar
$$\dfrac{AE}{AC}=\dfrac{DB}{BC}\Rightarrow \dfrac{1000\sqrt3}{1000+d}=\dfrac{1000}{d}\Rightarrow \sqrt3=1+\dfrac{1000}{d}\Rightarrow \sqrt3-1=\dfrac{1000}{d}\Rightarrow d=\dfrac{1000}{\sqrt3-1}=1366\ ft$$

Distance from point C to Point A $$=1000+d=1000\ ft+1366\ ft=2366\ ft$$
Question 10
1 / -0

At the foot of a mountain the angle of elevation of a summit is found to be $$45^{ 0}$$. After ascending $$1$$ km towards the mountain up. On a slope of inclination $$30^{0}$$, the angle of elevation is found to be $$60^{0}$$. The height of the mountain is

A
$$400(\sqrt{3}+2)$$ mt

B
$$500(\sqrt{3}+1)$$ mt

C
$$200(\sqrt{2}+1)$$ mt

D
$$100(\sqrt{6}+\sqrt{2})$$ mt