Some Applications of Trigonometry test

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Some Applications of Trigonometry test - 56

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Question 1
1 / -0

The angle of elevation of an aeroplane from a point $$30$$ metres above the water level is $$45^{0}$$, and the angle of depression of its reflection in the lake is $$60^{ }$$. The height of the aeroplane above the lake in metres is

A
$$30(2+\sqrt{3})$$

B
$$30(2-\sqrt{3})$$

C
$$30(\sqrt{5}+1)$$

D
$$30(\sqrt{5}-1)$$
Question 2
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A flag staff of height $$(a-b)$$ stands on the top of a tower subtends the same angle at point on the horizontal plane through the foot of the tower which are at distant $$a$$ and $$b$$ form the tower. The height of the tower is

A
$$b$$

B
$$a+b$$

C
$$a$$

D
$$a-b$$
Question 3
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From the top of a tree on one side of a street the angles of elevation and depression of the top and foot of a tower on the opposite side are respectively found to be $$\alpha$$ and $$\beta$$. If $$h$$ is the height of the tree, then the height of the tower is:

A
$$\displaystyle \frac{h\sin(\alpha+\beta)}{\cos\alpha\sin\beta}$$

B
$$\displaystyle \frac{h\sin(\alpha+\beta)}{\sin\alpha\cos\beta}$$

C
$$\displaystyle \frac{h\cos(\alpha-\beta)}{\cos\alpha\cos\beta}$$

D
$$\displaystyle \frac{h\cos(\alpha+\beta)}{\cos\alpha\cos\beta}$$

Solution

Let $$h_{t}+h=H $$ be the total height of tower.
Let $$x$$ be the distance between tower and tree.
$$\displaystyle \tan \beta= \frac{h}{x}$$

$$\displaystyle \tan \alpha= \frac{h_{t}}{x}$$

Now,
$$\displaystyle \tan \alpha= \frac{h_{t}}{h \cot \beta}$$

$$\displaystyle h_{t}=h \dfrac{\tan \alpha}{\tan \beta}$$

Total height of the tower,
$$\displaystyle H= h+h \dfrac{\tan \alpha}{tan \beta}$$

$$\displaystyle = \dfrac{h \tan \alpha + h \tan \beta }{\tan \beta }$$

Substituting for $$\tan \alpha$$ and $$\tan \beta$$
we get,
$$\displaystyle H= h \left ( \frac{\sin \beta \cos \alpha +\sin \alpha \cos \beta}{\sin \beta \cos \alpha } \right )$$
Hence, option 'A' is correct.
Question 4
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From the top of cliff $$300\ meters$$ high, the top of tower was observed at an angle of depression $${30}^{\circ}$$ and from the foot of the tower the top of the cliff was observed at an angle of elevation $${45}^{\circ}$$. The height of the tower is

A
$$50\left( 3-\sqrt { 3 } \right) m$$

B
$$200\left( 3-\sqrt { 3 } \right) m$$

C
$$100\left( 3-\sqrt { 3 } \right) m$$

D
$$None\ of\ these$$

Solution

Let $$AB$$ be the cliff with $$AB=300\ m$$
Let $$DE$$ be the tower with height $$h\ $$
Let $$BE=x\ m$$
In right angles $$\triangle ABE$$,
$$\displaystyle \frac { AB }{ BE } =\tan { { 45 }^{ 0 } }$$

$$\Rightarrow \dfrac { 300 }{ x } =1$$

$$\Rightarrow x=300\ m$$

In right angled $$\triangle ACD$$

$$\displaystyle \dfrac { AC }{ CD } =\tan { { 30 }^{ 0 } } $$

$$\Rightarrow \dfrac { 300-h }{ x } =\dfrac { 1 }{ \sqrt { 3 } } $$

$$\displaystyle \Rightarrow \frac { 300-h }{ 300 } =\frac { 1 }{ \sqrt { 3 }}$$

$$\Rightarrow 300\sqrt { 3 } -\sqrt { 3 } h=300$$

$$\Rightarrow 300\sqrt { 3 } -300=\sqrt { 3 } h$$

$$\displaystyle \Rightarrow h=\frac { 300\left( \sqrt { 3 } -1 \right) }{ \sqrt { 3 }}$$

By rationalize the denominator , we get
$$ \Rightarrow h=100\sqrt { 3 } \left( \sqrt { 3 } -1 \right) m\quad \\ \Rightarrow h=100\left( 3-\sqrt { 3 } \right)\ m$$
Question 5
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A $$25\ m$$ ladder is placed against a vertical wall of a building. The foot of the ladder is $$7\ m$$ from the base of the building. If the top of the ladder slips $$4\ m$$, then the foot of the ladder will slide by:

A
$$5\ m$$

B
$$8\ m$$

C
$$9\ m$$

D
$$15\ m$$

Solution

The ladder forms a right angled triangle with the wall and the base with length of the ladder being the length of hypotenuse of this triangle.
Hypotenuse $$= 25$$ m
Base $$=7$$ m
Height $$=25^2-7^2=24$$ m
Now the ladder slides. In the new position, the hypotenuse of the new right angled triangle will be the same.
The top slides down by a distance $$4m$$. Hence, the height is $$24-4=20$$ m. Let the base slide by a distance $$x$$. Therefore, hypotenuse $$= 25$$ m
Base $$=(7+x)$$ m
Height $$=20$$ m
By Pythagoras Theorem, we get
$$25^2-20^2 = (7+x)^2$$
$$\therefore (7+x) = 15$$
Hence, $$x=8$$
The correct answer is Option B.
Question 6
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A lamp post standing at a point $$A$$ on a circular path of radius $$r$$ subtends an angle $$\alpha$$ at some point $$B$$ on the path, and $$AB$$ subtends an angle of $$45^{\circ}$$ at any other point on the path, the height of the lamp post is

A
$$\sqrt{2}r\cot\alpha$$

B
$$\displaystyle \frac{r}{\sqrt{2}}\tan\alpha$$

C
$$\sqrt{2}r\tan\alpha$$

D
$$\displaystyle \frac{r}{\sqrt{2}}\cot\alpha$$

Solution

Let $$AM$$ is the lamp post at point $$A$$ of height $$h.$$

$$AB$$ subtends an angle of $$45^\circ$$ at any point on the circle. Therefore it will subtend an angle of $$90^\circ$$ at the center.

So, we can say that $$\triangle OAB$$ is a right-angled triangle with a right angle at $$O.$$

By applying pythagoras theorem,
$$\begin{aligned}{}O{A^2} + O{B^2}& = A{B^2}\\{r^2} + {r^2} &= A{B^2}\\A{B^2}& = 2{r^2}\\AB& = r\sqrt 2 \end{aligned}$$

In $$\triangle AMB,$$
$$\tan \alpha = \dfrac {AM}{AB}$$

$$\Rightarrow \tan \alpha = \dfrac {h}{r \sqrt 2}$$

$$\Rightarrow h = r \sqrt 2 \tan \alpha$$
Question 7
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A person walking along a canal observes that two objects are in the same line which is inclined at an angle $$\alpha$$ to the canal. He walks a distance $$C$$ for the and observes that the objects subtended their greatest angle $$\beta$$ then the distance between the object is

A
$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\beta-\cos\alpha}$$

B
$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha-\cos\beta}$$

C
$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$$

D
$$\displaystyle \frac{2C\cos\alpha\cos\beta}{\sin\alpha-\sin\beta}$$

Solution

$$\tan \alpha = \dfrac {h_1}{a} - (1), \tan \alpha = \dfrac {h_2}{a + b} - (2), \tan \beta = \dfrac {h_1}{a - c} - (3)$$
$$\tan \beta = \dfrac {h_2}{a + b - c} - (4)$$
$$\tan \beta (a + b -c) = \tan \alpha (a + b)$$ (From (1) & (3))
$$\tan \alpha a =\tan \beta (a - c)$$ (From (2) and (4))
$$a = \dfrac {\tan \beta c}{\tan \beta - \tan \alpha}$$
Question 8
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Vertical pole $$PO$$ stands at the centre $$O$$ of a square $$ABCD$$. If $$AC$$ subtends an angle at the top $$P$$ of the pole, then the angle $$90^{0}$$ subtended by a side of the square at $$p$$ is

A
$$30^{0}$$

B
$$45^{0}$$

C
$$60^{0}$$

D
$$90^{0}$$
Question 9
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At the foot of a mountain the elevation of its peak is found to be $$\displaystyle \frac{\pi }{4}.$$ After ascending $$ h $$ metres toward the mountain up a slope of $$\displaystyle \frac{\pi }{6}$$ inclination, the elevation is found to be $$\displaystyle \frac{\pi }{3}.$$ Height of the mountain is

A
$$\displaystyle \frac{h}{2}(\sqrt{3}+1)m$$

B
$$h(\sqrt{3}+1)m$$

C
$$\displaystyle \frac{h}{2}(\sqrt{3}-1)m$$

D
$$h(\sqrt{3}-1)m$$

Solution

Let 'A' be the top of hill and 'P' be a point on it's foot.
We have
$$\angle APB=\dfrac{\pi }{4}, \angle QPB=\dfrac{\pi }{6}, \angle AQR=\dfrac{\pi }{3}, PQ=h\ m$$
In $$\Delta APB,$$
$$PB=AB \cot \dfrac{\pi }{4}=AB$$
In $$\Delta PQ_{1}Q$$
$$QQ_{1}=PQ \sin \displaystyle \frac{\pi }{6}=\dfrac{h\sqrt{3}}{2}$$
In $$\Delta AQR, \tan \displaystyle \dfrac{\pi }{3}=\dfrac{AR}{QR}=\dfrac{AB-QQ_{1}}{PB-PQ_{1}}$$
$$\Rightarrow \displaystyle \dfrac{AB-\dfrac{h}{2}}{AB-\dfrac{h\sqrt{3}}{2}}=\sqrt{3}$$
$$\Rightarrow AB=\displaystyle \dfrac{h}{(\sqrt{3}-1)}=\dfrac{h(\sqrt{3}+1)}{2}\ m$$
Question 10
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The angle of elevation from a point on the bank of a river to the top of a temple on the other bank is $$45^o$$. Retreating $$50\ m$$, the observer finds the new angle of elevation as $$30^{\circ}$$. What is the width of the river ?

A
$$50\ m$$

B
$$50\sqrt { 3 }\ m $$

C
$$\dfrac {50}{(\sqrt { 3 } -1)} m$$

D
$$100\ m$$

Solution

Let the height of the temple be h.
And let length of the bank of the river be y.
Then
$$\tan45^{0}=\dfrac{h}{y}=1$$
Hence
$$h=y$$ ...(i)
and
$$\tan30^{0}=\dfrac{1}{\sqrt{3}}=\dfrac{h}{y+50}$$
Or
$$\sqrt{3}y=y+50$$ ...$$(h=y)$$ from $$(i)$$
$$y=\dfrac{50}{\sqrt{3}-1}m$$

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Some Applications of Trigonometry test - 56

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Some Applications of Trigonometry test - 56

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SHARING IS CARING

Question 1

$$30(2+\sqrt{3})$$

$$30(2-\sqrt{3})$$

$$30(\sqrt{5}+1)$$

$$30(\sqrt{5}-1)$$

Question 2

$$b$$

$$a+b$$

$$a$$

$$a-b$$

Question 3

$$\displaystyle \frac{h\sin(\alpha+\beta)}{\cos\alpha\sin\beta}$$

$$\displaystyle \frac{h\sin(\alpha+\beta)}{\sin\alpha\cos\beta}$$

$$\displaystyle \frac{h\cos(\alpha-\beta)}{\cos\alpha\cos\beta}$$

$$\displaystyle \frac{h\cos(\alpha+\beta)}{\cos\alpha\cos\beta}$$

Solution

Question 4

$$50\left( 3-\sqrt { 3 } \right) m$$

$$200\left( 3-\sqrt { 3 } \right) m$$

$$100\left( 3-\sqrt { 3 } \right) m$$

$$None\ of\ these$$

Solution

Question 5

$$5\ m$$

$$8\ m$$

$$9\ m$$

$$15\ m$$

Solution

Question 6

$$\sqrt{2}r\cot\alpha$$

$$\displaystyle \frac{r}{\sqrt{2}}\tan\alpha$$

$$\sqrt{2}r\tan\alpha$$

$$\displaystyle \frac{r}{\sqrt{2}}\cot\alpha$$

Solution

Question 7

$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\beta-\cos\alpha}$$

$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha-\cos\beta}$$

$$\displaystyle \frac{2C\sin\alpha\sin\beta}{\cos\alpha+\cos\beta}$$

$$\displaystyle \frac{2C\cos\alpha\cos\beta}{\sin\alpha-\sin\beta}$$

Solution

Question 8

$$30^{0}$$

$$45^{0}$$

$$60^{0}$$

$$90^{0}$$

Question 9

$$\displaystyle \frac{h}{2}(\sqrt{3}+1)m$$

$$h(\sqrt{3}+1)m$$

$$\displaystyle \frac{h}{2}(\sqrt{3}-1)m$$

$$h(\sqrt{3}-1)m$$

Solution

Question 10

$$50\ m$$

$$50\sqrt { 3 }\ m $$

$$\dfrac {50}{(\sqrt { 3 } -1)} m$$

$$100\ m$$

Solution

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