Tag: trigonometrical ratios

Questions Related to trigonometrical ratios

Find number of solutions to the equation:$[ \sin x + \cos x ] = 3 + [ - \sin x ] + [ - \cos x ]$

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. 0

  2. 1

  3. 2

  4. Infinite

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Correct Option: D
Explanation:
$\sin {x}+\cos x=3-\sin x-\cos x$
$2(\sin x+\cos x)=3$
$\sin x+\cos x=1.5$
so these are infinite value of $x$ for which this equation will satisfy

if $\displaystyle Sin\theta =\frac{3}{5}$ what is the value of $\displaystyle  \left ( \tan \theta +\sec \theta  \right )^{2}$?

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. $2$

  2. $3$

  3. $4$

  4. $-4$

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Correct Option: C
Explanation:

$Sin \theta=\dfrac{P}{H}=\dfrac{3}{5}$

According to the Pythagorean therom
$H^2=B^2+P^2$
$\Rightarrow B=\sqrt{h^2-p^2}$
$\Rightarrow B=\sqrt{5^2-3^2}$
$\Rightarrow B=\sqrt{16}$
$\Rightarrow B=4 cm$
$\therefore tan \theta= \dfrac{P}{B}=\dfrac{3}{4}$
$sec \theta=\dfrac{H}{B}=\dfrac{5}{4}$
$\therefore (tan \theta+sec \theta)^2=(\dfrac{3}{4}+\dfrac{5}{4})^2$
$\Rightarrow (\dfrac{8}{4})^2=(2)^2=4$

The side opposite to the right angle in a right angled triangle is called

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. Base

  2. Perpendicualr

  3. Hypotenuse

  4. None of these

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Correct Option: C
Explanation:

The side opposite to the right angle in a right angled triangle is called Hypotenuse.

The area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the difference of the areas of the semicircles drawn on the other two sides of the triangles.

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. True

  2. False

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Correct Option: B

If $E. \ tan(x -
30^{\circ}) = j. \ tan(x+120^{\circ})$, then $\frac{E + J}{E-J} =$

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. $\ sin 2x$

  2. $2 \ cos 2x$

  3. $\ tan2x$

  4. None of these.

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Correct Option: B
Explanation:
Given $E\tan (x-30^{\circ})=J\tan (x+120^{\circ})$

$\implies \dfrac{E}{J}=\dfrac{\tan (x+120^{\circ})}{\tan (x-30^{\circ})}$

Applying compoundo and dividendo rule

$\dfrac{E+J}{E-J}=\dfrac{\tan (x-30^{\circ})+\tan (x+120^{\circ})}{\tan (x+120^{\circ})-\tan (x-30^{\circ})}=\dfrac{\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}+\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}}{\frac{\sin (x+120^{\circ})}{\cos (x+120^{\circ})}-\frac{\sin (x-30^{\circ})}{\cos (x-30^{\circ})}}$

                                                        $=\dfrac{\sin (x-30^{\circ})\cos(x+120^{\circ})+\sin (x+120^{\circ})\cos(x-30^{\circ})}{\sin (x+120^{\circ})\cos(x-30^{\circ})-\sin (x-30^{\circ})\cos (x+120^{\circ})}$

                                                       $=\dfrac{\sin (x-30^{\circ}+x+120^{\circ})}{\sin (x+120^{\circ}-x+30^{\circ})}$

                                                      $=\dfrac{\sin (90^{\circ}+2 x)}{\sin 150^{\circ}}$

                                                      $=2\cos 2 x$

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height 5 meters. At point on the plane, the angle of elevation of the bottom and top of the flag staff are respectively 30$^{\circ}$ and 60$^{\circ}$. The height of tower is

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. 2m

  2. 5m

  3. 2.5m

  4. 3m

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Correct Option: C
Explanation:

Let AB be the tower of height h meter and BC be the height of flag staff surmounted on the tower. Let the point of the plane be D at a distance m meter from the foot of the tower.

In $\bigtriangleup$ ABD,

$\displaystyle tan 30^{\circ} = \frac{AB}{BD}$

$\displaystyle \Rightarrow \frac{1}{\sqrt3} = \frac{h}{x} \Rightarrow x = \sqrt 3 h$.......(1)

In $\displaystyle \bigtriangleup ADC, tan 60^{\circ} = \frac{AC}{AD}$

$\displaystyle \Rightarrow \sqrt3 = \frac{5+h}{x} \Rightarrow x = \frac{5 +h}{\sqrt3}$ .....(2)

From (1) and (2), $\displaystyle \sqrt 3h = \frac{5+h}{\sqrt 3}$

$\Rightarrow 3h = 5 + h \Rightarrow 2h = 5$

$\Rightarrow \displaystyle h = \frac{5}{2} = 2.5 m$

So, the height of tower = 2.5m

If the angle of elevation of a cloud from a point 200 meter above a lake is $\displaystyle 30^{\circ}$ and the angle of depression of its reflection in the lake is $\displaystyle 60^{\circ}$ then the height of the cloud (in meters )above the lake is 

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. $200$

  2. $300$

  3. $500$

  4. $None$

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Correct Option: D
Explanation:

$\tan { 60°=\sqrt { 3 }  } =\dfrac { p }{ b } $

$\tan { 30°=\dfrac { \sqrt { 3 }  }{ 3 }  } =\dfrac { p }{ b } $
$b=200\sqrt { 3 } $
$\tan { 60°=\sqrt { 3 }  } =\dfrac { p }{ 200 } $
$p=200\sqrt { 3 } $
height of the cloud (in meters )above the lake = $200+200\sqrt { 3 } $
Answer none 

If the distance between a 13-foot ladder and a vertical wall is $5$ feet along the ground, how high can a person climb if the ladder is inclined against wall?

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. $18$ feet

  2. $65$ feet

  3. $\cfrac{13}{5}$ feet

  4. $8$ feet

  5. $12$ feet

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Correct Option: E
Explanation:

Length of Ladder = 13 ft

Length between Ladder and Wall = 5 ft
by pythagoras theorum
(hypotenuse)^2 = (perpendicular)^2 + (base)^2
(13)^2 = (height of wall)^2+(5)^2
by solving
Height of Wall = 12 ft

If $sin\theta = 3sin(\theta +2\alpha)$, then the value of $tan(\theta+\alpha)+ 2tan\alpha$ is

physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

  1. 3

  2. 2

  3. 1

  4. 0

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Correct Option: D
Explanation:
$\sin \theta =3\sin \left( \theta +2\alpha\right)$

$\Rightarrow 3\sin \left( \theta +\alpha +\alpha \right)=\sin \theta$

      $3\sin \left(\theta+\alpha\right)\cos \alpha +3\cos \left(\theta+\alpha\right)\sin \alpha =\sin \left(\theta+\alpha-\alpha \right)$

      $3\sin \left(\theta+\alpha\right)\cos\alpha +3\cos \left(\theta+\alpha\right)\sin \alpha  =\sin \left(\theta+\alpha\right) \cos \alpha -\sin \alpha \cos \left(\theta+\alpha\right)$

      $2\sin\left(\theta+\alpha\right)\cos \alpha =-4\cos \left(\theta+\alpha\right)\sin \alpha$

      $2\tan \left(\theta+\alpha\right)=-4\tan \alpha$

      $\tan \left(\theta+\alpha\right)=-2\tan \alpha$

$\Rightarrow \tan \left(\theta+\alpha\right)+2\tan \alpha=0$

Hence, the answer is $0.$