Tag: angle and their measurement

Quizzes (6) Questions (67)

Questions Related to angle and their measurement

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

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.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Pythagorean theorem states that a^2 + b^2 = c^2. The area of a semicircle is (pi * r^2) / 2. For a triangle with sides a, b, and hypotenuse c, the areas are (pi * (a/2)^2) / 2, (pi * (b/2)^2) / 2, and (pi * (c/2)^2) / 2. Since a^2 + b^2 = c^2, the sum of the areas of the semicircles on the legs equals the area of the semicircle on the hypotenuse, not the difference.

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

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

  1. $\ sin 2x$
  2. $2 \ cos 2x$
  3. $\ tan2x$

  4. None of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
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

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

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

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

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

                                                      $=2\cos 2 x$
Multiple choice physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

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

  1. 2m

  2. 5m

  3. 2.5m

  4. 3m
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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 

  1. $200$
  2. $300$
  3. $500$
  4. $None$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 

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

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?

  1. $18$ feet
  2. $65$ feet
  3. $\cfrac{13}{5}$ feet
  4. $8$ feet
  5. $12$ feet
Reveal answer Fill a bubble to check yourself
E Correct answer
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

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

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

  1. 3

  2. 2

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $0.$