Tag: measurement of an angle

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Questions Related to measurement of an angle

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

Which is the greatest angle in the given set: $\dfrac{1}{3}$ of complete angle, $\dfrac{1}{3}$ of straight angle or a right angle?

  1. $\dfrac{1}{3}$ of complete angle
  2. $\dfrac{1}{3}$ of straight angle

  3. A right angle

  4. All are equal
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Complete angle $=360^{\circ}$

$\dfrac{1}{3}$ complete angle $=\dfrac { 1 }{ 3 } \times { 360 }^{ \circ  }={ 120 }^{ \circ  }$
Right angle $=90^{\circ}$
$\dfrac{1}{3}$ right angle $=\dfrac { 1 }{ 3 } \times { 90 }^{ \circ  }={ 30 }^{ \circ  }$
So $\dfrac{1}{3}$ complete angle is greater.

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

Rank the following angles in descending order. 
1. Straight angle
2. Reflex angle
3. Right angle 

  1. $2$, $1$, $3$
  2. $3$, $2$, $1$
  3. $2$, $3$, $1$
  4. $3$, $1$, $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
  1. Right angle $=90^{\circ}$

    2. Straight angle $=180^{\circ}$

    3. Reflex angle lies between $180^{\circ}$ and $360^{\circ}$

    So the descending order of angles is
    $2>1>3$
Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In a triangle, the angles are in ratio $1: 3: 2$. Find the difference between the greatest and smallest angle of the triangle.

  1. $10^o$
  2. $70^o$
  3. $60^o$
  4. $20^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Le the angles be $x,3x$ and $2x$

Using angle sum property of triangle 
$x+3x+2x={ 180 }^{ \circ  }\ 6x={ 180 }^{ \circ  }\ \Rightarrow x={ 30 }^{ \circ  }$
So the angles are 
$x={ 30 }^{ \circ  }\ 3x=3\times { 30 }^{ \circ  }={ 90 }^{ \circ  }\ 2x=2\times { 30 }^{ \circ  }={ 60 }^{ \circ  }$
Difference between largest and smallest $={ 90 }^{ \circ  }-{ 30 }^{ \circ  }={ 60 }^{ \circ  }$

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

If the difference of two supplementary angles is $40^{\circ}$, then the measurement of the greater angle is

  1. $65^{\circ}$
  2. $110^{\circ}$
  3. $130^{\circ}$
  4. $220^{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the two supplementary angles are $x^{\circ}$ and
$180^{\circ} - x$
By hypothesis, $x - (180^{\circ} - x) = 40^{\circ}$
or $2x - 180^{\circ} = 40^{\circ}$
or  $2x = 220^{\circ}$
or    $x = 110^{\circ}$

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In a $\Delta$ PQR, if $3\sin P+4\cos Q=6$ and $4 \sin Q+3\cos P=1$, then the angle $R$ is equal to :

  1. $\dfrac{3\pi}{4}$
  2. $\dfrac{5\pi}{6}$
  3. $\dfrac{\pi}{6}$
  4. $\dfrac{\pi}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given trignometric equations are:

$3 \sin{P} +4 \cos{Q} =6$ -------(1)
$4 \sin{Q} +3 \cos{P} =1$ -------(2)

Squaring both equations (1) and (2) and adding them, we get
$\Rightarrow 9\left( \sin ^{ 2 }{ P } +\cos ^{ 2 }{ P }  \right) +16\left( \sin ^{ 2 }{ Q } +\cos ^{ 2 }{ Q }  \right) +24\left( \sin { Q } \cos { P } +\cos { Q } \sin { P }  \right) =36+1$

$ \Rightarrow 9+16+24\sin { \left( P+Q \right)  } =37$

$ \therefore \sin { \left( P+Q \right)  } =\cfrac { 37-25 }{ 24 } =\cfrac { 12 }{ 24 } =\cfrac { 1 }{ 2 } $

$\therefore P+Q=30°$

Hence, angle $R=180°-30°=150°=\cfrac { 5\pi  }{ 6 } $radian

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In triangle $ABC,$ if $\dfrac { 1 }{ a+c } +\dfrac { 1 }{ b+c } =\dfrac { 3 }{ a+b+c } ,$ then $\angle c$  is equal to:

  1. $30^{\circ}$
  2. $45^{\circ}$
  3. $60^{\circ}$
  4. $90^{\circ}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given: $\cfrac { 1 }{ a+c } +\cfrac { 1 }{ b+c } =\cfrac { 3 }{ a+b+c } $


$\Rightarrow \quad \cfrac { a+b+2c }{ (a+c)(b+c) } =\cfrac { 3 }{ a+b+c } $

$ \therefore (a+b+2c)(a+b+c)=3(a+c)(b+c)$

$\Rightarrow$ $ { a }^{ 2 }+ab+ac+ab+{ b }^{ 2 }+bc+2ac+2bc+2{ c }^{ 2 }$$ =3(ab+ac+bc+{ c }^{ 2 })$

$ \therefore { a }^{ 2 }+2ab+3ac+{ b }^{ 2 }+3bc+2{ c }^{ 2 }$$ =3ab+3ac+3bc+3{ c }^{ 2 }$

$\Rightarrow$ ${ a }^{ 2 }+{ b }^{ 2 }=ab+{ c }^{ 2 }$

$\Rightarrow$ $ { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 }=ab$

$\Rightarrow$ $ \cfrac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ ab } =1$

$\Rightarrow \cfrac { { a }^{ 2 }+{ b }^{ 2 }-{ c }^{ 2 } }{ 2ab } =\cfrac { 1 }{ 2 } $

$\cos { C } =\dfrac{1}{2}\Rightarrow \angle C={ 60 }^{\circ}$

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In $\Delta ABC\,,\,if\,\,A\,\,:\,\,B\,:\,\,C\, = \,1\,\,:\,\,5\,\,:\,\,6\,\,then$ find the value of $\sin A: \sin B: \sin C$

  1. $\left( {\sqrt 3 \, - \,1} \right)\,:\,2\sqrt 2 \,:\,\left( {\sqrt 3 \, + \,1} \right)$
  2. $2\sqrt 2 \,:\,\left( {\sqrt 3 \, - \,1} \right)\,:\,\left( {\sqrt 3 \, + \,1} \right)$
  3. $ \,\left( {\sqrt 3 \, - \,1} \right)\,:\,\left( {\sqrt 3 \, + \,1} \right)\,:\,2\sqrt 2 $
  4. $ \,\left( {\sqrt 3 \, - \,1} \right)\,:\,\sqrt 3 :\,\sqrt 2 $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given that $ A:B:C = 1:5:6 $


We know that the sum of angles in a triangle is $180^0$

Let us first find each angle.

Total no.of parts= $1+5+6= 12$

$A$= $ \dfrac{1}{12} $ ($180^0$) =$15^0$

$B$= $ \dfrac{5}{12} $ ($180^0$) =$75^0$

$C$= $ \dfrac{6}{12} $ ($180^0$) =$90^0$

Hence, $ \sin A: \sin B: \sin C$ = $\sin 15^o$$:\sin 75^o$$:\sin 90^o$

$=\dfrac {\sqrt{3}-1} {2\sqrt 2} : \ \dfrac{\sqrt3+1}{2\sqrt2 } :1$ 

$= ({\sqrt{3}-1})  : ({\sqrt3+1} ): ({2\sqrt2 }) $ 

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

In $\Delta ABC$ and $\Delta DEF$, we have $\dfrac {AB}{DE}=\dfrac {BC}{FD}$. Triangles ABC and DEF will be similar if :

  1. $\angle A = \angle D$
  2. $\angle A = \angle F$
  3. $\angle B = \angle E$
  4. $\angle B = \angle D$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For two triangles to be similar by SAS similarity, the included angle between the proportional sides must be equal. Given AB/DE = BC/FD, the included angles are angle B and angle D, so angle B must equal angle D.

Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

Using ruler and compasses only, construct a triangle POR such that $\angle P = 120^{\circ}$, PO = 5 cm PR = 6 cm.In the same figure, find a point which is equidistant from its sides. Name this point With this point as centre draw a circle touching all the sides of the triangle.

  1. Circumcentre

  2. Incentre

  3. Mid point

  4. Data insufficient
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

(I) We draw a triangle POR with $\quad \angle RPO={ 120 }^{ O }.\quad $

(II) OI & OR are the angular bisectors of $\quad \angle RPO\quad & \angle ROP\quad $
The bisectors intersect at I.
(i) IM & IN  are drawn perpendiculars from i to PR & PO respectively. 
 (iii) The circle which touches the sides of the triangle POR.
has the radius  IM=IN.
Justification-
Between $\quad \Delta IPM\quad & \quad \Delta IPN,\ \angle IMP={ 90 }^{ o }=\angle INP,\ \angle IPM=\angle IPN\quad $
So the third angles  $\quad \angle PIN=\angle PIM\quad $
Also the side IP is common.
So, by ASA rule, $\quad \Delta IPM\equiv \Delta IPN.\quad $  
i.e IM=IN.
Similarly, by considering the triangles INO & ISO it can be shown that
IN=IS.
So IM=IN=IS.
i.e the circle with centre I touches the sides of the given triangle.
So  I is the INCENTRE of the triangle POR.
Ans- Option B.