Tag: maths

Questions Related to maths

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?

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

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

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

  3. A right angle

  4. All are equal

Show answer
Correct Option: A
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.

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

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

  1. $2$, $1$, $3$

  2. $3$, $2$, $1$

  3. $2$, $3$, $1$

  4. $3$, $1$, $2$

Show answer
Correct Option: A
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
    $2>1>3$

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

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

  1. $10^o$

  2. $70^o$

  3. $60^o$

  4. $20^o$

Show answer
Correct Option: C
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 $={ 90 }^{ \circ  }-{ 30 }^{ \circ  }={ 60 }^{ \circ  }$

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

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

  1. $65^{\circ}$

  2. $110^{\circ}$

  3. $130^{\circ}$

  4. $220^{\circ}$

Show answer
Correct Option: B
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    $x = 110^{\circ}$

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 :

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

  1. $\dfrac{3\pi}{4}$

  2. $\dfrac{5\pi}{6}$

  3. $\dfrac{\pi}{6}$

  4. $\dfrac{\pi}{4}$

Show answer
Correct Option: B
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

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

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

  1. $30^{\circ}$

  2. $45^{\circ}$

  3. $60^{\circ}$

  4. $90^{\circ}$

Show answer
Correct Option: C
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 } $

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

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

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

  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 $

Show answer
Correct Option: C
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 }) $ 

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

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

  1. $\angle A = \angle D$

  2. $\angle A = \angle F$

  3. $\angle B = \angle E$

  4. $\angle B = \angle D$

Show answer
Correct Option: D

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.

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

  1. Circumcentre

  2. Incentre

  3. Mid point

  4. Data insufficient

Show answer
Correct Option: B
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.

The measure of the trisected angle of $\displaystyle 162^{\circ}$ is:

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

  1. $\displaystyle 54^{\circ}$

  2. $\displaystyle 81^{\circ}$

  3. $\displaystyle 162^{\circ}$

  4. $\displaystyle 90^{\circ}$

Show answer
Correct Option: A
Explanation:

Trisected angle $\displaystyle =\frac{162^{\circ}}{3}=54^{\circ}$