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

The difference between two angles is $19$$\displaystyle ^{o}$ and their sum is $\displaystyle \frac{890}{9}^o$. Find the greater angle.

  1. $63^o$
  2. $35^o$
  3. $27^o$
  4. $59^o$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the two angles be $a$ and $b$

Then, 
$a  - b = 19$

and, $a + b = \dfrac{890}{9}$

Adding the two equations,

$2a = \dfrac{890 + 19\times 9}{9}$

$2a = \dfrac{1061}{9}$

Thus, $a = \dfrac{1061}{18}$

$a \approx 59^{\circ}$
Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

If two angles of a triangle are acute angles, the third angle:

  1. is less than the sum of the two angles

  2. is an acute angle

  3. is the largest angle of the triangle

  4. may be an obtuse angle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
For a triangle $ABC$, sum of angles is $180^0$
Two angles are given as acute, $\angle A, \angle B < 90^0$
$\angle A = 90^0 - x$
$\angle B = 90^0 - y$
where, $x,y < 90^0$
$\therefore \angle A + \angle B+ \angle C = 180^0$
$\Rightarrow \angle C = x+y $
$x+y$ can be $>90^0$ or $<90^0$
So, it may be obtuse or acute.
Hence, option D.
Multiple choice maths complementary angle, supplementary angles and adjcent angles acute and obtuse angles types of angle measurement of an angle

An angle which measures $\displaystyle 0^{o}$ is called:

  1. obtuse angle

  2. straight angle

  3. zero angle

  4. right angle
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An angle which measure $0^0$ is called zero angle.

An angle which measure more than $0^o$ and less than $90^o$ is an acute angle.
An angle which measures $90^o$ is called a right angle.
An angle which measure more than $90^o$ and less than $180^o$ is an obtuse angle.

Hence, the answer is zero angle.

Multiple choice intersection of a line and a parabola conic section maths

The length of the chord of the parabola $y^2 = 4x$ which passes through the vertex and makes $30^o$ angle with x-axis is

  1. $\dfrac{\sqrt{3}}{2}$
  2. $\dfrac{3}{2}$
  3. $8\sqrt{3}$
  4. $\sqrt{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The chord passes through the vertex (0,0) and makes 30 degrees with the x-axis, so its equation is y = tan(30)x = x/sqrt(3). Substituting into y^2 = 4x gives x^2/3 = 4x, so x = 12. The length is sqrt(x^2 + y^2) = sqrt(144 + 144/3) = sqrt(144 * 4/3) = 12 * 2/sqrt(3) = 8*sqrt(3).

Multiple choice intersection of a line and a parabola conic section maths

If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is 

  1. $\sqrt {{c^2} - {{\left( {a - b} \right)}^2}} $
  2. $\sqrt {4{c^2} - 2{{\left( {a - b} \right)}^2}} $
  3. $\sqrt {3{c^2} - {{\left( {a - b} \right)}^2}} $
  4. $\sqrt {2{c^2} - {{\left( {a - b} \right)}^2}} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$ 
${{S} _{1}}\equiv {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$               …….. (1)

 

${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$

${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$                        ……… (2)

 

Since, $a\ne b$

 

Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.

 

We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$

 

So, the equation is

$\left( b-a \right)x+\left( a-b \right)y=0$             …….. (3)

 

We know that the length of common chord is

$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$                      ………. (4)

Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.

 

So,

${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$

$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $

$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $

 

From equation (4),

The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $

$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $

$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $

 

Hence, this is the answer.

Multiple choice intersection of a line and a parabola conic section maths

If a$\ne $b then the length of common chord of the circles ${\left( {x - a} \right)^2} + {\left( {y - b} \right)^2} = {c^2}$ and ${\left( {x - {b^{}}} \right)^2} + {\left( {y - a} \right)^2} = c^2$ is 

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

We have,

${{\left( x-a \right)}^{2}}+{{\left( y-b \right)}^{2}}={{c}^{2}}$ 
${{S} _{1}}\equiv {{x}^{2}}+{{a}^{2}}-2ax+{{y}^{2}}+{{b}^{2}}-2by-{{c}^{2}}=0$               …….. (1)

 

${{\left( x-b \right)}^{2}}+{{\left( y-a \right)}^{2}}={{c}^{2}}$

${{S} _{2}}\equiv {{x}^{2}}+{{b}^{2}}-2bx+{{y}^{2}}+{{a}^{2}}-2ay-{{c}^{2}}=0$                        ……… (2)

 

Since, $a\ne b$

 

Centre of the circle ${{S} _{1}}=\left( a,b \right)$ and radius ${{r} _{1}}=c$.

 

We know that the equation of common chord is ${{S} _{1}}-{{S} _{2}}=0$

 

So, the equation is

$\left( b-a \right)x+\left( a-b \right)y=0$             …….. (3)

 

We know that the length of common chord is

$=2\sqrt{{{r} _{1}}^{2}-{{d} _{1}}^{2}}$                      ………. (4)

Where ${{r} _{1}}=$ radius and ${{d} _{1}}$ is the length of perpendicular drawn from the centre to the chord.

 

So,

${{d} _{1}}=\left| \dfrac{\left( b-a \right)a+\left( a-b \right)b}{\sqrt{{{\left( b-a \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right|$

$ {{d} _{1}}=\left| \dfrac{ab-{{a}^{2}}+ab-{{b}^{2}}}{\sqrt{{{\left( a-b \right)}^{2}}+{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{2ab-{{a}^{2}}-{{b}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-{{\left( a-b \right)}^{2}}}{\sqrt{2{{\left( a-b \right)}^{2}}}} \right| $

$ {{d} _{1}}=\left| \dfrac{-\left( a-b \right)}{\sqrt{2}} \right| $

$ {{d} _{1}}=\dfrac{\left( a-b \right)}{\sqrt{2}} $

 

From equation (4),

The length of common chord $ =2\sqrt{{{c}^{2}}-{{\left( \dfrac{a-b}{\sqrt{2}} \right)}^{2}}} $

$ =2\sqrt{{{c}^{2}}-{{\dfrac{\left( a-b \right)}{2}}^{2}}} $

$ =2\sqrt{{{\dfrac{2{{c}^{2}}-\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{{{\dfrac{8{{c}^{2}}-4\left( a-b \right)}{2}}^{2}}} $

$ =\sqrt{4{{c}^{2}}-2{{\left( a-b \right)}^{2}}} $

 

Hence, this is the answer.

Multiple choice intersection of a line and a parabola conic section maths

The length of normal chord to the parabola $y^{2} = 4x$ which subtends a right angle at the vertex is

  1. $6\sqrt {3}$
  2. $6\sqrt {2}$
  3. $7\sqrt {2}$
  4. $7\sqrt {3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A chord is a line segment that passes through any two points on the parabola. A normal chord is a chord that is perpendicular to a tangent of the parabola at the point of intersection of the chord with the parabola. 


Let $y=mx+c$   is the tangent of parabola.


As given parabola is $y^2=4x$

So any point on the parabola is $(t^2,2t)$

As $y=mx+\dfrac{a}{m}$ , is the tangent to the parabola $y^2=4ax$

then the tangent equation to this given parabola is $y=mx+\dfrac{1}{m}$

Let the tangent passes through point $P(t _1^{2},2t _1)$

On substituting P in parabola equation we get $m=\dfrac{1}{t _1}$

So the slope of the normal is $-t _1$

Let the chord joins $P(t _1^2,2t _1)$ and $Q(t _2^2,2t _2)$ 

On solving we will get slope of line PQ as $\dfrac{2}{t _1+t _2}$

So , $\dfrac{2}{{t _1}+{t _2}}= \dfrac{1}{t _1}$

$\Rightarrow t _1^{2}+{t _1}{t _2}=-2$

As from properties of a normal chord which subtends a right angle at the vertex, ${t _1}{t _2}=-4$

On solving above two equations we get $ t _1=\sqrt{2} , t _2=-2\sqrt{2} $

Hence the points are $P(2,2\sqrt{2})$ and $Q(8,-4\sqrt{2}) $

By applying distance formula we get the distance between P and Q as

$\Rightarrow PQ=\sqrt{(8-2)^2+(-4\sqrt{2}-2\sqrt{2})^2}$

$\Rightarrow PQ=\sqrt{108}=6\sqrt{3} units $