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Lines and Angles Test - 15

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Lines and Angles Test - 15
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  • Question 1
    1 / -0
    Find the angle which is $$30^o$$ more than its complement.
    Solution
    Let the required angle be $$ x$$.
    Then, its complement $$=(90^o-x)$$.
    Given, the angle is $$30^o$$ more than its complement.
    Then, $$ x=(90^o-x)+30^o$$
    $$\Rightarrow x+x=90^o+30^o$$
    $$ \Rightarrow 2x=120^o\\ \Rightarrow x=\dfrac { 120 }{ 2 } ^o\\ \Rightarrow x={ 60 }^{ o }$$.

    Therefore, option $$C$$ is correct.
  • Question 2
    1 / -0
    A ray stands on a line, then the sum of the two adjacent angles so formed is ______.
    Solution

    If a ray stands on a line, then the sum of two adjacent angles so formed is equal to $$180^\circ$$ which is also known as a linear pair.

  • Question 3
    1 / -0
    Find the angle which is equal to its supplement
    Solution
    Let the required angle be $$x$$, then its supplement $$=(180-x)\quad $$ 
    Given that $$ x=(180-x)$$
    $$ \Rightarrow 2x=180\\ \Rightarrow x={ 90 }^{ o }
    $$
  • Question 4
    1 / -0
    If a transversal intersects a pair of lines in such a way that the sum of interior angles on the same side of transversal is $$180^o$$, then the lines are 
    Solution
    If a transversal intersects a pair of lines in such a way that the sum of interior angles on the same side of transversal is $$180^{\circ}$$ then the lines are parallel.
    Option $$A$$ is correct.
  • Question 5
    1 / -0
    Find the angle which is $$20^o$$ more than its supplement.
    Solution
    Let the required angle be $$x$$, then its supplement $$=(180-x)\quad $$
    Given that $$ x=(180-x)+20$$
    $$ \Rightarrow 2x=200\\ \Rightarrow x={ 100 }^{ o }
    $$
  • Question 6
    1 / -0
    If an angle is $$28^o$$ less than its complement, find its measure.
    Solution
    Let the required angle be $$ x$$.
    Then, its complement $$=(90^o-x)$$.
    Given, the angle is $$28^o$$ less than its complement.
    Then, $$ x=(90^o-x)-28^o$$
    $$\Rightarrow x+x=90^o-28^o$$
    $$ \Rightarrow 2x=62^o\\ \Rightarrow x=\dfrac { 62 }{ 2 } ^o\\ \Rightarrow x={ 31 }^{ o }$$.

    Therefore, option $$A$$ is correct.
  • Question 7
    1 / -0
    Find the measure of the supplementary angle of $$54^o$$
    Solution
    $$Two\quad angles\quad are\quad supplementary\quad when\quad they\quad add\quad upto\quad form\quad 180\quad degrees\quad .\\ If\quad one\quad angle\quad =\quad 54\\ Let\quad the\quad other\quad angle\quad be\quad x\\ Hence\quad x\quad =\quad 180-54\\ =126\\ Hence\quad supplementary\quad angle\quad of\quad the\quad following\quad angle\quad is\quad 126$$
  • Question 8
    1 / -0
    Find the measure of the complementary angle of $$77^o$$.
    Solution

    We know, two angles are complementary when they add up to $$90^o.$$

    Given, measure of one complementary angle is $$ 77^{\circ}$$.

    $$\Rightarrow$$ Measure of other complementary angle $$=90^o-77^o$$ $$=13^o.$$

    $$\therefore$$ Measure of a complementary angle of $$ 77^{o}=13^o$$.

    Therefore, option $$D$$ is correct.

  • Question 9
    1 / -0
    Find the measure of the supplementary angle of $$138^\circ$$.
    Solution
    Two angles are supplementary when they add upto form $$180^o$$.
    Given, one angle $$= 138^o$$.
    Let the supplement angle be $$x$$
    Hence, $$x= 180^o-138^o$$ $$=42^o$$ 
    Hence, supplementary angle of the following angle is $$42^o$$.
  • Question 10
    1 / -0
    Find the measure of the complementary angle of each of $$20^o$$.
    Solution

    We know, two angles are complementary when they add up to $$90^o.$$

    Given, measure of one complementary angle is $$= 20^{\circ}$$.

    $$\Rightarrow$$ Measure of other complementary angle $$=90^o-20^o$$ $$=70^o.$$

    $$\therefore$$ Measure of a complementary angle of $$20^{\circ}$$ $$= 70^{\circ}$$.

    Therefore, option $$B$$ is correct.

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