3 Mark Question

Question 13 Marks

In the given figure, AOB is a straight line and the rays OC and OD stands on it. If $\angle\text{AOC} = 65^\circ, \angle\text{BOD} = 70^\circ$ and $\angle\text{COD} = \text{x}^\circ $find the value of x.

Answer

Since AOB is a straight line, we have:
$\angle\text{AOC}+ \angle\text{BOD} +\angle\text{COD} = 180^\circ$
$65^\circ+ 70^\circ + \text{x}^\circ = 180^\circ(\text{given)}$
$135^\circ + \text{x}^\circ = 180^\circ$
$\text{x}^\circ = 45^\circ$
Thus, the value of x is 45

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Question 23 Marks

Among two supplementary angles, the measure of the larger angle is $36^{\circ}$ more than the measure of the smaller. Find their measures.

Answer

Let the two supplementary angles be $x^{\circ}$ and $(180-x)^{\circ}$
Since it is given that the measure of the larger angle is $36^{\circ}$ more than the smaller angle.
Let the larger angle be $x^{\circ}$
$\therefore(180-x)^{\circ}+36^{\circ}=x^{\circ}$
Or $216=2 x$
Or $108= x$
Larger angle $=108^{\circ}$
Smaller angle $=(108-36)^{\circ}$
$=72^{\circ}$

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Question 33 Marks

Find the angle which is equal to its supplement.

Answer

Let the measure of the required angle be x.
Since it is its own supplement:
x + x = 180º
2x = 180º
x = 90º x + x
= 180º
2x = 180º
x = 90º
Therefore, the required angle is 90°.

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Question 43 Marks

In the given figure, AOB is a straight line and the ray OC stands on it. If $\angle\text{AOC} = (2\text{x} -10)^\circ$ and $\angle\text{BOC} = (3\text{x} + 20)^\circ,$ find the value of x. Also, find $\angle\text{AOC}$ and $\angle\text{BO}.$

Answer

By linear pair property,
$\angle\text{AOC}+\angle\text{BOC}= 180^\circ$
$(2\text{x}-10)^\circ + (3\text{x}+20)^\circ = 180^\circ(\text{given)}$
$5\text{x} + 10 =180^\circ$
$5\text{x} = 170^\circ$
$\text{ x} = 34^\circ$
$\therefore\angle\text{AOC} =(2\text{x}-10)^\circ = (2\times34-10)^\circ = 58^\circ$
$\angle\text{BOC} = (3\text{x}+20)^\circ = (3\times34+20)^\circ= 122^\circ$

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Question 53 Marks

In the given figure, rays OA, OB, OC and OD are such that $\angle\text{AOB} = 56^\circ,\angle\text{BOC} =100^\circ, \angle\text{COD} = \text{x}^\circ$ and $\angle\text{DOA} = 74^\circ.$ Find the value of x.

Answer

Sum of all the angles around a point is 360°.
$\therefore\angle\text{AOB} +\angle\text{BOC} +\angle\text{COD} + \angle\text{DOA} = 360^\circ$
$56^\circ + 100^\circ + \text{x}^\circ + 74^\circ = 360^\circ(\text{given)}$
$230^\circ + \text{x}^\circ = 360^\circ$
$\text{x}^\circ = 130^\circ$
$\text{x} = 130$

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Question 63 Marks

In the given figure, AOB is a straight line and the ray OC stands on it. If $\angle\text{AOC} = 64^\circ$ and $\angle\text{BOC} = \text{x}^\circ,$ find the value of x.

Answer

By linear pair property:
$\angle\text{AOC} +\angle\text{COB}= 180^\circ$
$64^\circ + \angle\text{COB} = 180^\circ$
$\angle\text{COB} = \text{x}^\circ= 180^\circ - 64^\circ = 116^\circ$
$\therefore\text{x} = 116^\circ$

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