Tag: complementary angle, supplementary angles and adjcent angles

Questions Related to complementary angle, supplementary angles and adjcent angles

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. $\angle B = \angle D$


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.

  1. Circumcentre

  2. Incentre

  3. Mid point

  4. Data insufficient


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:

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

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

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

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


Correct Option: A
Explanation:

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

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. $59^o$


Correct Option: D
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, $a = \dfrac{1061}{18}$

$a \approx 59^{\circ}$

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


Correct Option: D
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.

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

  1. obtuse angle

  2. straight angle

  3. zero angle

  4. right angle


Correct Option: C
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.

Find the supplement  of the following angle.
$40^{\circ}$

  1. $140$

  2. $40$

  3. $10$

  4. None of these


Correct Option: A
Explanation:

We know that the supplement angle

$=180^0-\theta$

So,
The supplement angle of $40^0$ will be
$=180^0-40^0=140^0$ 

Hence, this is the answer.

Two triangles are similar, if their corresponding angles are ________.

  1. Proportional

  2. Equal

  3. A & B

  4. None of the above


Correct Option: B
Explanation:

Two triangles are similar, if their corresponding angles are equal.

(Two triangles are similar, if their corresponding angles are equal and corresponding sides are proportional.)