Question 13 Marks
The angles of a triangle are in the ratio $3 : 4 : 5$. Find the smallest angle.
AnswerGiven that Angles of a triangle are in the ratio: $3 : 4 : 5$
Measure of the angles be $3x, 4x, 5x$
Sum of the angles of a triangle $= 180^\circ$ $3\text{x}+4\text{x}+5\text{x}=180^\circ$
$12\text{x}=180^\circ$
$\text{x}=\frac{180}{12}$
$\text{x}=15^\circ$
Smallest angle $= 3x = 3 \times 15^\circ = 45^\circ$
View full question & answer→Question 23 Marks
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
AnswerLet the three angles of the given triangle be $\angle \text{a},\angle \text{b}$ and $\angle \text{c}$
We know: $\angle \text{a}>\angle \text{b}+\angle \text{c}...(\text{i})$
(Given) We also know that the sum of all the angles of a triangle is equal to $180^\circ$
$\therefore\angle \text{a}+\angle \text{b}+\angle \text{c}=180^\circ$
$\Rightarrow \angle \text{b}+\angle \text{c}=180^\circ-\angle \text{a}$
Putting the value of $\angle \text{b}+\angle \text{c}$ from equation $(i)$: $\angle \text{a}>180^\circ-\angle \text{a}$
$\Rightarrow 2\angle \text{a}>180^\circ$
$\Rightarrow \angle \text{a}>90^\circ$
Thus, the angle is more than $90^\circ$
Hence, we can conclude by saying that the given triangle is an abtuse triangle.
View full question & answer→Question 33 Marks
Is it possible to have a triangle, in which. Each angle is less than $60^\circ $?
AnswerGive reasons in support of your answer in.
No, because if each angle is less than $60^\circ$ , then the sum of all three angles will be less than $180^\circ$ ,
which is not possible in case of a triangle.
Proof:
Let the three angles of the triangle be $\angle \text{A},\angle \text{B}$ and $\angle \text{C}$ As per the given information,
$\angle \text{A}<60^\circ...(\text{i})$
$\angle \text{B}<60^\circ...(\text{ii})$
$\angle \text{C}<60^\circ...(\text{iii})$ On adding $(i), (ii)$ and $(iii)$,
we get: $\angle \text{A}+\angle \text{B}+\angle \text{C}<60^\circ+60^\circ+60^\circ$ $\angle \text{A}+\angle \text{B}+\angle \text{C}<180^\circ$ We can see that the sum of all three angles is less than $180^\circ$ , which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be less than $60^\circ$
View full question & answer→Question 43 Marks
Compute the value of $x$ in the following figure.
AnswerIn the given figure, we have a quadrilateral whose sum of all angles is $360^\circ$
Thus, $35^\circ+45^\circ+50^\circ+\text{reflex}\ \angle \text{ADC}=360^\circ$ Or,
Reflex $\angle \text{ADC}=230^\circ$
$230^\circ+\text{x}=360^\circ$ (A complete angle)
$\text{x}=130^\circ$
View full question & answer→Question 53 Marks
If one angle of a triangle is equal to the sum of the other two, show that the triangle is a right triangle.
AnswerOne angle of a triangle is equal to the sum of the other two
$x = y + z$
Let the measure of angles be $x, y, z$
$x + y + z = 180^\circ $
$x + x = 180^\circ $
$2x = 180^\circ $
$\text{x}=\frac{180}{2}$
$\text{x}=90^\circ$
If one angle is $90^\circ $ then the given triangle is a right angled triangle.
View full question & answer→Question 63 Marks
A man goes $15m$ due west and then $8m$ due north. How far is he from the starting point?
Answer
Let $O$ be the starting point and $P$ be the final point.
By using the Pythagoras theorem, we can find the distance $OP$.
$\mathrm{OP}^2=15^2+8^2$
$\mathrm{OP}^2=225+64$
$\mathrm{OP}^2=289$
$\mathrm{OP}=17$
Hence, the required distance is $17\ m$ . View full question & answer→Question 73 Marks
The exterior angles, obtained on producing the base of a triangle both ways are $104^\circ $ and $136^\circ $. Find all the angles of the triangle.
Answer
In the given figure, $\angle \text{ABE}$ and $\angle \text{ABC}$ form a linear pair.
$\angle \text{ABE}+\angle \text{ABC}=180^\circ$
$\angle \text{ABC}=180^\circ-136^\circ$
$\angle \text{ABC}=44^\circ$
We can also see that $\angle \text{ACD}$ and $\angle \text{ACB}$ form a linear pair.
$\angle \text{ACD}+\angle \text{ACB}=180^\circ$
$\angle \text{AUB}=180^\circ-104^\circ$
$\angle \text{ACB}=76^\circ$
We know that the sum of interior opposite angles is equal to the exterior angle.
Therefore, we can say that:
$\angle \text{BAC}+\angle \text{ABC}=104^\circ$
$\angle \text{BAC}=104^\circ-44^\circ$
$=60^\circ$
Thus,
$\angle \text{ACE}=76^\circ$ and $\angle \text{BAC}=60^\circ$ View full question & answer→Question 83 Marks
Compute the value of $x$ in the following figure.
AnswerFrom the given figure,
we can say that: $\angle \text{ACD}+\angle \text{ACB}=180^\circ$ (Linear pair) Or,
$\angle \text{ACB}=180^\circ-112^\circ=68^\circ$
We can also say that: $\angle \text{BAE}+\angle \text{BAC}=180^\circ$ (Linear pair) Or,
$\angle \text{BAC}=180^\circ-120^\circ=60^\circ$
We know that the sum of all angles of a triangle is $180^\circ$
Therefore, for $\triangle \text{ABC}:$
$\text{x}+\angle \text{BAC}+\angle \text{ACB}=180^\circ$
$\text{x}=180^\circ-60^\circ-68^\circ=52^\circ$
$\text{x}=52^\circ$
View full question & answer→Question 93 Marks
In $\triangle \text{ABC},\angle \text{A}=100^\circ,$ AD bisects $\angle \text{A}$ and $\text{AD}\bot\text{BC}.$ Find $\angle \text{B}$
Answer
Consider $\triangle \text{ABD}$
$\angle \text{BAD}=\frac{100}{2}$ $($AD bisects $\angle \text{A})$
$\angle \text{BAD}=50^\circ$
$\angle \text{ADB}=90^\circ$ $(AD$ perpendicular to $BC)$
We know that the sum of all three angles of a triangle is $180^\circ $
Thus,
$\angle \text{ABD}+\angle \text{BAD}+\angle \text{ADB}=180^\circ$ $($Sum of angles of $\triangle \text{ABD})$
Or,
$\angle \text{ABD}+50^\circ+90^\circ=180^\circ$
$\angle \text{ABD}=180^\circ-140^\circ$
$\angle \text{ABD}=40^\circ$ View full question & answer→Question 103 Marks
$ABC$ is a triangle in which $\angle \text{B} = \angle \text{C}$ and ray $AX$ bisects the exterior angle $DAC$. If $\angle \text{DAX}=70^\circ.$ Find $\angle \text{ACB}$
AnswerHere, $\angle \text{CAX}=\angle \text{DAX}$
$($AX bisects $\angle \text{CAD})$
$\angle \text{CAX}=70^\circ$
$\angle \text{CAX}+\angle \text{DAX}+\angle \text{CAB}=180^\circ$
$70^\circ+70^\circ+\angle \text{CAB}=180^\circ$
$\angle \text{CAB}=180^\circ-140^\circ$
$\angle \text{CAB}=40^\circ$
$\angle \text{ACB}+\angle \text{CBA}+\angle \text{CAB}=180^\circ$
$($Sum of the angles of $\triangle \text{ABC})$
$\angle \text{ACB}+\angle \text{ACB}+40^\circ=180^\circ$
$(\angle \text{C}=\angle \text{B})$
$2\angle \text{AVB}=180^\circ-40^\circ$
$\angle \text{ACB}=\frac{140}{2}$
$\angle \text{ACB}=70^\circ$
View full question & answer→Question 113 Marks
Line segments $AB$ and $CD$ intersect at $O$ such that $AC$ perpendicular $DB$. It $\angle \text{CAB}=35^\circ$ and$ \angle \text{CDB}=55^\circ$. Find $\angle \text{BOD}$
Answer
We know that $AC$ parallel to $BD$ and $AB$ cuts $AC$ and $BD$ at $A$ and $B$, respectively.
$\angle \text{CAB}=\angle \text{DBA}$ (Alternate interior angles) $\angle \text{DBA}=35^\circ$
We also know that the sum of all three angles of a triangle is $180^\circ $
Hence, for $\triangle \text{OBD},$
we can say that: $\angle \text{DBO}+\angle \text{ODB}+\angle \text{BOD}=180^\circ$
$35^\circ+55^\circ+\angle \text{BOD}=180^\circ$
$(\angle \text{DBO}=\angle \text{DBA}$ and $\angle \text{ODB}=\angle \text{CDM})$
$\angle \text{BOD}=180^\circ-90^\circ$ $\angle \text{BOD}=90^\circ$ View full question & answer→Question 123 Marks
Two angles of a triangle are of measures $150^\circ $ and $30^\circ $. Find the measure of the third angle.
AnswerLet the third angle be $x$ Sum of all the angles of a triangle
$= 180^\circ\ 105^\circ + 30^\circ + x$
$= 180^\circ\ 135^\circ + x $
$= 180^\circ\ x $
$= 180^\circ - 135^\circ x $
$= 45^\circ $
Therefore the third angle is $45^\circ $
View full question & answer→Question 133 Marks
A ladder $50\ dm$ long when set against the wall of a house just reaches a window at a height of $48\ dm$. How far is the lower end of the ladder from the base of the wall?
Answer
Let the distance of the lower end of the ladder from the wall be $x \mathrm{~m}$.
On using the Pythagoras theorem, we get:
$x^2+48^2=50^2$
$x^2=50^2-48^2$
$=2500-2304$
$=196$
$H=14\ d m$
Hence, the distance of the lower end of the ladder from the wall is $14\ dm$ . View full question & answer→Question 143 Marks
If the angles of a triangle are in the ratio $1 : 2 : 3$, determine three angles.
AnswerIf angles of the triangle are in the ratio $1 : 2 : 3$
then take first angle as $‘x’$, second angle as $‘2x’$ and third angle as $‘3x’$
Sum of all the angles of a triangle $= 180^\circ$
$\text{x}+2\text{x}+3\text{x}=180^\circ$
$6\text{x}=180^\circ$
$\text{x}=\frac{180}{6}$
$\text{x}=30^\circ$
$2\text{x}=30^\circ\times 2=60^\circ$
$3\text{x}=30^\circ\times 3=90^\circ$
Therefore the first angle is $30^\circ$ , second angle is $60^\circ$ and third angle is $90^\circ$
View full question & answer→Question 153 Marks
A student when asked to measure two exterior angles of $\triangle \text{ABC}$ observed that the exterior angles at $A$ and $B$ are of $103^\circ $ and $74^\circ $ respectively. Is this possible? Why or why not?
AnswerInternal angle at $A +$ External angle at $A= 180^\circ $
Internal angle at $A + 103º=180º$
Internal angle at $A= 77^\circ $
Internal angle at $B +$ External angle at $B= 180^\circ $
Internal angle at $B + 74^\circ = 180^\circ $
Internal angle at $B = 106^\circ $
Sum of internal angles at $A$ and $B = 77^\circ + 106^\circ = 183^\circ $
It means that the sum of internal angles at $A$ and $B$ is greater than $180^\circ $, which cannot be possible.
View full question & answer→Question 163 Marks
$O$ is a point in the exterior of $△ABC$. What symbol $‘>’, ’<’$ or $‘=’$ will you see to complete the statement $OA + OB…. AB?$ Write two other similar statements and show that. $\text{OA}+\text{OB}+\text{OC}>\frac{1}{2}(\text{AB}+\text{BC}+\text{CA})$
AnswerBecause the sum of any two sides of a triangle is always greater than the third side, in triangle $OAB$,
we have: $OA + OB > AB... (i) $
$OB + OC > BC... (ii) $
$OA + OC > CA... (iii)$
On adding equations $(i), (ii)$ and $(iii)$ we get: $OA + OB + OB + OC + OA + OC > AB + BC + CA 2(OA + OB + OC) > AB + BC + CA$ $\text{OA}+\text{OB}+\text{OC}>\frac{(\text{AB}+\text{BC}+\text{CA})}{2}$
View full question & answer→Question 173 Marks
Draw a triangle $A B C$, with $A C=4 \mathrm{~cm}, B C=3 \mathrm{~cm}$ and $\angle C=80^{\circ}$. Measure $A B$. Is $(A B)^2=(A C)^2+$ $(B C)^2$ ? If not which one of the following is true:
$(A B)^2>(A C)^2+(B C)^2 \text { or }(A B)^2<(A C)^2+(B C)^2$
Answer
Draw $\triangle \mathrm{ABC}$,
Draw a line $B C=3 \mathrm{~cm}$.
At point $C$, draw a line at $80^{\circ}$ angle with $BC$ .
Take an arc of $4\ cm$ from point $C$, which will cut the line at point $A$.
Now, join $A B$; it will be approximately $4.5\ cm$ .
$A C^2+B C^2$
$=4^2+3^2$
$=9+16$
$=25$
$\mathrm{AB}^2=4.5^2=20.25$
$A B^2$ not equal to $A C^2+B C^2$
Here,
$A B^2$ View full question & answer→Question 183 Marks
A ladder $3.7m$ long is placed against a wall in such a way that the foot of the ladder is $1.2m$ away from the wall. Find the height of the wall to which the ladder reaches.
Answer
Let the hypotenuse be $h$ .
Using the Pythagoras theorem, we get:
$3.72=1.22+h^2$
$h^2=13.69-1.44=12.25$
$h=3.5 \mathrm{~m}$
Hence, the height of the wall is $3.5 m$ . View full question & answer→Question 193 Marks
Compute the value of $x$ in the following figure.
AnswerFrom the given figure, we can say that:
$\angle \text{ABC}+120^\circ=180^\circ$ (Linear pair)
$\angle \text{ABC}=60^\circ$
We can also say that:
$\angle \text{ACB}+110^\circ=180^\circ$ (Linear pair)
$\angle \text{ACB}=70^\circ$
We know that the sum of all angles of a triangle is $180^\circ $
Therefore, for $\triangle \text{ABC}:$
$\text{x}+\angle \text{ABC}+\angle \text{ACB}=180^\circ$
$\text{x}=50^\circ$
View full question & answer→Question 203 Marks
In the fig., two of the angles are indicated. What are the measures of $\angle \text{ACX}$ and $\angle \text{ACB}$?
AnswerIn $\triangle \text{ABC}, \angle \text{A}=50^\circ$ and $\angle \text{B}=55^\circ$
Because of the angle sum property of the triangle,
we can say that $\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$
$55^\circ+55^\circ+\angle \text{C}=180^\circ$ Or,
$\angle \text{C}=75^\circ$ $\angle \text{ACB}=75^\circ$
$\angle \text{ACX}=180^\circ-\angle \text{ACB}=180^\circ-75^\circ$
$=105^\circ$
View full question & answer→Question 213 Marks
If one angle of a triangle is $60^\circ $ and the other two angles are in the ratio $1 : 2$, find the angles.
AnswerWe know that one of the angles of the given triangle is $60^\circ $ (Given)
We also know that the other two angles of the triangle are in the ratio $1 : 2.$
Let one of the other two angles be $x$.
Therefore, the second one will be $2x.$
We know that the sum of all the three angles of a triangle is equal to $180^\circ $.
$60^\circ + x + 2x = 180^\circ\ 3x = 180^\circ - 60^\circ\ 3x = 120^\circ $
$\text{x}=\frac{120}{3}$ $x = 40^\circ 2x = 2 \times 40\ 2x = 80^\circ $
Hence, we can conclude that the required angles are $40^\circ $ and $80^\circ $
View full question & answer→Question 223 Marks
Two acute angles of a right triangle are equal. Find the two angles.
Answer
Given acute angles of a right angled triangle are equal.
Right triangle: whose one of the angle is a right angle.
Measured angle be $x, x, 90^\circ $
$\text{x}+\text{x}+180^\circ=180^\circ$
$2\text{x}=90^\circ$
$\text{x}=\frac{90}{2}$
$\text{x}=45^\circ$
The two angles are $45^\circ $ and $45^\circ $ View full question & answer→Question 233 Marks
In Figure, $\text{AC}\bot\text{CE}$ and C $\angle \text{A}:\angle \text{B}:\angle \text{C}=3:2:1.$ Find the value of $\angle \text{ECD}$ 
AnswerIn the given triangle, the angles are in the ratio $3 : 2 : 1$.
Let the angles of the triangle be $3x, 2x$ and $x$.
Because of the angle sum property of the triangle, we can say that:
$3x + 2x + x = 180^\circ\ 6x = 180º$ Or, $x = 30^\circ …(i)$
Also, $\angle \text{ACB}+\angle \text{ACE}+\angle \text{ECD}=180^\circ$
$\text{x}+90^\circ+\angle \text{ECD}=180^\circ$ $(\angle \text{ACE}=90^\circ)$
$\angle \text{ECD}=60^\circ$ $[$From $(i)]$
View full question & answer→Question 243 Marks
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is $10^\circ $. Find the three angles.
AnswerLet the first angle be $x$
Second angle be $x + 10^\circ $
Third angle be $x + 10^\circ + 10^\circ $
Sum of all the angles of a triangle $= 180^\circ x + x + 10^\circ + x + 10^\circ + 10^\circ = 180^\circ $
$3x + 30 = 180$ $3x = 180 - 30$ $3x = 150$
$\text{x}=\frac{150}{3}$ $\text{x}=50^\circ$
First angle is $50^\circ $
Second angle $x + 10^\circ = 50 + 10 = 60^\circ $
Third angle $x + 10^\circ + 10^\circ = 50 + 10 + 10 = 70^\circ $
View full question & answer→Question 253 Marks
Compute the value of $x$ in the following figure.
AnswerFrom the given figure,
we can see that: $\angle \text{BAD}=\angle \text{ADC}=52^\circ$ (Alternate angles)
We know that the sum of all the angles of a triangle is $180^\circ $
Therefore, for $\triangle \text{DEC}:$
$\text{x}+40^\circ+52^\circ=180^\circ$
$\text{x}=88^\circ$
View full question & answer→Question 263 Marks
Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^\circ $. Determine all the angles of the triangle.
AnswerLet the first and second angle be $x$
The third angle is greater than the first and second by $30^\circ = x + 30^\circ $
The first and the second angles are equal
Sum of all the angles of a triangle $= 180^\circ $
$x + x + x + 30^\circ = 180^\circ $
$3x + 30 = 180$
$3x = 180 - 30$
$3x = 150$
$\text{x}=\frac{150}{3}$
$\text{x}=50^\circ$
Third angle $= x + 30^\circ = 50^\circ + 30^\circ = 80^\circ $
The first and the second angle is $50^\circ $ and the third angle is $80^\circ $
View full question & answer→Question 273 Marks
In a triangle, an exterior angle at a vertex is $95^\circ $ and its one of the interior opposite angles is $55^\circ $. Find all the angles of the triangle.
Answer
We know that the sum of interior opposite angles is equal to the exterior angle.
Hence, for the given triangle,
we can say that: $\angle \text{ABC}+\angle \text{BAC}=\angle \text{BCO}$
$55^\circ+\angle \text{BAC}=95^\circ$ Or, $\angle \text{BAC}=95^\circ-95^\circ$
$=\angle \text{BAC}=40^\circ$
We also know that the sum of all angles of a triangle is $180^\circ$
Hence, for the given $\triangle \text{ABC},$
we can say that: $\angle \text{ABC}+\angle \text{BAC}+\angle \text{BCA}=180^\circ$
$55^\circ+40^\circ+\angle \text{BCA}=180^\circ$ Or, $\angle \text{BCA}=180^\circ-95^\circ$
$=\angle \text{BCA}=85^\circ$ View full question & answer→Question 283 Marks
One of the exterior angles of a triangle is $80^\circ$, and the interior opposite angles are equal to each other. What is the measure of each of these two angles?
AnswerLet us assume that $A$ and $B$ are the two interior opposite angles.
We know that $\angle \text{A}$ is equal to $\angle \text{B}$
We also know that the sum of interior opposite angles is equal to the exterior angle.
Hence, we can say that:
$\angle \text{A}+\angle \text{B}=80^\circ$ Or, $\angle \text{A}+\angle \text{A}=80^\circ$
$(\angle \text{A}=\angle \text{B})$
$2\angle \text{A}=80^\circ$ $\angle \text{A}=\frac{40}{2}=40^\circ$
$\angle \text{A}=\angle \text{B}=40^\circ$
Thus, each of the required angles is of $40^\circ $
View full question & answer→Question 293 Marks
One of the angles of a triangle is $130^\circ $, and the other two angles are equal. What is the measure of each of these equal angles?
AnswerLet the second and third angle be $x$ Sum of all the angles of a triangle
$= 180^\circ$ $130^\circ + x + x = 180^\circ$ $130^\circ + 2x= 180^\circ$ $2x= 180^\circ - 130^\circ$ $2x = 50^\circ $
$\text{x}=\frac{50}{2}$
$\text{x}=25^\circ$
Therefore the two other angles are $25^\circ $ each.
View full question & answer→Question 303 Marks
The three angles of a triangle are equal to one another. What is the measure of each of the angles?
AnswerLet the each angle be $x$ Sum of all the angles of a triangle $= 180^\circ $
$\text{x}+\text{x}+\text{x}=180^\circ$
$3\text{x}=180^\circ$ $\text{x}= \frac{180}{3}$
$\text{x}=60^\circ$
Therefore angle is $60^\circ $ each.
View full question & answer→Question 313 Marks
In Fig, the sides $BC, CA$ and $BA$ of a $\triangle \text{ABC}$ have been produced to $D, E$ and $F$ respectively. If $\angle \text{ACD}=105^\circ$ and $\angle \text{EAF}=45^\circ,$ find all the angles of the $\triangle \text{ABC}$
Answer$\angle \text{EAF}$In a $\triangle \text{ABC}, \angle \text{BAC}$ and are vertically opposite angles.
Hence, we can say that:
$\angle \text{BAC}=\angle \text{EAF}=45^\circ$
Considering the exterior angle property, we can say that:
$\angle \text{BAC}+\angle \text{ABC}=\angle \text{ACD}=105^\circ$
$\angle \text{ABC}=105^\circ - 45^\circ =60^\circ$
Because of the angle sum property of the triangle, we can say that:
$\angle \text{ABC}+\angle \text{ACS}+\angle \text{BAC}=180^\circ$
$\angle \text{ACB}=75^\circ$
Therefore, the angles are $45^\circ , 65^\circ $ and $75^\circ $.
View full question & answer→Question 323 Marks
Two poles of heights $6m$ and $11m$ stand on a plane ground. If the distance between their feet is $12m$. Find the distance between their tops.
(Hint: Find the hypotenuse of a right triangle having the sides $(11 - 6)m = 5m$ and $12m$)
Answer
The distance between the tops of the poles is the distance between points $A$ and $B$ .
We can see from the given figure that points $\mathrm{A}, \mathrm{B}$ and $C$ form a right triangle, with $AB$ as the hypotenuse.
On using the Pythagoras Theorem in $\triangle \mathrm{ABC}$, we get:
$(11-6)^2+12^2=A B^2$
$A B^2=25+144$
$A B^2=169$
$A B=13$
Hence, the distance between the tops of the poles is $13\ m$ . View full question & answer→Question 333 Marks
The angles of a triangle are $(x − 40)^\circ , (x − 20)^\circ $ and $\Big(\frac{1}{2}\text{x}-10\Big)^\circ.$ Find the value of $x$.
AnswerSum of all the angles of a triangle $= 180^\circ $
$(\text{x}-40)^\circ+(\text{x}-20)^\circ+\Big(\frac{\text{x}}{2}-10\Big)^\circ=$
$\text{x}+\text{x}+\frac{\text{x}}{2}-40^\circ-20^\circ-10^\circ=180^\circ$
$\text{x}+\text{x}+\frac{\text{x}}{2}-70^\circ=180^\circ$
$\text{x}+\text{x}+\frac{\text{x}}{2}=180^\circ+70^\circ$
$\frac{5\text{x}}{2}=250^\circ$
$\text{x}=\frac{2}{5}\times 250^\circ$
$\text{x}=100^\circ$
Hence, we can conclude that $x$ is equal to $100^\circ $
View full question & answer→Question 343 Marks
It one angle of a triangle is $100^\circ $ and the other two angles are in the ratio $2 : 3$. find the angles.
AnswerWe know that one of the angles of the given triangle is $100^\circ $
We also know that the other two angles are in the ratio $2 : 3$.
Let one of the other two angles be $2x$
Therefore, the second angle will be $3x$
We know that the sum of all three angles of a triangle is
$180^\circ $ $100^\circ + 2x + 3x = 180^\circ $
$5x = 180^\circ - 100^\circ $
$5x = 80^\circ $
$\text{x}=\frac{80}{5}$
$2x = 2 \times 16 2x = 32^\circ $
$3x = 3 \times 16$
$3x = 48^\circ $
Thus, the required angles are $32^\circ $ and $48^\circ $
View full question & answer→Question 353 Marks
Draw a triangle $ABC$ , with $\mathrm{AC}=4 \mathrm{~cm}, \mathrm{BC}=3 \mathrm{~cm}$ and $\angle C=105^{\circ}$. Measure AB . Is. $(\mathrm{AB})^2=(\mathrm{AC})^2+$ $(B C)^2$ ? If not which one of the following is true:
$(A B)^2>(A C)^2+(B C)^2 \text { or }(A B)^2<(A C)^2+(B C)^2$
Answer
Draw $\triangle \mathrm{ABC}$,
Draw a line $B C=3 \mathrm{~cm}$.
At point $C$, draw a line at $105^{\circ}$ angle with $BC$ .
Take an arc of $4\ cm$ from point $C$, which will cut the line at point $A$.
Now, join $A B$, which will be approximately $5.5\ cm$ .
$\mathrm{AC}^2+\mathrm{BC}^2$
$=4^2+3^2$
$=9+16$
$=25$
$A B^2=5.52=30.25$
$A B^2$ not equal to $A C^2+B C^2$
Here,
$\mathrm{AB}^2>\mathrm{AC}^2+\mathrm{BC}^2$ View full question & answer→Question 363 Marks
State Pythagoras theorem and its converse.
AnswerThe Pythagoras Theorem: In a right triangle, the square of the hypotenuse is always equal to the sum of the squares of the other two sides.
Converse of the Pythagoras Theorem: If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle, with the angle opposite to the first side as right angle.
View full question & answer→Question 373 Marks
Is it possible to have a triangle, in which.
Each angle is greater than $60^\circ $?
AnswerGive reasons in support of your answer in. No, because if each angle is greater than $60^\circ $,
then the sum of all three angles will be greater than $180^\circ $, which is not possible.
Proof:
Let the three angles of the triangle be $\angle \text{A},\angle \text{B}$ and $\angle \text{C}.$
As per the given information,
$\angle \text{A}>60^\circ...(\text{i})$
$\angle \text{B}>60^\circ...(\text{ii})$
$\angle \text{C}>60^\circ...(\text{iii})$ On adding $(i), (ii)$ and $(iii)$,
we get: $\angle \text{A}+\angle \text{B}+\angle \text{C}>60^\circ+60^\circ+60^\circ$
$\angle \text{A}+\angle \text{B}+\angle \text{C}>180^\circ$
We can see that the sum of all three angles of the given triangle are greater than $180^\circ$ ,
which is not possible for a triangle.
Hence, we can say that it is not possible for each angle of a triangle to be greater than $60^\circ$
View full question & answer→Question 383 Marks
In Figure, $AD$ and $CF$ are respectively perpendiculars to sides $BC$ and $AB$ of$ \triangle \text{ABC}.$ If $\angle \text{FCD}=50^\circ,$ find $\angle \text{BAD}$
AnswerWe know that the sum of all angles of a triangle is $180^\circ$
Therefore, for the given $\triangle \text{FCB},$
we can say that: $\angle \text{FCB}+\angle \text{CBF}+\angle \text{BFC}=180^\circ$
$50^\circ+\angle \text{CBF}+90^\circ=180^\circ$ Or, $\angle \text{CBF}=180^\circ-50^\circ-90^\circ=40^\circ...(\text{i})$
Using the above rule for $\triangle \text{ABD},$
we can say that: $\angle \text{ABD}+\angle \text{BDA}+\angle \text{BAD}=180^\circ$
$\angle \text{BAD}=180^\circ-90^\circ-40^\circ=50^\circ$ [from $(i)$]
View full question & answer→Question 393 Marks
Is it possible to have a triangle, in which. Each angle is equal to $60^\circ $
AnswerGive reasons in support of your answer in. Yes, if each angle of the triangle is equal to $60^\circ $,
then the sum of all three angles will be $180^\circ $ , which is possible in case of a triangle.
Proof:
Let the three angles of the triangle be $\angle \text{A},\angle \text{B}$ and $\angle \text{C}.$
As per the given information,
$\angle \text{A}=60^\circ...(\text{i})$
$\angle \text{B}=60^\circ...(\text{ii})$
$\angle \text{C}=60^\circ...(\text{iii})$ On adding $(i), (ii)$ and $(iii),$
we get: $\angle \text{A}+\angle \text{B}+\angle \text{C}=60^\circ+60^\circ+60^\circ$
$\angle \text{A}+\angle \text{B}+\angle \text{C}=180^\circ$
We can see that the sum of all three angles of the given triangle is equal to $180^\circ$ ,
which is possible in case of a triangle.
Hence, we can say that it is possible for each angle of a triangle to be equal to $60^\circ$
View full question & answer→Question 403 Marks
The foot of a ladder is 6 m away from a wall and its top reaches a window 8 m above the ground. If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?
Question 413 Marks
Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between their feet is 12 m , find the distance between their tops.
[Hint: Find the hypotenuse of a right triangle having the sides $(11-6) m=5 m$ and 12 m ]
View full question & answer→Question 423 Marks
A ladder 3.7 m long is placed against a wall in such a way that the foot of the ladder is 1.2 m away from the wall. Find the height of the wall to which the ladder reaches.
Question 433 Marks
The hypotenuse of a triangle is 2.5 cm . If one of the sides is 1.5 cm , find the length of the other side.
Question 443 Marks
In a $\triangle A B C, A D$ is the altitude from $A$ such that $A D=12 cm, B D=9 cm$ and $D C=16 cm$. Examine if $\triangle A B C$ is right angled at $A$.
View full question & answer→Question 453 Marks
The two legs of a right triangle are equal and the square of the hypotenuse is 50 . Find the length of each leg.
Question 463 Marks
A ladder 50 dm long when set against the wall of a house just reaches a window at a height of 48 dm . How far is the lower end of the ladder from the base of the wall?
Question 473 Marks
In $\triangle A B C, \angle A=100^{\circ}, \angle B=30^{\circ}, \angle C=50^{\circ}$. Name the smallest and the largest sides of the triangle.
View full question & answer→Question 483 Marks
$P$ is a point in the interior of $\triangle A B C$ as shown in Fig. . State which of the following statements are true (T) or false ( F ):
(i) $A P+P B<A B$ $\quad$ (ii) $A P+P C>A C$ $\quad$ (iii) $B P+P C=B C$
Answer(i) false $\quad$ (ii) true $\quad$ (iii) false
View full question & answer→Question 493 Marks
In Fig. $P$ is the point on the side $B C$. Complete each of the following statements using symbol ' $=$ ',' $>$ ' or ' <' so as to make it true:
(i) $A P \ldots A B+B P$ $\quad$ (ii) $A P \ldots . A C+P C$ $\quad$ (iii) $A P \ldots \cdot \frac{1}{2}(A B+A C+B C)$
Answer(i) $<$ $\quad$ (ii) $<$ $\quad$ (iii) $<$
View full question & answer→Question 503 Marks
In Fig. $A C \perp C E$ and $\angle A: \angle B: \angle C=3: 2: 1$, find the value of $\angle E C D$.
View full question & answer→Question 513 Marks
The exterior angles, obtained on producing the base of a triangle both ways are $104^{\circ}$ and $136^{\circ}$. Find all the angles of the triangle.
Answer$60^{\circ}, 76^{\circ}, 44^{\circ}$
View full question & answer→Question 523 Marks
In Fig., two of the angles are indicated. What are the measures of $\angle A C X$ and $\angle A C B$ ?
Answer$\angle A C X=105^{\circ}, \angle A C B=75^{\circ}$
View full question & answer→Question 533 Marks
$A B C$ is a triangle in which $\angle B=\angle C$ and ray $A X$ bisects the exterior angle $D A C$. If $\angle D A X=70^{\circ}$, find $\angle A C B$.
View full question & answer→Question 543 Marks
In Fig. measures of some angles are indicated. Find the value of $x$.
View full question & answer→Question 553 Marks
In Fig., $\angle C B X$ is an exterior angle of $\triangle A B C$ at $B$. Name:
(i) the interior adjacent angle
(ii) the interior opposite angles to exterior $\angle C B X$.
Also, ame the interior opposite angles to an exterior angle at $A$.
Answer(i) $\angle A B C$
(ii) $\angle B A C, \angle A C B ; \angle A B C, \angle A C B$
View full question & answer→Question 563 Marks
Two angles of a triangle are equal and the third angle is greater than each of those angles by $30^{\circ}$. Determine all the angles of the triangle.
Answer$50^{\circ}, 50^{\circ}, 80^{\circ}$
View full question & answer→Question 573 Marks
The angles of a triangle are arranged in ascending order of magnitude. If the difference between two consecutive angles is $10^{\circ}$, find the three angles.
Answer$50^{\circ}, 60^{\circ}, 70^{\circ}$
View full question & answer→Question 583 Marks
The angles of a triangle are $(x-40)^{\circ},(x-20)^{\circ}$ and $\left(\frac{1}{2} x-10\right)^{\circ}$. Find the value of $x$.
View full question & answer→Question 593 Marks
In Fig., $\triangle A B C$ is right angled at $A . Q$ and $R$ are points on line $B C$ and $P$ is a point such that $Q P \| A C$ and $R P \| A B$. Find $\angle P$.
View full question & answer→Question 603 Marks
Line segments $A B$ and $C D$ intersect at $O$ such that $A C \| D B$. If $\angle C A B=35^{\circ}$ and $\angle C D B=55^{\circ}$, find $\angle B O D$.
View full question & answer→Question 613 Marks
In $\triangle A B C, \angle B=60^{\circ}, \angle C=40^{\circ}, A L \perp B C$ and $A D$ bisects $\angle A$ such that $L$ and $D$ lie on side $B C$. Find $\angle L A D$.
View full question & answer→Question 623 Marks
In $\triangle A B C, \angle A=50^{\circ}$ and $B C$ is produced to a point $D$. The bisectors of $\angle A B C$ and $\angle A C D$ meet at $E$. Find $\angle E$.
View full question & answer→Question 633 Marks
The bisectors of the acute angles of a right triangle meet at $O$. Find the angle at $O$ between the two bisectors.
View full question & answer→Question 643 Marks
In $\triangle A B C, \angle A=100^{\circ}, A D$ bisects $\angle A$ and $A D \perp B C$. Find $\angle B$.
View full question & answer→Question 653 Marks
In a $\triangle A B C$, if $3 \angle A=4 \angle B=6 \angle C$, calculate the angles.
Answer$80^{\circ}, 60^{\circ}, 40^{\circ}$
View full question & answer→Question 663 Marks
If one angle of a triangle is $100^{\circ}$ and the other two angles are in the ratio $2: 3$, find the angles.
Answer$32^{\circ}, 48^{\circ}$
View full question & answer→Question 673 Marks
If one angle of a triangle is $60^{\circ}$ and the other two angles are in the ratio $1: 2$, find the angles.
Answer$40^{\circ}, 80^{\circ}$
View full question & answer→Question 683 Marks
One angle of a triangle is greater than the sum of the other two. What can you say about the measure of this angle? What type of a triangle is this?
AnswerMore than $90^{\circ}$, Obtuse triangle
View full question & answer→Question 693 Marks
What is the difference between a triangle and triangular region?
Question 703 Marks
Distinguish between a triangle and its triangular region.