Question 1011 Mark
In $\triangle\text{ABC},$ BC = AB and $\angle\text{B} = 80^\circ.$ Then $\angle\text{A}$ is equal to:
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M.C.Q
Question 1021 Mark
If the altitudes from two vertices of a triangle to the opposite sides are equal then the triangle is:
Question 1031 Mark
In $\triangle\text{ABC},$ if $\angle\text{A} = 45^\circ$ and $\angle\text{B} = 70^\circ,$ then the shortest and the longest sides of the triangle are ________ .
Answer
View full question & answer→- BC, AC
Solution:
Smallest angle is A and greatest angle is B and hence sides opposite to these angles are BC and AC and they are shortest and longest respectively.
Question 1041 Mark
Answer
View full question & answer→- 105º
Solution:
Join AC. We get two isosceles triangles. $\triangle\text{ABC}$ and $\triangle\text{ACD}.$
In $\triangle\text{ABC},\ \angle\text{ABC}= 108^\circ$
$\therefore\angle\text{BAC} = \angle\text{BCA} = \frac{(180^\circ -108^\circ)}{2} = \frac{72^\circ}{2} = 36^\circ$
In $\triangle\text{ACD},\ \angle\text{ADC}= 42^\circ$
$\therefore\angle\text{DAC} = \angle\text{DCA} = \frac{(180^\circ -42^\circ)}{2} = \frac{138^\circ}{2} = 69^\circ$
Now, $\angle\text{BCD} = \angle\text{BCA} + \angle\text{DCA} = 36^\circ + 69^\circ = 105^\circ$
Question 1051 Mark
In $\angle\text{ABC}, \angle\text{A}=40^\circ$ and $\angle\text{B}=60^\circ$ Then, the longest side of $\triangle\text{ABC}$ is:
Answer
View full question & answer→- AB
Solution:
The third angle $\angle\text{C}=80^\circ$ [Angle sum property]
The side opposite to the longest angle is the longest side.
Therefore, AB is the longest side of the triangle.
Question 1061 Mark
In $\triangle\text{ABC,}$ if $\angle\text{C}>\angle\text{B},$ then:
Answer
View full question & answer→- AB > AC
Solution:
We know that in a triangle, the greater angle has the longer side opposite to it.
In $\triangle\text{ABC},$
$\angle\text{C}>\angle\text{B}$
$\Rightarrow\text{AB > AC}$
Question 1071 Mark
If one angle of a triangle is equal to the sum of the other two angles, then the triangle is:
Answer
View full question & answer→- A right triangle
Solution:
Let the three angles of a triangle be A, B and C.
Now, A + B + C = 180°
If A = B + C
Then A + (A) = 180°
i.e. 2A = 180°
i.e. A = 90°
Since, one of the angle is 90°, the triangle is a Right triangle.
Question 1081 Mark
Answer
View full question & answer→- 8cm
Solution:
Using relation
perimeter, $\triangle\text{DEF}=\frac{1}{2}$
perimeter, $\triangle\text{ABC}$
$=\frac{1}{2}×16=8\text{cm}$
Question 1091 Mark
Answer
View full question & answer→- SSS
Solution:
Given that two sides are equal and third side is common I.e. AD hence all three corresponding sides are equal.
Question 1101 Mark
In a $\triangle\text{ABC},$ if $3\angle\text{A}=4\angle\text{B}=6\angle\text{C}$ then A : B : C = ?
Answer
View full question & answer→- 3 : 4 : 6
Solution:
In the given figure, $\angle\text{ACB}+\angle\text{ACD}=180^\circ$ (Linear pair of angles)
$∴$ 5yº + 7yº = 180º
⇒ 12yº = 180º
⇒ y = 15
In $\triangle\text{ABC}$
$\angle\text{A}+\angle\text{B}+\angle\text{ACB}=180^\circ$ (Angle sum property)
$∴$ 3yº + xº + 5yº = 180º
⇒ xº + 8yº = 180º
⇒ xº + 8 × 15º = 180º
⇒ xº + 120º = 180º
⇒ xº = 180º − 120º = 60º
Thus, the value of x is 60º.
Hence the correct answer is 3 : 4 : 6
Question 1111 Mark
Answer
View full question & answer→- 360º
Solution:
We have:
$\angle1+\angle\text{BAE}=180^\circ\ ...\ \text{(i)}$
$\angle2+\angle\text{CBF}=180^\circ ...\ \text{(ii)}$
$\angle\text{3}+\angle\text{ACD}=180^\circ ...\ \text{(iii)}$
Adding (i),(ii) and (iii), we get:
$(\angle1+\angle2+\angle3)+(\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD})=540^\circ$
$⇒180^\circ+\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD}=540^\circ$ $ [∵ \angle1+\angle2+\angle3=180^\circ]$
$⇒\angle\text{BAE}+\angle\text{CBF}+\angle\text{ACD}=360^\circ.$
Question 1121 Mark
Answer
View full question & answer→- RHS
Solution:
In $\triangle\text{ABD}$ and $\triangle\text{ACD},$ we have
$\angle\text{ADB} = \angle\text{ADC}$ (Right angles)
AB = AC (Given and hypotenuses)
AD = AD (common in both)
Therefore, $\triangle\text{ABD}\cong\triangle\text{ACD}$ by RHS.
Question 1131 Mark
If two acute angles of a right triangle are equal, then each acute is equal to:
Answer
View full question & answer→- 45º
Solution:
Let the measure of each acute angle of a triangle be xº.
Then, we have
xº + xº + 90º = 180º
i.e. 2xº = 90º
i.e. xº = 45º
Question 1141 Mark
If $\triangle\text{PQR}≅\triangle\text{EFD},$ then ED =
Answer
View full question & answer→- PR
Solution:
Since, by corresponding part of congruent triangle ED of $\triangle\text{EFD}$ is equal to the PR of $\triangle\text{PQR}.$
Question 1151 Mark
Answer
View full question & answer→- 40º
Solution:
Since, It is given that PQ = QR, then $\angle\text{Q} = \angle\text{R}$ (Isosceles trangle property)
As $\angle\text{Q} = 70^\circ,$ therefore $\angle\text{R} = 70^\circ,$
Sum of all the three angles of triangle = 180º, therefore $\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ$
$\angle\text{P} = 180 - 70 - 70\ = \ 40^\circ$
Question 1161 Mark
Answer
View full question & answer→- $\angle\text{BDA}$
Solution:
In triangle ABD and CBD
AB = BC and $\angle\text{ABD} = \angle\text{CBD},$ (Given)
BD (Common)
Therefore In triangle ABD and CBD are congruent by SAS criteria.
Therefore, $\angle\text{BDA}=30^\circ$ (by CPCT)
Question 1171 Mark
IF $\text{AB}⊥\text{BC}$ and $\angle\text{BAC} = \angle\text{BCA},$ then which of the following statements is true?
Answer
View full question & answer→- AB = BC
Solution:
ABC is a right-angled isosceles triangle with $\angle\text{BAC} = \angle\text{BCA},$ and hence sides opposite to equal angles must be equal. Hence, we can say that AB = BC.
Question 1181 Mark
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Question 1191 Mark
Two sides of a triangle are of lengths 5cm and 1.5cm. The length of the third side of the triangle cannot be.
Answer
View full question & answer→- 3.4cm
Solution:
Given that: Two sides of triangle are 5cm and 1.5cm. We know that the sum of two sides of the triangle is always greater than the third side. Hence, 3.4cm cannot be the third side. If it is the third side the sum of 3.4cm and 1.5cm will be smaller than 5cm, so, the triangle will not be possible.
Question 1201 Mark
In a $\triangle\text{ABC},$ if $\text{AB}=\text{AC}$ and BC is produced to D such that $\angle\text{ACD}=100^\circ$ then $\angle\text{A}=$
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Question 1211 Mark
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Question 1221 Mark
In a $\triangle\text{ABC},\angle\text{A}=50^\circ$ and BC is produced to a point D. If the bisectors of $\angle\text{ABC}$ and $\angle\text{ACD}$ meet at E, then $\angle\text{E}=$
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Question 1231 Mark
Answer
View full question & answer→- 57.5º
Solution:
As BC = AC, therefore triangle ABC is an isoscelestriangle.
Given $\angle\text{ACD} = 115^\circ,\ \angle\text{ACB} = 180-115=65^\circ$ (Linear Pair)
As AC = BC, therefore $\angle\text{A} =\angle\text{B}$
As sum of all the three angles of atriangle is 180°
Therefore, $\angle\text{A} + \angle\text{B} + \angle\text{ACB} = 180^\circ$
$\angle\text{A} = \angle\text{B} = 57.5$
Question 1241 Mark
In a $\triangle\text{ABC},$ If $\angle\text{A}= 45^\circ$ and $\angle\text{B}= 70^\circ.$ Determine the shortest sides of the triangles.
Answer
View full question & answer→- BC
Solution:
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\angle\text{A}=45^\circ$
$\angle\text{B}=70^\circ$
$\angle\text{C}+45^\circ+70^\circ=180^\circ$
$\angle\text{C}+115^\circ=180^\circ$
$\angle\text{C}=180^\circ-115^\circ$
$\angle\text{C}=65^\circ$
$\angle\text{A}$ is shortest angle and the side opp to shortest angle is shortest.
So, BC is the shotest side.
Question 1251 Mark
In $\triangle\text{ABC}$ and $\triangle\text{DEF}$ its is given that $\angle\text{B}=\angle\text{E}$ and $\angle\text{C}=\angle\text{F}$ in order that $\triangle\text{ABC}≅\triangle\text{DEF}$ we must have,
Answer
View full question & answer→- BC = EF
Solution:
In $\triangle\text{ABC}$ and $\triangle\text{DEF}$
$\angle\text{B}=\angle\text{E}$ and $\angle\text{C}=\angle\text{F}$
For congruence, BC = EF
Therefore by AAS axiom
$\triangle\text{ABC}≅\triangle\text{DEF}$
Question 1261 Mark
Answer
View full question & answer→- AB + BC + AC > 2AD
Solution:
In triangle ADB
AB + BD > AD
In triangle ADC
AC + DC > AD
Adding both
AB + AC + BD + DC > 2AD
Now BD + DC = BC
So, AB + AC + BC > 2AD
Question 1271 Mark
In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, $\angle\text{B} = \angle\text{P},$ BC = PR. State which of the congruence conditions applies:
Answer
View full question & answer→- SAS
Solution:
Two sides and included angle are equal and is SAS axiom.
Question 1281 Mark
In a $\triangle\text{ABC},$ if $\angle\text{B} = \angle\text{C} = 45^\circ,$ which is the longest side?
Answer
View full question & answer→- BC
Solution:
We know that sum of all angles of a triangle is 180°
$\angle\text{A} + \angle\text{B} + \angle\text{C} = 180^\circ$
$\angle\text{B} = \angle\text{C} = 45^\circ$
$\angle\text{A} + 45^\circ + 45^\circ = 180^\circ$
$\angle\text{A} + 90^\circ = 180^\circ$
$\angle\text{A} = 180^\circ - 90^\circ$
$\angle\text{A} = 90^\circ$
So, angle A is the largest and the side opposite to the greatest angle is the longest so, side BC is the longest.
Question 1291 Mark
Answer
View full question & answer→- RHS
Solution:
In $\triangle\text{ABC}$ and $\angle\text{BAC}=\angle\text{ABD}$
BAD, we have (Right angles)
BC = AD (Hypotentuses and Given)
AB = AB (common in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{BAD}$ by RHS criterion.
Question 1301 Mark
Answer
View full question & answer→- 75º
Solution:
We have,
$∴\ \angle\text{ABD}+\angle\text{ABC}=180^\circ$ [$∵$ E is a straight line]
$⇒125^\circ+\angle\text{ABC}=180^\circ$
$⇒\angle\text{ABC}=55^\circ$
Also,
$\angle\text{ACE}+\angle\text{ACB}=180^\circ$
$⇒130^\circ+\angle\text{ACB}=180^\circ$
$⇒\angle\text{ACB}=50^\circ$
$∴\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ [Sum of the angles of a triangle]
$⇒\angle\text{BAC}+55^\circ+50^\circ=180^\circ$
$⇒\angle\text{BAC}=75^\circ$
$⇒\angle\text{A}=75^\circ.$
Question 1311 Mark
In the adjoining figure, $\angle\text{B} = \angle\text{C}$ and $\text{AD}\bot\text{BC}.$ The rule by which $\triangle\text{ABD}\cong\triangle\text{ADC}.$
Answer
View full question & answer→In $\triangle\text{ABD}$ and $\triangle\text{ADC},$ we have
$\angle\text{ABD} = \angle\text{ACD} \ ($given$)$
$\angle\text{BDA} = \angle\text{CDA} \ (90^\circ)$
$AD = AD ($common in both$)$
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by $\text{AAS}$ criterion.
$\angle\text{ABD} = \angle\text{ACD} \ ($given$)$
$\angle\text{BDA} = \angle\text{CDA} \ (90^\circ)$
$AD = AD ($common in both$)$
Hence, $\triangle\text{ABD}\cong\triangle\text{ADC}$ by $\text{AAS}$ criterion.
Question 1321 Mark
Answer
View full question & answer→- 65º
Solution:
Since BAE is a straight line, we have
$\angle\text{BAC}+\angle\text{CAE}=180^\circ$ .....(Supplementary angles)
$\Rightarrow\angle\text{BAC}+135^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=45^\circ$
Since CBD is a straight line, we have
$\angle\text{ABD}+\angle\text{ABC}=180^\circ$ ....(Supplmenetary angles)
$\Rightarrow110^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=70^\circ$
In $\triangle\text{ABC},$ we have
$\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ$ .....(Angle sum property)
$\Rightarrow45^\circ+70^\circ+\angle\text{ACB}=180^\circ$
$\Rightarrow\angle\text{ACB}=65^\circ$
Question 1331 Mark
Two sides of a triangle are of length 4cm and 2.5cm. The length of the third side of the triangle cannot be:
Answer
View full question & answer→- 6.5cm
Solution:
The sum of any two sides of a triangle is greater than the third side.
Since, 4cm + 2.5cm = 6.5cm
The length of third side of a triangle cannot be 6.5cm.
Question 1341 Mark
In $\triangle\text{AOC}$ and $\triangle\text{XYZ},\ \angle\text{A}=\angle\text{X},\ \angle\text{AO} = \angle\text{XY},\ \angle\text{AC} = \angle\text{XZ},$ then by which congruence rule $\triangle\text{AOC}\cong\triangle\text{XYZ}?$
Answer
View full question & answer→- SAS
Solution:
According to SAS criterion, if the corresponding sides and their included angles are equal, then the triangles are congruent. Here, in $\triangle\text{AOC}$ and $\triangle\text{XYZ},$ AO = XY, and AC = XZ are the corresponding sides and $\angle\text{A}=\angle\text{X}$ are included angles, Hence, $\triangle\text{AOC}\cong\triangle\text{XYZ},$ by SAS.
Question 1351 Mark
The triangle is not possible with the given measurements:
Answer
View full question & answer→- 5.4cm, 2.3cm, 3.1cm
Solution:
In a triangle, the sum of any two sides must be greater than the third side and here 2.3 + 3.1 = 5.4 and hence, the triangle is not possible with the given measurements.
Question 1361 Mark
If all the altitudes from the vertices to the opposite sides of a triangle are equal, then the triangle is:
Answer
View full question & answer→- Equilateral
Solution:
In an equilateral triangle all the altitudes,sides, angles, perpendicular bisectors, medians and angular bisectors are equal.
Question 1371 Mark
In a $\triangle\text{ABC},$ if $\angle\text{A}−\angle\text{B}=42^\circ$ and $\angle\text{B}−\angle\text{C}=21^\circ$ then $\angle\text{B} = ?$
Answer
View full question & answer→- 53º
Solution:
Let,
$\angle\text{A}−\angle\text{B}=42°...\ \text{(i)}$ and
$\angle\text{B}−\angle\text{C}=21^\circ ...\ \text{(ii)}$
Adding (i) and (ii),we get
$\angle\text{A}−\angle\text{C}=63° ...\ \text{(iii)}$
$\angle\text{B}=\angle\text{A}−42^\circ$ [using (i)]
$\angle\text{C}=\angle\text{A}−63^\circ$ [Using (iii)]
$∴\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ [Sum of the angles of a triangle]
$⇒\angle\text{A}+\angle\text{A}−42^\circ+\angle\text{A}−63^\circ=180^\circ$
$⇒3\angle\text{A}−105^\circ=180^\circ$
$⇒3\angle\text{A}=285^\circ$
$∴\angle\text{B}=(95−42)^\circ$
$⇒\angle\text{B}=53^\circ$
Question 1381 Mark
What is the sum of the angles of a quadrilateral?
Answer
View full question & answer→- 360º
Solution:
For a quadrilateral
Number of sides (n) = 4
Sum of interior angles = (n - 2) × 180º
p = (4 - 2) × 180º
p = 2 × 180º
p = 360º
Question 1391 Mark
Answer
View full question & answer→- SSS
Solution:
In $\triangle\text{ABC} $ and $\triangle\text{ADC}, $ we have
AB = AD (4cm)
BC = DC (2.7cm)
AC = AC (commom in both)
Hence, $\triangle\text{ABC}\cong\triangle\text{ADC}$ by SSS criterion.
Question 1401 Mark
If $\triangle\text{ABC}≅\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ},$ then which of the following is not true:
Answer
View full question & answer→- BC = PQ
Solution:
According to the condition given in the question,
If $\triangle\text{ABC}≅\triangle\text{PQR}$ and $\triangle\text{ABC}$ is not congruent to $\triangle\text{RPQ}.$
Then, clearly BC ≠ PQ
$\therefore$ It is false
Question 1411 Mark
Answer
View full question & answer→- 60º
Solution:
Let $ \angle\text{A}=(3\text{x})^\circ,\ \angle\text{B}=(2\text{x})^\circ$ and $\angle\text{C}=\text{x}^\circ$
Then,
3x + 2x + x = 180º [Sum of the angles of a triangle]
⇒ 6x = 180º
⇒ x = 30º
Hence, the angles are
$\angle\text{A}=3×30^\circ=90^\circ,\ \angle\text{B}=2×30^\circ=60^\circ$ and $\angle\text{C}=30^\circ$
Side BC of triangle ABC is produced to E.
$\therefore\ \angle\text{ACE}=\angle\text{A}+\angle\text{B}$
$⇒\angle\text{ACD}+\angle\text{ECD}=90^\circ+60^\circ$
$⇒90^\circ+\angle\text{ECD}=150^\circ$
$⇒\angle\text{ECD}=60^\circ$
Question 1421 Mark
In figure, $AB = AC$ and $\angle\text{ACD} = 115^\circ.$ Find $\angle\text{A}?$
Answer
View full question & answer→$C = 180^\circ - 115^\circ $
$= 65^\circ AB = AC$ And hence
$\angle\text{B} =\angle\text{C} = 65^\circ.$
Then $A = 180^\circ$
$ - 2 \times 65^\circ $
$= 50^\circ$
$= 65^\circ AB = AC$ And hence
$\angle\text{B} =\angle\text{C} = 65^\circ.$
Then $A = 180^\circ$
$ - 2 \times 65^\circ $
$= 50^\circ$
Question 1431 Mark
In the following, write the correct answer.
If AB = QR, BC = PR and CA = PQ, then:
If AB = QR, BC = PR and CA = PQ, then:
Answer
View full question & answer→- $\triangle\text{CBA}\cong\triangle\text{PRQ}$
Solution:
We know that, if is congruent to, then sides of $\triangle\text{RST}$ fall on corresponding equal sides of angles of fall on corresponding equal angles.
Here, given AB = QR, BC = PR and CA = PQ, which shows that AB covers QR, BC covers PR and CA covers PQ i.e., A correspond to Q, B correspond to R and C correspond to P.
Question 1441 Mark
Answer
View full question & answer→- AB > AD
Solution:
Given $\text{AB > AC}$
$\therefore\angle\text{ACB}>\angle\text{ABC}$ ...(We know that, if two sides of a triangle unequal, then the longer side has the greater angle opposite to it.)
$\angle\text{ADB}>\angle\text{ACD}$ ...(Exterior angle of a triangle is greater that the interior opposite angles)
$\therefore\angle\text{ADB}>\angle\text{ACB}>\angle\text{ABC}$
$\therefore\angle\text{ADB}>\angle\text{ABD}$
$\therefore\text{AB > AD}$
Question 1451 Mark
Answer
View full question & answer→- 60º
Solution:
In $\triangle\text{ABC},$
$\angle\text{BCA}+\angle\text{CAB}+\angle\text{ABC}=180^\circ$
$\Rightarrow3\text{y}^\circ+\text{x}^\circ+5\text{y}^\circ=180^\circ$
$\Rightarrow8\text{y}^\circ+\text{x}^\circ=180^\circ\dots(1)$
Also, $5\text{y}^\circ=180^\circ-7\text{y}^\circ$
$\Rightarrow12\text{y}^\circ=180^\circ$
$\Rightarrow\text{y}^\circ=15^\circ$
From (1), $\text{x}^\circ=180^\circ-8\text{y}^\circ$
$\Rightarrow\text{x}^\circ=180^\circ-8\times15^\circ$
$\Rightarrow\text{x}^\circ=60^\circ$
Question 1461 Mark
In $\angle\text{ABC}$ and $\angle\text{DEF},$ AB = DE and $\angle\text{A} = \angle\text{D}.$ Then, the two triangles will be congruent by SAS axiom if:
Answer
View full question & answer→- AC = DF
Solution:
The SAS rule states that:
If two sides and the included angle of one triangle are equal to two sides and included angle of another triangle, then the triangles are congruent.
Here, in $\angle\text{ABC},$ the two sides are AB and AC and the included angle is $\angle\text{A}.$ For $\angle\text{DEF},$ the two corresponding sides are DE and DF and the included angle is $\angle\text{D}.$
Hence, the two triangles will be congruent by SAS axiom if AC = DF.
Question 1471 Mark
Answer
View full question & answer→- 55º
Solution:
Since, It is given that AB = AC, then $\angle\text{B}=\angle\text{C}$ (Isosceles triangle property)
Given $\angle\text{A}=70^\circ$ Let angle B and C be xº.
Sum of all the three angles of triangle = 180º, therefore $\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
70 + x + x = 180º
x = 55º
$\angle\text{C}=55^\circ$
Question 1481 Mark
Answer
View full question & answer→- 50º
Solution:
EC || AB And, CD is transverse to it.
Now $\angle\text{ECD}=\angle\text{AOD}=70^\circ$ (Corresponding angles)
In $\triangle\text{OBD}$
$\angle\text{OBD}+\angle\text{BOD}+\angle\text{ODB}=180^\circ$
$\angle\text{BOD}=180^\circ-\angle\text{AOD}=180^\circ-70^\circ=110^\circ$
So $\angle\text{OBD}=180^\circ-\angle\text{BOD}-\angle\text{ODB}$
$=180^\circ-110^\circ-20^\circ$
$=50^\circ$
Question 1491 Mark
In $\triangle\text{ABC}$ and 6PQR, AB = PR and $\angle\text{A}=\angle\text{P}.$ Then, the two triangles will be congruent by SAS axiom if:
Answer
View full question & answer→- AC = PQ
Solution:
$\angle\text{A}$ is included between AB and AC and LP is included between PQ and PR and corresponding sides must be equal. Since AB = PR, hence AC = PQ for the given triangles to be congruent by SAS axiom.
Question 1501 Mark
In the following, write the correct answer.
In $\triangle\text{ABC}$ if AB = AC and $\angle\text{B}=50^{\circ}$ then is equal to:
In $\triangle\text{ABC}$ if AB = AC and $\angle\text{B}=50^{\circ}$ then is equal to:
Answer
View full question & answer→- 50º
Solution: Given $\triangle\text{ABC}$ such that AB = AC and $\angle\text{B}=50^{\circ}$