[4 marks sum]

Question 14 Marks

Construct an isosceles ∆ ABC such that:
BC = AB = 5.8 cm and ZB = 30°. Measure ∠A and ∠C.

Answer

Steps of Construction :
(i) Draw a line segment $BC =5.8 cm$

(ii) At $B$, draw a ray making an angle of $30^{\circ}$.
(iii) Cut off $B A=5.8 cm$
(iv) Join $AC$.
$\triangle ABC$ is the required triangle On measuring $\angle C$ and $\angle A$, each is equal to $75^{\circ}$.

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Question 24 Marks

Construct an isosceles ∆ ABC such that:
One of the equal sides = 6 cm and vertex angle = 45°. Measure the base angles.

Answer

Steps of Construction :
(i) Draw a line segment $A B=6 cm$

(ii) At A, construct an angle equal to $45^{\circ}$
(iii) Cut off $AC =6 cm$
(iv) Join $B C$.
$\triangle A B C$ is the required triangle.
On measuring, $\angle B$ and $\angle C$, each is equal $1^{\circ}$ to, $67 \frac{1}{2}^{\circ}$

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Question 34 Marks

Construct an isosceles ∆ ABC such that:
AB = AC = 6.5 cm and ∠A = 60°

Answer

Steps of Construction :

(i) Draw a line segment $A B=6.5 cm$.
(ii) At $A$, draw a ray making an angle of $60^{\circ}$.
(iii) Cut off $A C=6.5 cm$
(iv) Join $B C$.
$\triangle ABC$ is the required triangle.

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Question 44 Marks

Construct an isosceles Δ ABC such that:
Base AB = 6.2 cm and base angle = 45°. Measure the other two sides of the triangle.

Answer

Steps of Construction:
We know that in an isosceles triangle, base angles are equal.

(i) Draw a line segment $A B=6.2 cm$
(ii) At $A$ and $B$, draw rays making an angle of $45^{\circ}$ each which intersects each other at $C$.
$\triangle ABC$ is the required triangle.
On measuring the equal sides, each is $4.3 cm$ (approx.) in length.

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Question 54 Marks

Construct a ∆ PQR such that:
PR = 5.8 cm, ∠P = 60° and ∠R = 45°. Measure ∠Q and verify it by calculations

Answer

Steps of Construction :
(i) Draw a line segment $P R=5.8 cm$
(ii) At P, construct an angle of $60^{\circ}$
(iii) At $R$, draw another angle of $45^{\circ}$ meeting each other at $Q$.

$\triangle PQR$ is the required triangle. On measuring $\angle Q$, it is $75^{\circ}$
Verification: We know that sum of angles of a triangle is $180^{\circ}$
$
\begin{aligned}
& \therefore \angle P +\angle Q +\angle R =180^{\circ} \\
& \Rightarrow 60^{\circ}+\angle Q +45^{\circ}=180^{\circ}
\end{aligned}
$
$
\Rightarrow \angle Q +105^{\circ}=180^{\circ}
$
$
\Rightarrow \angle Q=180^{\circ}-105^{\circ}=75^{\circ} \text {. }
$

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Question 64 Marks

The base angle of an isosceles triangle is 15° more than its vertical angle. Find its each angle.

Answer

Let the vertical angle of the isosceles triangle $=x^{\circ}$
$\therefore$ Each base angle $=x+15^{\circ}$
$\therefore \mathrm{x}+15^{\circ}+\mathrm{x}+15^{\circ}+\mathrm{x}^{\circ}=180^{\circ}$ ......... (Sum of angles of a triangle)
$ \Rightarrow 3 \mathrm{x}+30^{\circ}=180^{\circ} $
$ \Rightarrow 3 \mathrm{x}=180^{\circ}-30^{\circ}=150^{\circ} $
$ \therefore x=\frac{150^{\circ}}{3}=50^{\circ} $
Hence vertical angle $=50^{\circ}$
and each base angle $=50^{\circ}+15^{\circ}=65^{\circ}$

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Question 74 Marks

One of the base angles of an isosceles triangle is 52°. Find its angle of the vertex.

Answer

Each of the base angles of an isosceles
$
\triangle A B C=52^{\circ}
$
$
\therefore \angle B=\angle C=52^{\circ}
$

But $\angle A+\angle B+\angle C=180^{\circ}$. .(Angles of a triangle)
$
\Rightarrow \angle A +52^{\circ}+52^{\circ}=180^{\circ}
$
$
\Rightarrow \angle A +104^{\circ}=180^{\circ}
$
$
\Rightarrow \angle A =180^{\circ}-104^{\circ}=76^{\circ}
$
Hence $\angle A=76^{\circ}$

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Question 84 Marks

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figure:

Answer

In fig.,
$a+b+40^{\circ}=180^{\circ}$ (Angles of a triangle)
$ \Rightarrow \mathrm{a}+\mathrm{b}=180^{\circ}-40^{\circ}=140^{\circ} $
But $\mathrm{a}=\mathrm{b}$ (Angles opposite to equal sides)
$ \therefore \mathrm{a}=\mathrm{b}=\frac{140^{\circ}}{2}=70^{\circ} $
$\therefore \mathrm{x}=\mathrm{b}+40^{\circ}=70^{\circ}+40^{\circ}=110^{\circ}$ ... (Exterior angle of a triangle is equal to the sum of its interior opposite angles)
Similarly $y=a+40^{\circ}$
$ =70^{\circ}+40^{\circ}=110^{\circ} $
Hence $x=110^{\circ}, y=110^{\circ}$

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Question 94 Marks

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figure:

Answer

In fig.,
$ x=a+b $
But $b=y$ ......... (Angles opposite to equal sides)
Similarly $\mathrm{a}=\mathrm{c}$
But a $+\mathrm{c}+30^{\circ}=180^{\circ}$
$ \Rightarrow \mathrm{a}+\mathrm{a}+30^{\circ}=180^{\circ} $
$ \Rightarrow 2 \mathrm{a}+30^{\circ}=180^{\circ} $
$\Rightarrow 2 \mathrm{a}=180^{\circ}-30^{\circ}=150^{\circ}$
$\Rightarrow \mathrm{a}=\frac{150^{\circ}}{2}=75^{\circ}$ and $\mathrm{b}+\mathrm{y}=90^{\circ}$
$ \Rightarrow \mathrm{y}+\mathrm{y}=90^{\circ} $
$ \Rightarrow 2 \mathrm{y}=90^{\circ} $
$ \Rightarrow \mathrm{y}=\frac{90^{\circ}}{2}=45^{\circ} \Rightarrow \mathrm{b}=45^{\circ} $
Hence $x=a+b=75^{\circ}+45^{\circ}=120^{\circ}$ and $y=45^{\circ}$

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Question 104 Marks

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figure:

Answer

In fig.,
$ \mathrm{a}=70^{\circ} $........(Angles opposite to equal sides)
But $a+70^{\circ}+x=180^{\circ}$ ......(Angles of a triangle)
$ \Rightarrow 70^{\circ}+70^{\circ}+\mathrm{x}=180^{\circ} $
$ \Rightarrow 140^{\circ}+\mathrm{x}=180^{\circ} $
$ \Rightarrow \mathrm{x}=180^{\circ}-140^{\circ}=40^{\circ} $
$ y=b $ .......... (Angles opposite to equal sides)
But $a=y+b$ ......... (Exterior angle of a triangle is equal to sum of its interior opposite angles)
$ \Rightarrow 70^{\circ}=\mathrm{y}+\mathrm{y} $
$ \Rightarrow 2 \mathrm{y}=70^{\circ} $
$ \Rightarrow \mathrm{y}=\frac{70^{\circ}}{2}=35^{\circ} $
Hence $x=40^{\circ}, y=35^{\circ}$

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Question 114 Marks

Find the unknown angles in the given figure:

Answer

In the figure,
$a=b$ (Angles opposite to equal sides)
But $a+b+80^{\circ}=180^{\circ}$ (Angles of a triangle)
$ \Rightarrow \mathrm{a}+\mathrm{a}+80^{\circ}=180^{\circ} $
$\Rightarrow 2 \mathrm{a}+80^{\circ}=180^{\circ}$
$\Rightarrow 2 \mathrm{a}=180^{\circ}-80^{\circ}=100^{\circ}$
$\Rightarrow \mathrm{a}=\frac{100^{\circ}}{2}=50^{\circ}$
$ \therefore \mathrm{b}=\mathrm{a}=50^{\circ} $
$x=a+80^{\circ}$ (Exterior angle of a triangle is equal to sum of its opposite interior angles)
$ =50^{\circ}+80^{\circ}=130^{\circ} $
Hence $\mathrm{a}=50^{\circ}, \mathrm{b}=50^{\circ}$ and $\mathrm{x}=130^{\circ}$

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Question 124 Marks

In the figure, given below, $A B C D$ is a square, and $\triangle B E C$ is an equilateral triangle. Find, the case: $\angle B A E$

Answer

We know that the sides of a square are equal and each angle is of $90^{\circ}$
Three sides of an equilateral triangle are equal and each angle is of 60 .
In fig.,
$\therefore A B C D$ is a square and $\triangle B E C$ is an equilateral triangle,
$ \begin{aligned} & \text { (i) } \angle \mathrm{ABE}=\angle \mathrm{ABC}-\angle \mathrm{CBE} \\ & =90^{\circ}-60^{\circ}=30^{\circ} \end{aligned} $
(ii) In $\triangle \mathrm{ABE}$,
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{AEB}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . . .(\text { Angles of a triangle }) \\ & \Rightarrow 30^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 30^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-30^{\circ}=150^{\circ} \\ & \Rightarrow \angle \mathrm{BAE}=\frac{150^{\circ}}{2}=75^{\circ} \end{aligned} $

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Question 134 Marks

In the figure, given below, $A B C D$ is a square, and $\triangle B E C$ is an equilateral triangle. Find, the case: $\angle A B E$

Answer

We know that the sides of a square are equal and each angle is of $90^{\circ}$
Three sides of an equilateral triangle are equal and each angle is of 60 .
Therefore, In fig., $A B C D$ is a square and $\triangle B E C$ is an equilateral triangle.
$ \begin{aligned} & \text { (i) } \angle \mathrm{ABE}=\angle \mathrm{ABC}+\angle \mathrm{CBE} \\ & =90^{\circ}+60^{\circ}=150^{\circ} \end{aligned} $
(ii) But in $\triangle \mathrm{ABE}$
$ \begin{aligned} & \angle \mathrm{ABE}+\angle \mathrm{BEA}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots \ldots . .(\text { Angles of a triangle }) \\ & \Rightarrow 150^{\circ}+\angle \mathrm{BAE}+\angle \mathrm{BAE}=180^{\circ} \ldots \ldots \ldots . .(\because \mathrm{AB}=\mathrm{BE}) \\ & \Rightarrow 150^{\circ}+2 \angle \mathrm{BAE}=180^{\circ} \\ & \Rightarrow 2 \angle \mathrm{BAE}=180^{\circ}-150^{\circ}=30^{\circ} \\ & \therefore \angle \mathrm{BAE}=\frac{30^{\circ}}{2}=15^{\circ} \end{aligned} $

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Question 144 Marks

In the given figure, express $a$ in terms of $b$

Answer

In $\triangle A B C$,
$B C=B A$
$\therefore \angle B C A=\angle B A C$
and Ext. $\angle C B E=\angle B C A+\angle B A C$
$\Rightarrow a=\angle B C A+\angle B C A$

$\Rightarrow a=2 \angle B C A$ .........(i)
But $\angle A C B=180^{\circ}-b$ $\ldots \ldots . .(\because \angle A C D$ and $\angle A C B$ are linear pair $)$
$\Rightarrow \angle B C A=180^{\circ}-b$ ..........(ii)
$\begin{aligned} & \therefore a=2 \angle B C A=2\left(180^{\circ}-b\right) \\ & =360^{\circ}-2 b\end{aligned}$

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Question 154 Marks

Calculate the angles of a triangle if they are in the ratio 4: 5: 6.

Answer

We know that sum of angles of a triangle is $180^{\circ}$

$\begin{aligned} & \therefore \angle A +\angle B +\angle C =180^{\circ} \\ & \text { But } \angle A : \angle B : \angle C =4: 5: 6 \\ & \text { Let } \angle A=4 x, \angle B=5 x \text { and } \angle C=6 x, \\ & \text { then } 4 x+5 x+6 x=180^{\circ} \\ & \Rightarrow 15 x =180^{\circ} \\ & \Rightarrow x =\frac{180^{\circ}}{15}=12^{\circ} \\ & \therefore \angle A =4 x =4 \times 12^{\circ}=48^{\circ} \\ & \angle B=5 x=5 \times 12^{\circ}=60^{\circ} \\ & \angle C=6 x=6 \times 12^{\circ}=72^{\circ}\end{aligned}$

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Question 164 Marks

Find the unknown marked angles in the given figure:

Answer

$\angle B+\angle A B C=180^{\circ} \ldots$ (Linear pair)
$
\begin{aligned}
& \angle B+140^{\circ}=180^{\circ} \\
& \angle B=180^{\circ}-140^{\circ} \\
& \angle B=40^{\circ}
\end{aligned}
$
Exterior angle $=$ sum of its interior opposite angles
$
\begin{aligned}
& 2 m+40^{\circ}=4 m \\
& 2 m-4 m=-40^{\circ} \\
& -2 m=-40^{\circ}
\end{aligned}
$
$
\begin{aligned}
& 2 m=40^{\circ} \\
& m=\frac{40}{2} \\
& m =20^{\circ}
\end{aligned}
$
$
\begin{aligned}
& \angle A =2 m =2 \times 20^{\circ}=40^{\circ} \\
& \angle B=40^{\circ} \\
& \angle D =4 m =4 \times 20^{\circ}=80^{\circ}
\end{aligned}
$

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Question 174 Marks

Construct a ∆ PQR such that PQ = 6 cm, ∠QPR = 45° and angle PQR = 60°. Locate its incentre and then draw its incircle.

Answer

Steps of Construction :
(i) Draw a line segment $P Q=6 cm$.
(ii) At $P$ draw rays making an angle of $45^{\circ}$ and at $Q$, making an angle of $60^{\circ}$, intersecting each other at $R$.
(iii) Draw the bisectors of $\angle P$ and $\angle Q$ intersecting each other at I.
(iv) From I, draw $IL \perp PQ$.

(v) With centre I and radius IL, draw a circle which touches the sides of $\triangle PQR$ internally. This is the required incircle whose $I$ is incentre.

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Question 184 Marks

Construct an equilateral ∆ DEF whose one side is 5.5 cm. Construct an incircle to this triangle.

Answer

Steps of Construction :
(i) Draw a line segment $BC =5.5 cm$.

(ii) With centres $B$ and $C$ and radius $5.5 cm$ each draw two arcs intersecting each other at $A$.
(iii) Join $A B$ and $A C$.
(iv) Draw the perpendicular bisectors of $\angle B$ and $\angle C$ intersecting each other at $I$.
(v) From I, draw IL $\perp B C$
(vi) With centre I and radius IL, draw a circle which touches the sides of the $\triangle A B C$ internally.
This is the required incircle.

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Question 194 Marks

Construct an isosceles ∆ MNP such that base MN = 5.8 cm, base angle MNP = 30°. Construct an incircle to this triangle and measure its radius.

Answer

Steps of Construction :
(i) Draw a line segment $MN =5.8 cm$.

(ii) At $M$ and $N$, draw two rays making an angle of $30^{\circ}$ each which intersect each other at $P$.
(iii) Now draw the angle bisectors of $\angle M$ and $\angle N$ which intersects each other at I.
(iv) From I, draw the perpendicular IL on MN.
(v) With centre I and radius IL, draw a circle which touches the sides of the $\triangle$ PMN internally.
On measuring the required incircle and its radius is $0.6 cm$.

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Question 204 Marks

Construct a ∆ ABC such that AB = 6 cm, BC = 5.6 cm and CA = 6.5 cm.
Inscribe a circle to this triangle and measure its radius.

Answer

Steps of Construction :
(i) Draw a line segment $A B=6 cm$.

(ii) With centre $A$ and radius $6.5 cm$ and with centre $B$ and radius 5.6 $cm$, draw arcs intersecting each other at $C$.
(iii) Join $AC$ and $BC$.
(iv) Draw the angle bisector of $\angle A$ and $\angle B$ intersecting each other at I.
(v) From I, draw IL $\perp AB$
(vi) With centre I and radius IL, draw a circle which touches the sides of $\triangle ABC$ internally.
On measuring the required incircle whose radius is $1.6 cm$.

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Question 214 Marks

Construct an equilateral triangle ABC such that it's one side = 5.5 cm. Construct a circumcircle to this triangle.

Answer

Steps of Construction :
(i) Draw a line segment $A B=5.5 cm$

(ii) With centres $A$ and $B$ and radius $5.5 cm$, draw two arcs intersecting each other at $C$.
(iii) Join $A C$ and $B C$.
$\triangle A B C$ is the required triangle.
(iv) Draw perpendicular bisectors of sides $A C$ and $B C$ which intersect each other at $O$.
(v) Join $O A, O B$ and $O C$.
(vi) With centre $O$ and radius $OA$ or $OB$ or $OC$, draw a circle which passes through $A, B$ and $C$. This is the required circumcircle.

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Question 224 Marks

Construct an isosceles ∆ PQR such that PQ = PR = 6.5 cm and ∠PQR = 75°. Using a ruler and compasses only constructs a circumcircle to this triangle.

Answer

Steps of Construction :
(i) Draw a line segment $PQ =6.5 cm$
(ii) At $Q$, draw a ray making an angle of $75^{\circ}$.

(iii) Through P, with a radius of $6.5 cm$, draw an arc which intersects the angle ray at $R$.
(iv) Join PR,
$\triangle P Q R$ is the required triangle.
(v) Draw the perpendicular bisectors of sides PQ and PR intersecting each other at $O$.
(vi) Join $OP , OQ$ and $OR$.
(vii) With centre $O$ and radius equal to $O P$ or $O Q$ or $O R$ draw a circle which passes through $P, Q$ and $R$. This is the required circumcircle of $\triangle PQR$

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Question 234 Marks

Construct a ∆ ABC such that AB = 6 cm, BC = 4.5 cm and AC = 5.5 cm. Construct a circumcircle of this triangle.

Answer

Steps of Construction :
(i) Draw a line segment $BC =4.5 cm$

(ii) With centre $B$ and radius $6 cm$, draw are arc
(iii) With centre $C$ and radius $5.5 cm$, draw another arc intersecting the first arc at $A$.
(iv) Join $A B$ and $A C$.
$\triangle ABC$ is the required triangle.
(v) Draw the perpendicular bisectors of $A B$ and $A C$. Which intersects each other at $O$.
(vi) Join $O B, O C$ and $O A$.
(vii) With centre $O$ and radius $OA$, draw a circle which passes through A, B and C.
This is the required circumcircle of $\triangle A B C$.

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Question 244 Marks

Construct an equilateral Δ ABC such that:
AB = 5 cm. Draw the perpendicular bisectors of BC and AC. Let P be the point of intersection of these two bisectors. Measure PA, PB, and PC.

Answer

Steps of Construction :
(i) Draw a line segment $A B=5 cm$.

(ii) With centres $A$ and $B$ and radius $5 cm$ each, draw two arcs intersecting each other at $C$.
(iii) Join $A C$ and $B C \triangle A B C$ is the required triangle.
(iv) Draw the perpendicular bisectors of sides $A C$ and $B C$ which intersect each other at $P$.
(v) Join PA, PB, and PC.
On measuring, each is $2.8 cm$.

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Question 254 Marks

Apply the properties of isosceles and equilateral triangles to find the unknown angles in the given figure:

Answer

In the fig.,
a = b ..........(Angles opposite to equal sides)
∴ y = 120°
But a + 120° = 180° ..............(Linear pair)
⇒ a = 180°− 120° = 60°
∴ b = 60°
But x + a + b = 180° ...............(Angles of a triangle)
⇒ x + 60° + 60° = 180°
⇒ x + 120° = 180°
∴ x = 180°− 120° = 60°
b = z + 25 .........(Exterior angle of a triangle is equal to the sum of its interior opposite angles)
⇒ 60° = z + 25°
⇒ z = 60°− 25° = 35°
Hence x = 60°, y = 120° and z = 35°

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Question 264 Marks

In $\triangle A B C, B A$ and $B C$ are produced. Find the angles $a$ and $h$. if $A B=$ BC.

Answer

In $\triangle A B C$, sides $B A$ and $B C$ are produced
$ \angle A B C=54^{\circ} ; A B=B C $
Now in $\triangle \mathrm{ABC}$,
$ \angle \mathrm{BAC}+\angle \mathrm{BCA}+\angle \mathrm{ABC}=180^{\circ} . $........ (Angles of a triangle)
$ \Rightarrow \angle \mathrm{BAC}+\angle \mathrm{BAC}+54^{\circ}=180^{\circ} $
$ (\because A B=B C) $
$ \Rightarrow 2 \angle \mathrm{BAC}=180^{\circ}-54^{\circ} $
$ \Rightarrow 2 \angle \mathrm{BAC}=126^{\circ} $
$\therefore \angle \mathrm{BAC}=\frac{126^{\circ}}{2}=63^{\circ}$ and $\angle \mathrm{BCA}=63^{\circ}$
$ \angle \mathrm{BAC}+\mathrm{b}=180^{\circ} $.......... (Linear pair)
$ \Rightarrow 63^{\circ}+\mathrm{b}=180^{\circ} $
$ \Rightarrow \mathrm{b}=180^{\circ}-63^{\circ}=117^{\circ} $
and $\angle \mathrm{BCA}+\mathrm{a}=180^{\circ}$. (Linear pair)
$ \therefore 63^{\circ}+\mathrm{a}=180^{\circ} $
$ \Rightarrow \mathrm{a}=180^{\circ}-63^{\circ}=117^{\circ} $
Hence $\mathrm{a}=117^{\circ}, \mathrm{b}=117$

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Question 274 Marks

Find $x$ in Figure Given: $D A=D B=D C, B D$ bisects $\angle A B C$ and $\angle A D B=$$70^{\circ}$.

Answer

In the figure,
$D A=D B=D C$
$B D$ bisects $\angle A B C$
and $\angle ADB =70^{\circ}$

But $\angle ADB +\angle DAB +\angle DBA =180^{\circ} \ ($Angles of a triangle$)$
$\Rightarrow 70^{\circ}+\angle DBA +\angle DBA =180^{\circ} \ldots \ldots . .(\because DA = DB )$
$\Rightarrow 70^{\circ}+2 \angle DBA =180^{\circ}$
$\Rightarrow 2 \angle DBA =180^{\circ}-70^{\circ}$
$=110^{\circ}$
$\therefore \angle DBA =\frac{110^{\circ}}{2}=55^{\circ}$
$\because B D$ is the bisector of $\angle A B C,$
$\therefore \angle DBA =\angle DBC =55^{\circ}$
But in $\triangle DBC$,
$D B=D C$
$\therefore \angle DCB =\angle DBC$
$\Rightarrow x=55^{\circ}$

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Question 284 Marks

In Figure, $B P$ bisects $\angle A B C$ and $A B=A C$. Find $x$

Answer

In the figure,
$A B=A C$ and $B P$ bisects $\angle A B C$
$A P \| B C$ is drawn.
Now $\angle \mathrm{PBC}=\angle \mathrm{PBA}$ ( $\because P B$ is the bisector of $\angle A B C$ )
$\because A P \| B C$
$\therefore \angle \mathrm{APB}=\angle \mathrm{PBC}$ ....... (Alternate angles)
$ \Rightarrow x=\angle \mathrm{PBC} $
In $\triangle \mathrm{ABC}, \angle \mathrm{A}=60^{\circ}$
and $\angle B=\angle C$. $\ldots(\because A B=A C)$
But $\angle A+\angle B+\angle C=180^{\circ}$......... (Angles of a triangle)
$\Rightarrow 60^{\circ}+\angle B+\angle C=180^{\circ}$
$\Rightarrow 60^{\circ}+\angle B+\angle B=180^{\circ}$
$\Rightarrow 2 \angle \mathrm{B}=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore \angle B=\frac{120^{\circ}}{2}=60^{\circ}$
$\Rightarrow \frac{1}{2} \angle \mathrm{B}=\frac{60^{\circ}}{2}=30^{\circ}$
$\Rightarrow \angle \mathrm{PBC}=30^{\circ}$
$\therefore$ From (i),
$ x=30^{\circ} $

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Question 294 Marks

In the given figure, BI is the bisector of ∠ABC and Cl is the bisector of ∠ACB. Find ∠BIC.

Answer

$
\text { In } \triangle A B C \text {, }
$
$B I$ is the bisector of $\angle A B C$ and $C I$ is the bisector of $\angle A C B$.

$\begin{aligned} & \because A B=A C \\ & \therefore \angle B=\angle C\end{aligned}$ ......(Angles opposite to equal sides)
But $\angle A =40^{\circ}$
and $\angle A+\angle B+\angle C=180^{\circ}$. (Angles of a triangle)
$
\begin{aligned}
& \Rightarrow 40^{\circ}+\angle B+\angle B=180^{\circ} \\
& \Rightarrow 40^{\circ}+2 \angle B =180^{\circ} \\
& \Rightarrow 2 \angle B=180^{\circ}-40^{\circ}=140^{\circ} \\
& \Rightarrow \angle B=\frac{140^{\circ}}{2}=70^{\circ}
\end{aligned}
$
$
\therefore \angle ABC =\angle ACB =70^{\circ}
$
But $BI$ and $Cl$ are the bisectors of $\angle ABC$ and $\angle ACB$ respectively.
$
\therefore \angle BC =\frac{1}{2} \angle ABC =\frac{1}{2}\left(70^{\circ}\right)=35^{\circ}
$
and $\angle ICB =\frac{1}{2} \angle ACB =\frac{1}{2} \times 70^{\circ}=35^{\circ}$
Now in $\triangle I B C$
$
\begin{aligned}
& \angle BIC +\angle IBC +\angle ICB =180^{\circ} . \ldots \ldots . . \text { (Angles of a triangle) } \\
& \Rightarrow \angle B I C+35^{\circ}+35^{\circ}=180^{\circ} \\
& \Rightarrow \angle BIC +70^{\circ}=180^{\circ} \\
& \Rightarrow \angle BIC =180^{\circ}-70^{\circ}=110^{\circ}
\end{aligned}
$
Hence $\angle BIC =110^{\circ}$

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Question 304 Marks

In the given figure, show that: $\angle a =\angle b +\angle c$

(i) If $\angle b =60^{\circ}$ and $\angle c =50^{\circ}$; find $\angle a$.
(ii) If $\angle a =100^{\circ}$ and $\angle b =55^{\circ}$ : find $\angle c$.
(iii) If $\angle a =108^{\circ}$ and $\angle c =48^{\circ}$; find $\angle b$.

Answer

∵ AB || CD
∴ b = c and ∠A = ∠C ........(Alternate angles)
Now in Δ PCD,
Ext. ∠APC = ∠C + ∠D
⇒ a = b + c
(i) If b = 60°, c = 50°, then
a = b + c = 60° + 50° = 110°
(ii) If a = 100° and b = 55°,
then a = b + c
⇒ 100° = 55°+ c
⇒ c = 100°− 55° = 45°
(iii) If a = 108° and c = 48°
then a = b + c
⇒ 108° = b + 48°
⇒ b = 108°− 48° = 60°

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Question 314 Marks

Find the unknown marked angles in the given figure:

Answer

We know that in a triangle if one side of it is produced, then
Exterior angle = sum of its interior opposite angles.
In fig., 125° = a + c .......(i)
and 140° = a + b ............(ii)
Adding, we get
a + c + a + b = 125° + 140°
⇒ a + a + b + c = 265°
But a + b + c = 180° ........(sum of angles of a triangle)
∴ a + 180° = 265°
⇒ a = 265° − 180° = 85°
But a + b = 140°
⇒ 85° + b = 140°
⇒ b = 140°− 85° = 55°
and a + c = 125°
⇒ 85° + c = 125°
⇒ c = 125° − 85° = 40°
Hence a = 85°, b = 55° and c = 40°

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Question 324 Marks

One angle of a triangle is $61^{\circ}$ and the other two angles are in the ratio $1 \frac{1}{2}: 1 \frac{1}{3}$. Find these angles.

Answer

In $\triangle A B C$
Let $\angle A=61^{\circ}$
But $\angle A+\angle B+\angle C=180^{\circ}$ ...........(Angles of a triangle)

$
\begin{aligned}
& \Rightarrow 61^{\circ}+\angle B+\angle C=180^{\circ} \\
& \Rightarrow \angle B+\angle C=180^{\circ}-61^{\circ}=119^{\circ} \\
& \text { But } \angle B: \angle C=1 \frac{1}{2}: 1 \frac{1}{3}=\frac{3}{2}: \frac{4}{3} \\
& =\frac{9: 8}{6} \\
& =9: 8
\end{aligned}
$
Let $\angle B=9 x$ and $\angle C=8 x$,
then, $9 x+8 x=119^{\circ}$
$
\begin{aligned}
& \Rightarrow 17 x=119^{\circ} \\
& \Rightarrow x=\frac{119^{\circ}}{17}=7^{\circ} \\
& \therefore \angle B=9 x=9 \times 7^{\circ}=63^{\circ} \\
& \angle C=8 x=8 \times 7^{\circ}=56^{\circ}
\end{aligned}
$

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Question 334 Marks

One angle of a triangle is 60°. The other two angles are in the ratio of 5: 7. Find the two angles.

Answer

$
\text { In } \triangle A B C \text {, }
$
Let $\angle A=60^{\circ}$ and then $\angle B: \angle C=5: 7$
But $\angle A +\angle B +\angle C =180^{\circ}$ ..........(Angles of a triangle)

$\begin{aligned} & \Rightarrow 60^{\circ}+\angle B+\angle C=180^{\circ} \\ & \Rightarrow \angle B+\angle C=180^{\circ}-60^{\circ}=120^{\circ} \\ & \text { Let } \angle B=5 x \text { and } \angle C=7 x \\ & \therefore 5 x+7 x=120^{\circ} \\ & \Rightarrow 12 x =120^{\circ} \\ & \Rightarrow x =\frac{120^{\circ}}{12}=10^{\circ} \\ & \therefore \angle B =5 x =5 \times 10^{\circ}=50^{\circ} \\ & \angle C =7 x =7 \times 10^{\circ}=70^{\circ}\end{aligned}$

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