RS AGGARWAL CLASS 9 CHAPTER 8 TRIANGLES EXERCISE 8

 EXERCISE 8

PAGE NO-252

Question 1:

In ∆ABC, if ∠B = 76° and ∠C = 48°, find ∠A.

Answer 1:

$$\begin{gathered} \mathrm{In}∆ABC, \\ \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle A+76°+48°=180° \\ \Rightarrow \angle A+124°=180° \\ \Rightarrow \angle A=56° \end{gathered}$$

Question 2:

The angles of a triangle are in the ratio 2 : 3 : 4. Find the angles.

Answer 2:

Let the angles of the given triangle measure $\left(2x\right)°, \left(3x\right)° \text{and} \left(4x\right)°$, respectively.
Then,
$$\begin{gathered} 2x+3x+4x=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 9x=180° \\ \Rightarrow x=20° \end{gathered}$$
Hence, the measures of the angles are $2\times 20°=40°, 3\times 20°=60° \text{and 4}\times 20°\text{=}80°$.

Question 3:

In ∆ABC, if 3∠A = 4 ∠B = 6 ∠C, calculate ∠A, ∠B and ∠C.

Answer 3:

Let $3\angle A=4\angle B=6\angle C=x°$.
Then,
$$\begin{gathered} \angle A={\left(\frac{x}{3}\right)}^{°}, \angle B={\left(\frac{x}{4}\right)}^{°} \text{and}\angle C={\left(\frac{x}{6}\right)}^{°} \\ \therefore \frac{x}{3}+\frac{x}{4}+\frac{x}{6}=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 4x+3x+2x=2160° \\ \Rightarrow 9x=2160° \\ \Rightarrow x=240° \end{gathered}$$
Therefore,
$\angle A={\left(\frac{240}{3}\right)}^{°}=80°, \angle B={\left(\frac{240}{4}\right)}^{°}=60° \text{and}\angle C={\left(\frac{240}{6}\right)}^{°}=40°$

Question 4:

In ∆ABC, if ∠A + ∠B = 108° and ∠B + ∠C = 130°, find ∠A, ∠B and ∠C.

Answer 4:

Let $\angle A+\angle B=108° \text{and} \angle B+\angle C=130°$.
$$\begin{gathered} \Rightarrow \angle A+\angle B+\angle B+\angle C=\left(108+130\right)° \\ \Rightarrow \left(\angle A+\angle B+\angle C\right)+\angle B=238° \left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow 180°+\angle B=238° \\ \Rightarrow \angle B=58° \end{gathered}$$

$\therefore \angle C=130°-\angle B$
          $$\begin{gathered} =\left(130-58\right)° \\ =72° \end{gathered}$$

$\therefore \angle A=108°-\angle B$
         $$\begin{gathered} =\left(108-58\right)° \\ =50° \end{gathered}$$

Question 5:

In ∆ABC, ∠A + ∠B = 125° and ∠A + ∠C = 113°. Find ∠A, ∠B and ∠C.

Answer 5:

Let $\angle A+\angle B=125° \text{and} \angle A+\angle C=113°$.
Then,
$$\begin{gathered} \angle A+\angle B+\angle A+\angle C=\left(125+113\right)° \\ \Rightarrow \left(\angle A+\angle B+\angle C\right)+\angle A=238° \\ \Rightarrow 180°+\angle A=238° \\ \Rightarrow \angle A=58° \end{gathered}$$

$\therefore \angle B=125°-\angle A$
           $$\begin{gathered} =\left(125-58\right)° \\ =67° \end{gathered}$$

$\therefore \angle C=113°- \angle A$
         $$\begin{gathered} =\left(113-58\right)° \\ =55° \end{gathered}$$

Question 6:

In ∆PQR, if ∠P − ∠Q = 42° and ∠Q − ∠R = 21°, find ∠P, ∠Q and ∠R.

Answer 6:

 $\mathrm{Given}: \angle P-\angle Q=42° \text{and} \angle Q-\angle R=21°$
Then,
$$\begin{gathered} \angle P=42°+\angle Q \text{and} \angle R=\angle Q-21° \\ \therefore 42°+\angle Q+\angle Q+\angle Q-21°=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 3\angle Q=159° \\ \Rightarrow \angle Q=53° \end{gathered}$$

$\mathbf{\therefore }\angle P=42°+\angle Q$
         $$\begin{gathered} =\left(42+53\right)° \\ =95° \end{gathered}$$

$\therefore \angle R=\angle Q-21°$
         $$\begin{gathered} =\left(53-21\right)° \\ =32° \end{gathered}$$

Question 7:

The sum of two angles of a triangle is 116° and their difference is 24°. Find the measure of each angle of the triangle.

Answer 7:

Let $\angle A+\angle B=116° \text{and} \angle A-\angle B=24°$
Then,
$$\begin{gathered} \therefore \angle A+\angle B+\angle A-\angle B=\left(116+24\right)° \\ \Rightarrow 2\angle A=140° \\ \Rightarrow \angle A=\mathbf{70}\mathbf{°} \end{gathered}$$

$\therefore \angle B=116°-\angle A$
          $$\begin{gathered} =\left(116-70\right)° \\ =\mathbf{46}\mathbf{°} \end{gathered}$$

Also, in ∆ ABC:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 70°+46°+\angle C=180° \\ \Rightarrow \angle C=\mathbf{64}\mathbf{°} \end{gathered}$$

Question 8:

Two angles of a triangle are equal and the third angle is greater than each one of them by 18°. Find the angles.

Answer 8:

Let $\angle A=\angle B \text{and} \angle C=\angle A+18°$.
Then,
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \angle A+\angle A+\angle A+18°=180° \\ \Rightarrow 3\angle A=162° \\ \Rightarrow \angle A=54° \end{gathered}$$
Since,
$$\begin{gathered} \angle A=\angle B \\ \Rightarrow \angle B=54° \\ \therefore \angle C=\angle A+18° \end{gathered}$$

        $$\begin{gathered} =\left(54+18\right)° \\ =72° \end{gathered}$$

Question 9:

Of the three angles of a triangle, one is twice the smallest and another one is thrice the smallest. Find the angles.

Answer 9:

Let the smallest angle of the triangle be $\angle C$ and let $\angle A=2\angle C \text{and} \angle B=3\angle C$.
Then,
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 2\angle C+3\angle C+\angle C=180° \\ \Rightarrow 6\angle =180° \\ \Rightarrow \angle C=30° \end{gathered}$$

$\therefore \angle A=2\angle C$
         $$\begin{gathered} =2\left(30\right)° \\ =60° \end{gathered}$$
Also,
$$\begin{gathered} \angle B=3\angle C \\ =3\left(30\right)° \\ =90° \end{gathered}$$

Question 10:

In a right-angled triangle, one of the acute angles measures 53°. Find the measure of each angle of the triangle.

Answer 10:

Let ABC be a triangle right-angled at B.
Then, $\angle B=90° \text{and let }\angle A=53°.$
$$\begin{gathered} \therefore \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 53°+90°+\angle C=180° \\ \Rightarrow \angle C=37° \end{gathered}$$
Hence, $\angle A=53°, \angle B=90° \text{and} \angle C=37°$.

Question 11:

If one angle of a triangle is equal to the sum of the other two, show that the triangle is right-angled.

Answer 11:

Let ABC be a triangle.
Then,$\angle A=\angle B+\angle C$
$$\begin{gathered} \therefore \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle B+\angle C+\angle B+\angle C=180° \\ \Rightarrow 2\angle B+\angle C=180° \\ \Rightarrow \angle B+\angle C=90° \\ \Rightarrow \angle A=\mathbf{90}\mathbf{°} \left[\because \angle A=\angle B+\angle C\right] \end{gathered}$$
This implies that the triangle is right-angled at A.

Question 12:

If each angle of a triangle is less than the sum of the other two, show that the triangle is acute-angled.

Answer 12:

Let ABC be the triangle.
Let $\angle A<\angle B+\angle C$
Then,
$$\begin{gathered} 2\angle A<\angle A+\angle B+\angle C \left[\text{Adding }\angle A \text{to both sides}\right] \\ \Rightarrow 2\angle A<180° \left[\because \angle A+\angle B+\angle C =180°\right] \\ \Rightarrow \angle A<\mathbf{90}\mathbf{°} \end{gathered}$$

Also, let $\angle B<\angle A+\angle C$
Then,
$$\begin{gathered} 2\angle B<\angle A+\angle B+\angle C \left[\text{Adding }\angle B \text{to both sides}\right] \\ \Rightarrow 2\angle B<180° \left[\because \angle A+\angle B+\angle C =180°\right] \\ \Rightarrow \angle B<\mathbf{90}\mathbf{°} \end{gathered}$$

And let $\angle C<\angle A+\angle B$
Then,
$$\begin{gathered} 2\angle C<\angle A+\angle B+\angle C \left[\text{Adding }\angle C \text{to both sides}\right] \\ \Rightarrow 2\angle C<180° \left[\because \angle A+\angle B+\angle C =180°\right] \\ \Rightarrow \angle C<\mathbf{90}\mathbf{°} \end{gathered}$$

Hence, each angle of the triangle is less than $90°$.
Therefore, the triangle is acute-angled.

Question 13:

If one angle of a triangle is greater than the sum of the other two, show that the triangle is obtuse-angled.

Answer 13:

Let ABC be a triangle and let $\angle C>\angle A+\angle B$.
Then, we have:
$$\begin{gathered} 2\angle C>\angle A+\angle B+\angle C \left[\text{Adding }\angle C \text{to both sides}\right] \\ \Rightarrow 2\angle C>180° \left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow \angle C>\mathbf{90}\mathbf{°} \end{gathered}$$
Since one of the angles of the triangle is greater than $90°$, the triangle is obtuse-angled.

Question 14:

In the given figure, side BC of ∆ABC is produced to D. If ∠ACD = 128° and ∠ABC = 43°, find ∠BAC and ∠ACB.

Answer 14:

Side BC of triangle ABC is produced to D.
$$\begin{gathered} \therefore \angle ACD=\angle A+\angle B [\mathrm{Exterior} \mathrm{angle} \mathrm{property}] \\ \Rightarrow 128°=\angle A+43° \\ \Rightarrow \angle A=\left(128-43\right)° \\ \Rightarrow \angle A=85° \\ \Rightarrow \angle BAC=85° \end{gathered}$$
Also, in triangle ABC,
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 85°+43°+\angle ACB=180° \\ \Rightarrow 128°+\angle ACB=180° \\ \Rightarrow \angle ACB=52° \end{gathered}$$

Question 15:

In the given figure, the side BC of ∆ ABC has been produced on both sides−on the left to D and on the right to E. If ∠ABD = 106° and ∠ACE = 118°, find the measure of each angle of the triangle.

Answer 15:

Side BC of triangle ABC is produced to D.

$$\begin{gathered} \therefore \angle ABC=\angle A+\angle C \\ \Rightarrow 106°=\angle A+\angle C ...\left(i\right) \end{gathered}$$

Also, side BC of triangle ABC is produced to E.

$$\begin{gathered} \angle ACE=\angle A+\angle B \\ \Rightarrow 118°=\angle A+\angle B ...\left(ii\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Adding} \left(i\right) \mathrm{and} \left(ii\right), \mathrm{we} \mathrm{get}: \\ \angle A+\angle A+\angle B+\angle C=\left(106+118\right)° \end{gathered}$$
$$\begin{gathered} \Rightarrow \left(\angle A+\angle B+\angle C\right)+\angle A=224° \left[\because \angle A+\angle B+\angle C=180°\right] \\ \Rightarrow 180°+\angle A=224° \\ \Rightarrow \angle A=44° \end{gathered}$$

$$\begin{gathered} \therefore \angle B=118°-\angle A \left[\text{Using }\left(ii\right)\right] \\ \Rightarrow \angle B=\left(118-44\right)° \\ \Rightarrow \angle B=74° \end{gathered}$$

And,

$$\begin{gathered} \angle C=106°-\angle A \left[\text{Using }\left(i\right)\right] \\ \Rightarrow \angle C=\left(106-44\right)° \\ \Rightarrow \angle C=62° \end{gathered}$$

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Question 16:

Calculate the value of x in each of the following figures.

Answer 16:

(i)
Side AC of triangle ABC is produced to E.
$$\begin{gathered} \therefore \angle EAB=\angle B+\angle C \\ \Rightarrow 110°=x+\angle C ...\left(i\right) \end{gathered}$$
Also,
$$\begin{gathered} \angle ACD+\angle ACB=180° \left[\text{linear pair}\right] \\ \Rightarrow 120°+\angle ACB=180° \\ \Rightarrow \angle ACB=60° \\ \Rightarrow \angle C=60° \end{gathered}$$
Substituting the value of $\angle C \text{in} \left(i\right), \text{we get} x=50$

(ii)
From $∆ABC$ we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 30°+40°+\angle C=180° \\ \Rightarrow \angle C=110° \\ \Rightarrow \angle ACB=110° \end{gathered}$$
Also,
$$\begin{gathered} \angle ECB+\angle ECD=180° \left[\text{linear pair}\right] \\ \Rightarrow 110°+\angle ECD=180° \\ \Rightarrow \angle ECD=70° \\ \mathrm{Now}, \text{in} ∆ECD, \\ \therefore \angle AED=\angle ECD+\angle EDC [\mathrm{exterior} \mathrm{angle} property] \\ \Rightarrow x=70°+50° \\ \Rightarrow x=120° \end{gathered}$$

(iii)
$$\begin{gathered} \angle ACB+\angle ACD=180° \left[\text{linear pair}\right] \\ \Rightarrow \angle ACB+115°=180° \\ \Rightarrow \angle ACB=65° \end{gathered}$$
Also,
$$\begin{gathered} \angle EAF=\angle BAC \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle BAC=60° \\ \therefore \angle BAC+\angle ABC+\angle ACB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 60°+x+65°=180° \\ \Rightarrow x=55° \end{gathered}$$

(iv)
$$\begin{gathered} \angle BAE=\angle CDE \left[\text{Alternate angles}\right] \\ \Rightarrow \angle CDE=60° \\ \therefore \angle ECD+\angle CDE+\angle CED=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 45°+60°+x=180° \\ \Rightarrow x=75° \end{gathered}$$

(v)
From $∆ABC$, we have:
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 40°+\angle ABC+90°=180° \\ \Rightarrow \angle ABC=50° \end{gathered}$$
Also, from $∆EBD$, we have:
$$\begin{gathered} \angle BED+\angle EBD+\angle BDE=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 100°+50°+x=180° \left[\because \angle ABC=\angle EBD\right] \\ \Rightarrow x=30° \end{gathered}$$

(vi)
From $∆ABE$, we have:
$$\begin{gathered} \angle BAE+\angle ABE+\angle AEB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 75°+65°+\angle AEB=180° \\ \Rightarrow \angle AEB=40° \\ \therefore \angle AEB=\angle CED \left[\text{Vertically-opposite angles}\right] \\ \Rightarrow \angle CED=40° \end{gathered}$$

Also, From $∆CDE$, we have
$$\begin{gathered} \angle ECD+\angle CDE+\angle CED=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 110°+x+40°=180° \\ \Rightarrow x=30° \end{gathered}$$

Question 17:

In the figure given alongside, AB || CD, EF || BC, ∠BAC = 60º and ∠DHF = 50º. Find ∠GCH and ∠AGH. 

Answer 17:

In the given figure, AB || CD and AC is the transversal.

∴ ∠ACD = ∠BAC = 60º     (Pair of alternate angles)

Or ∠GCH = 60º

Now, ∠GHC = ∠DHF = 50º      (Vertically opposite angles)

In ∆GCH,

∠AGH = ∠GCH + ∠GHC       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ ∠AGH = 60º + 50º = 110º
 

Question 18:

Calculate the value of x in the given figure.

Answer 18:

Join A and D to produce AD to E.
Then,
$$\begin{gathered} \angle CAD+\angle DAB=55° \text{and} \\ \angle CDE+\angle EDB=x° \end{gathered}$$
Side AD of triangle ACD is produced to E.
$\therefore \angle CDE=\angle CAD+\angle ACD ...\left(i\right)$   (Exterior angle property)
Side AD of triangle ABD is produced to E.
$\therefore \angle EDB=\angle DAB+\angle ABD ...\left(ii\right)$   (Exterior angle property)
$$\begin{gathered} \mathrm{Adding} \left(\mathrm{i}\right) \mathrm{and} \left(\mathrm{ii}\right)\mathrm{we} \mathrm{get}, \\ \angle CDE+\angle EDB=\angle CAD+\angle ACD+\angle DAB+\angle ABD \end{gathered}$$
 
                 $$\begin{gathered} \Rightarrow x°=\left(\angle CAD+\angle DAB\right)+30°+45° \\ \Rightarrow x°=55°+30°+45° \\ \Rightarrow x°=130° \\ \Rightarrow x=130 \end{gathered}$$

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Question 19:

In the given figure, AD divides ∠BAC in the ratio 1 : 3 and AD = DB. Determine the value of x.

Answer 19:

$$\begin{gathered} \angle BAC+\angle CAE=180° \left[\because BE \text{is a straight line}\right] \\ \Rightarrow \angle BAC+108°=180° \\ \Rightarrow \angle BAC=72° \end{gathered}$$

Now, divide $72°$ in the ratio 1 : 3.
$$\begin{gathered} \therefore a+3a=72° \\ \Rightarrow a=18° \\ \therefore a=18° \text{and} 3a=54° \end{gathered}$$
Hence, the angles are 18^o and 54^o
$\therefore \angle BAD=18° \text{and} \angle DAC=54°$

Given,
$$\begin{gathered} AD=DB \\ \Rightarrow \angle DAB=\angle DBA=18° \end{gathered}$$

In $∆ABC$, we have:
$$\begin{gathered} \angle BAC+\angle ABC+\angle ACB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 72°+18°+x°=180° \\ \Rightarrow x°=90° \\ \therefore x=90 \end{gathered}$$

Question 20:

If the sides of a triangle are produced in order, prove that the sum of the exterior angles so formed is equal to four right angles.

Answer 20:

Side BC of triangle ABC is produced to D.
$\angle ACD=\angle B+\angle A ...\left(i\right)$
Side AC of triangle ABC is produced to E.
$\angle BAC=\angle B+\angle C ...\left(i\right)$
And side AB of triangle ABC is produced to F.
$\angle CBF=\angle C+\angle A ...\left(iii\right)$

$$\begin{gathered} \mathrm{Adding} \left(i\right), \left(ii\right) \mathrm{and} \left(iii\right), \text{we get:} \\ \angle ACD+\angle BAE+\angle CBF=2\left(\angle A+\angle B+\angle C\right) \end{gathered}$$
                               $$\begin{gathered} =2\left(180\right)° \\ =360°=4\times 90°=4 \mathrm{right} \mathrm{angle}s \end{gathered}$$
Hence, the sum of the exterior angles so formed is equal to four right angles.

Question 21:

In the adjoining figure, show that
∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360°.

Answer 21:

In $∆ACE$ , we have :
$\angle A+\angle C+\angle E=180° ...\left(i\right)$   [Sum of the angles of a triangle]
In $∆BDF$, we have :
$\angle B+\angle D+\angle F=180° ...\left(ii\right)$   [Sum of the angles of a triangle]​

$$\begin{gathered} \mathrm{Adding} \left(i\right) \mathrm{and} \left(ii\right), \text{we get:} \\ \angle A+\angle C+\angle E+\angle B+\angle D+\angle F=\left(180+180\right)° \end{gathered}$$
$\Rightarrow \angle A+\angle B+\angle C+\angle D+\angle E+\angle F\mathbf{=}\mathbf{360}\mathbf{°}$

Question 22:

In the given figure, AM ⊥ BC and AN is the bisector of ∠A. If ∠ABC = 70° and ∠ACB = 20°, then ∠MAN = ?

Answer 22:

In $∆ABC$, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle A+70°+20°=180° \\ \Rightarrow \angle A=90° \\ \Rightarrow \frac{1}{2}\angle A=45° \\ \Rightarrow \angle BAN=45° \end{gathered}$$
In $∆ABM$, we have:
$$\begin{gathered} \angle ABM+\angle AMB+\angle BAM=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 70°+90°+\angle BAM=180° \\ \Rightarrow \angle BAM=20° \\ \therefore \angle MAN=\angle BAN-\angle BAM \\ \Rightarrow \angle MAN=45°-20° \\ \Rightarrow \angle MAN=25° \end{gathered}$$

Question 23:

In the given figure, BAD || EF, ∠AEF = 55° and ∠ACB = 25°, find ∠ABC.

Answer 23:

In the given figure, EF || BD and CE is the transversal.

∴ ∠CAD = ∠AEF         (Pair of corresponding angles)

⇒ ∠CAD = 55°

In ∆ABC,

∠CAD = ∠ABC + ∠ACB       (Exterior angle of a triangle is equal to the sum of the two interior opposite angles)

⇒ 55° = ∠ABC + 25°

⇒ ∠ABC = 55° − 25° = 30°

Thus, the measure of ∠ABC is 30°.

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Question 24:

In the given figure, ABC is a triangle in which ∠A : ∠B : ∠C = 3 : 2 : 1 and AC ⊥ CD. Find the measure of ∠ECD.

Answer 24:

Let $\angle A=\left(3x\right)°, \angle B=\left(2x\right)° \text{and} \angle C=x°$
From $∆ABC$, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 3x+2x+x=180° \\ \Rightarrow 6x=180° \\ \Rightarrow x=30° \\ \therefore \angle A=3\left(30\right)°=60° \\ \angle B=2\left(30\right)°=60° \\ \text{and} \angle C=30° \end{gathered}$$
Side BC of triangle ABC is produced to E.
$$\begin{gathered} \therefore \angle ACE=\angle A+\angle B \\ \Rightarrow \angle ACD+\angle ECD=\left(90+60\right)° \\ \Rightarrow 90°+\angle ECD=150° \\ \Rightarrow \angle ECD=60° \end{gathered}$$

Question 25:

In the given figure, AB || DE and BD || FG such that ∠ABC = 50° and ∠FGH = 120°. Find the values of x and y.

Answer 25:

We have, ∠FGE + ∠FGH = 180°       (Linear pair of angles)

∴ y + 120° = 180°

⇒ y = 180° − 120° = 60°

Now, AB || DF and BD is the transversal.

∴ ∠ABD = ∠BDF           (Pair of alternate angles)

⇒ ∠BDF = 50°

Also, BD || FG and DF is the transversal.

∴ ∠BDF = ∠DFG           (Pair of alternate angles)

⇒ ∠DFG = 50°       .....(1)

In ∆EFG,

∠FGH = ∠EFG + ∠FEG       (Exterior angle of a triangle is equal to the sum of the two opposite interior angles)

⇒ 120° = 50° + x                   [Using (1)]

⇒ x = 120° − 50° = 70°

Thus, the values of x and y are 70° and 60°, respectively.

Question 26:

In the given figure, AB || CD and EF is a transversal. If ∠AEF = 65°, ∠DFG = 30°, ∠EGF = 90° and ∠GEF = x°, find the value of x.

Answer 26:

It is given that, AB || CD and EF is a transversal.

∴ ∠EFD = ∠AEF          (Pair of alternate angles)

⇒ ∠EFD = 65°

⇒ ∠EFG + ∠GFD = 65° 

⇒ ∠EFG + 30° = 65°

⇒ ∠EFG = 65° − 30° = 35°

In ∆EFG, 

∠EFG + ∠GEF + ∠EGF = 180°           (Angle sum property)

⇒ 35° + x + 90° = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the value of x is 55°.

Question 27:

In the given figure, AB || CD, ∠BAE = 65° and ∠OEC = 20°. Find ∠ECO.

Answer 27:

In the given figure, AB || CD and AE is the transversal.

∴ ∠DOE = ∠BAE         (Pair of corresponding angles)

⇒ ∠DOE = 65°

In ∆COE,

∠DOE = ∠OEC + ∠ECO          (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ 65° = 20° + ∠ECO

⇒ ∠ECO = 65° − 20° = 45°

Thus, the measure of ∠ECO is 45°.

Question 28:

In the given figure, AB || CD and EF is a transversal, cutting them at G and H respectively. If ∠EGB = 35° and QP ⊥ EF, find the measure of ∠PQH.

Answer 28:

In the given figure, AB || CD and EF is a transversal.

∴ ∠PHQ = ∠EGB          (Pair of alternate exterior angles)

⇒ ∠PHQ = 35°

In ∆PHQ,

∠PHQ + ∠QPH + ∠PQH = 180°     (Angle sum property)

⇒ 35° + 90° + x = 180°

⇒ 125° + x = 180°

⇒ x = 180° − 125° = 55°

Thus, the measure of ∠PQH is 55°.

Question 29:

In the given figure, AB || CD and EF ⊥ AB. If EG is the transversal such that ∠GED = 130°, find ∠EGF.
 

Answer 29:

In the given figure, AB || CD and GE is the transversal.

∴ ∠GED + ∠EGF = 180°       (Sum of adjacent interior angles on the same side of the transversal is supplementary)

⇒ 130° + ∠EGF = 180°

⇒ ∠EGF = 180° − 130° = 50°

Thus, the measure of ∠EGF is 50°.