Chapter 6 Rectilinear Figures Ex 6.1

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 6 Rectilinear Figures Ex 6.1

Question 1.
From the given figure, find the three angles of the triangle ABC.

Solution.
Here ∠DCE = 53°
∴∠ACB = 53°
(Vertically opposite angle)
In ∆ABC
∠ABC + ∠ACB = ∠CAF
(By exterior angle property)
⇒ ∠ABC + 53°= 112°
⇒ ∠ABC = 112° – 53°
⇒ ∠ABC = 59°
Now in ∆ABC
∠BAC + ∠ABC + ∠BCA = 180°
(by angle sum property)
⇒ ∠BAC + 59° + 53° = 180°
⇒ ∠BAC = 68°
Hence, ∠BAC = 68°, ∠ABC = 59° and ∠ACB = 53°.

Question 2.
In figure, ∆ABC is an equilateral triangle. Find the values of ∠x, ∠y and ∠z from the given figure.

Solution.
∵ ∆ABC is an equilateral triangle
∴AB = BC = CA
⇒ ∠ABC = ∠ACB = ∠BAC = 60°
60° = x + 22°
(exterior angle property)
x = 38°
Also
38° + 22° + ∠z = 180°
(by angle sum property of a ∆)
⇒ ∠z = 180° – 60° = 120°
Again by exterior angle property
⇒ ∠ACB = ∠y + 38°
⇒ 60° = ∠y + 38°
⇒ ∠y = 22°
Hence, ∠x = 38°, ∠y = 22° and ∠z = 120°

Question 3.
In the given figure, the sides AB and AC of ∆ABC are produced to point E and D respectively. If the bisectors BO and CO of ∠CBE and ∠BCD respectively meet at point O, then prove
∠BOC = 90°- ∠x.

Solution.
Ray BO is the bisector of ∠CBE.
Therefore, ∠CBO = ∠CBE
⇒ ∠CBO = (180° – y) = 90° – …(i)
Similarly, ray CO is the bisector of ∠BCD
Therefore, ∠BCO = ∠BCD
= (180° – z)
= 90°-  …(ii)
Now in ∆BOC,
∠CBO + ∠BCO + ∠BOC = 180° …(iii)
(angle sum property of a triangle)
Using (i) and (ii) in (iii), we get
⇒ 90°-  + 90°-  + ∠BOC = 180°

⇒ ∠BOC =  – (y + z) …(iv)
But in ∆ABC,
x + y + z = 180°
(angle sum property of a triangle)
⇒ y + z = 180° – x
⇒ 2∠BOC = 180° – x
[using relation (iv)]
⇒ ∠BOC = 90° –  ∠x
Hence proved.

Question 4.
In figure, ∠P = 52°, ∠PQR = 64°. If QO and RO are the bisectors of ∠PQR and ∠PRQ respectively of ∆PQR, find ∠x and

Solution.
Given: QO and RO are the bisectors of ∠PQR and ∠PRQ respectively of ∆PQR and ∠P = 52°, ∠PQO = 64°.
To find: ∠x and ∠y
In ∆PQR,
∠P + ∠PQR + ∠PRQ = 180°
(angle sum property of a triangle)
⇒ 52° + 64° + ∠PRQ = 180°
⇒ ∠PRQ = 180° – 116°
⇒ ∠PRQ = 64°
⇒ ∠y = 32°
(as RD is bisector of ∠PRQ)
In ∆OQR,
∠OQR + ∠ORQ + ∠x = 180°
(reason as above)
⇒ 32° + 32° + ∠x= 180°
[QO and RO are bisector of ∠PQR and ∠PRQ]
⇒ ∠x = 180° – 64°
⇒ ∠x = 116°
Hence, ∠x = 116°, ∠y = 32°.

Question 5.
In figure, AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.

Solution.
We are given that AB || DE
and ∠BAC = 35° and ∠CDE = 53°
AB || DE (given)
⇒ ∠BAE = ∠AED = 35°
(alternate angles)
Now in ∆CDE,
∠CDE + ∠E + ∠DCE = 180°
(angle sum property of a triangle)
⇒ 53° + 35° + ∠DCE = 180°
⇒ ∠DCE = 180° – 88° = 92°
⇒ ∠DCE = 92°
Hence, ∠DCE = 92°.

Question 6.
In the adjoining figure if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.

Solution.
In ∆PRT,
∠RPT + ∠PRT + ∠RTP = 180°
(angle sum property of a triangle)
⇒ 95° + 40° + ∠RTP = 180°
⇒ 135° + ∠RTP = 180°
⇒ ∠RTP = 180° – 135°
= 45°
Now ∠RTP = ∠STQ = 45°
(vertically opposite angles)
In ∆STQ,
∠STQ + ∠TSQ + ∠SQT = 180°
(angle sum property of a triangle)
⇒ 45° + 75° + ∠SQT = 180°
⇒ ∠SQT = 180° – (45° + 75°)
= 180° – 120° = 60°
Hence, ∠SQT = 60°.

Question 7.
In figure, sides QP and RQ of ∆PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.

Solution.
We are given that ∠SPR = 135° and ∠PQT =110°.
At point Q:
∠PQT + ∠PQR = 180°
(linear pair of angles)
⇒ 110° + ∠PQR = 180°
⇒ ∠PQR = 180° – 110°
= 70° …(i)
At point P:
∠QPR + ∠SPR = 180°
(linear pair of angles)
⇒ ∠QPR + 135°= 180°
⇒ ∠QPR = 45° …(ii)
In ∆PQR,
∠PQR + ∠PRQ + ∠QPR = 180°
⇒ 70° + ∠PRQ + 45° = 180°
⇒ ∠PRQ = 180° – 115°
⇒ ∠PRQ = 65°
Hence, ∠PRQ = 65°.

Question 8.
In figure, if PQ ⊥ PS, PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.

Solution.
We are given that PQ ⊥ PS, PQ || SR
∠SQR = 28° and ∠QRT = 65°
SR and a transversal QR intersect them
∠PQR = ∠QRT = 65°
(alternate angles)
x + 28° = 65°
x = 65° – 28°
x = 37°
Now, in right angled triangle SPQ
x + y + ∠SPQ = 180°
(by angle sum property of a triangle)
x + y + 90° = 180°
⇒ 37° + y + 90°= 180°
⇒ 127° + y = 180°
⇒ y = 180 – 127°
⇒ y = 53°
Hence, x = 37° and y = 53°.

Question 9.
In the given figure, the side QR and APQR is produced to a point S. If the bisector of ∠PQR and ∠PRS meet at a point T, then prove that ∠QTR = ∠QPR.

Solution.
Side QR of ∆PQR is produced to S,
Exterior ∠PRS = ∠P + ∠Q …(i)
(Exterior angle is equal to sum of its opposite interior angles)

Dividing of both side relation (i) by 2, we get
∠PRS = ∠P + ∠Q
⇒ x = ∠P + y …(ii)
[∵ RT and QT are bisectors of ∠PRS and ∠PQS respectively]
Also, in ∆QRT,
Exterior ∠TRS = ∠T + ∠y
⇒ x = ∠T + y …(iii)
From (i) and (iii), we get
 ∠P + y = ∠T + y
∠P = ∠T
∠QTR = ∠QPR
Hence proved.

Question 10.
In ∆ABC, ∠A = 90°, AL ⊥ BC, prove that ∠BAL = ∠ACB.
Solution.
Given: In ∆ABC, ∠A = 90° and AL ⊥ BC
To prove: ∠BAL = ∠ACB

Proof: Suppose ∠BAL = ∠1, ∠CAL = ∠2, ∠ABL = 3 and ∠ACL = ∠4
Now in ∆ABC
∠A + ∠B + ∠C = 180°
(angle sum property of a triangle)
⇒ ∠90° + ∠3 + ∠4 = 180°
(∵ ∠A = 90 given)
⇒ ∠3 + ∠4 = ∠90°
⇒ ∠4 = ∠90° – ∠3 …(i)
Now in ∠BAL
∠1 + ∠3 + ∠ALB = 180°
(Angle sum property of a triangle)
⇒ ∠1 + ∠3 + 90° = 180°
⇒ ∠1 + ∠3 = 90°
⇒ ∠1 = 90° – ∠3 …(ii)
From (i) and (ii), we get
∠1 = ∠4 => ∠BAL = ∠ACB
Hence proved.

Question 11.
The angles of a triangle are the ratio 2 : 3 : 4. Find all the three angles of a triangle.
Solution.
Let angles of a triangle are 2x, 3x, 4x
2x + 3x + 4x = 180°
⇒ 9x = 180°
⇒ x = 20°
Angles are
2x = 2 x 20° = 40°,
3x = 3 x 20° = 60°,
4x = 4 x 20° = 80°.