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In the figure given below, ABC is a triangle. BC is parallel to AE. If BC = AC, then what is the value of ∠CAE?
Given that, BC  AE
∠CBA + ∠EAB = 180°
⇒ ∠EAB = 180° – 65° = 115°
∵ BC = AC
Hence, ΔABC is an isosceles triangle.
⇒ ∠CBA = ∠CAB = 65°
Now, ∠EAB = ∠EAC + ∠CAB
⇒ 115° = x + 65° ⇒ x = 50°.
Hence, option D is correct.
In the figure given below, AB is parallel to CD. ∠ABC = 65°, ∠CDE = 15° and AB = AE. What is the value of ∠AEF?
Given that,
∠ABC = 65° and ∠CDE = 15°
Here, ∠ABC + ∠TCB = 180°
(∵ AB  CD)
∠TCB = 180° – ∠ABC
∴ ∠TCB = 180° – 65° = 115°
∵ ∠TCB + ∠DCB = 180°
(Linear pair)
∴ ∠DCB = 65°
Now, in ΔCDE
∠CED = 180° – (∠ECD + ∠EDC)
(∵ ∠ECD = ∠BCD)
= 180° – (– 65° + 15° ) = 100°
∵ ∠DEC + ∠FEC = 180°
⇒ ∠FEC = 180° – 100° = 80°
Given that, AB = AE.
i.e. ΔABE an isosceles triangle.
∴ ∠ABE = ∠AEB = 65°
∵ ∠AEB + ∠AEF + ∠FEC = 180°
(straight line)
⇒ 65° + x° + 80° = 180°
∴ x° = 180° – 145° = 35°.
Hence, option B is correct.
The angles x°, a°, c° and (π – b)° are indicated in the figure given below. Which one of the following is correct?
∠PCT + ∠PCB = π
(Linear pair)
∠PCB = π – (π – b°) = b° ..... (i)
In ΔBPC,
∠PCB + ∠BPC + ∠PBC = π
∠PBC = π – ∠PCB – ∠BPC = π – b° – a° ..... (ii)
∵ ∠ABE + ∠EBC = π
(∵ ∠PBC = ∠EBC) (linear pair)
∠ABE = π – ∠PBC = π – (π – b° – a°) = a° + b° ...(iii)
Now, in ΔABE
Sum of two interior angles = Exterior angle
∠EAB + ∠ABE = ∠BES ⇒ c° + b° + a° = x°
∴ x° = a° + b° + c°.
Hence, option C is correct.
Consider the following statements
I. The locus of points which are equidistant from two parallel lines is a line parallel to both of them and drawn mid way between them.
II. The perpendicular distance of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.
Which of the above statements is/are correct?
Statements I and II are both true, because the locus of points which are equidistant from two parallel lines is a line parallel to both of them and draw mid way between them.
Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.
Hence, option C is correct.
A wheel makes 12 revolutions per min. The angle in radian described by a spoke of the wheel in 1 s is:
= 12 × Its circumference
= 12 × 2πr
∴ In 1 s distance travelled by the wheel =
(12 × 2πr) / 60 = (2 / 5) πr
∴ Angle = (Arc / Radius)
= (2 / 5πr) / r = 2π / 5
Which is the required angle.
Hence, option B is correct.
If one arm of an angle is respectively parallel to the arm of another angle formed with a common line between them, then the two angles are
If one arm of an angle is respectively parallel to the arm of another angle formed with a common line between them, then the two angles are not equal but supplementary.
If l_{1}  l_{2} ⇒ ∠1 + ∠2 = 180°
(Supplementary)
Hence, option B is correct.
In a Δ ABC, (1 / 2) ∠A + (1 / 2) ∠C + (1 / 2) ∠B = 80°, then what is the value of ∠C?
Given that,
(1 / 2) ∠A + (1 / 2) ∠C + (1 / 2) ∠B = 80°
⇒ 3∠A + 2∠C + 3∠B = 480°
⇒ 3(∠A + ∠B) + 2∠C = 480° ....(i)
Also, in ΔABC,
∠A + ∠B + ∠C = 180°
On multiplying both sides L.H.S. & R.H.S. by 3, we get
3(∠A + ∠B) + 3∠C = 540° ..... (ii)
On subtracting Eq. (i) from Eq. (ii), we get
∠C = 60°.
Hence, option C is correct.
Complementary angles: Complementary angles are angle pairs whose measures sum to one right angle (90°).
So, the required angle will be 10°
180° = π radian
1º = (π / 180)
∴ 10º = {(π × 10) / 180} = (π / 18)
Hence, option C is correct.
In the given figure AB  CD, ∠ALC = 60° and EC is the bisector of ∠LCD. If EF  AB then the value of ∠CEF is
∠ALC = ∠LCD = 60°
[∵ Alternate angles]
EC is the bisector of ∠LCD
∴ ∠ ECD = (1 / 2) × ∠ LCD
= (1 / 2) × 60º = 30º
∠CEF + ∠ECD = 180°
[∵ Pair of interior angles]
∠CEF + 30° = 180°
∠CEF = 180° – 30° = 150°
Hence, option C is correct.
In the given figure lines AP and OQ intersect at G. If ∠AGO + ∠PGF = 70° and ∠PGQ = 40°. Find the angle value of ∠PGF.
As AP is a straight line and rays GO and GF stands on it.
∴ ∠AGO + ∠OGF + ∠PGF = 180°
⇒ (∠AGO + ∠PGF) + ∠OGF = 180°
⇒ 70° + ∠OGF = 180°
⇒ ∠OGF = 180° – 70°
⇒ ∠OGF = 110°
As, OQ is a straight line, rays GF and GP stands on it.
∠OGF + ∠PGF + ∠PGQ = 180°
Putting the value of ∠OGF & ∠PGQ
110° + ∠PGF + 40° = 180°
∠PGF = 180° – 150° = 30°
Hence, option C is correct.
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