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# In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree

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Math Expert
Joined: 02 Sep 2009
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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 00:44
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Question Stats:

84% (00:58) correct 16% (01:12) wrong based on 78 sessions

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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 00:49
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Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Let $$∠C = x$$

$$∠B = x + 10$$

$$∠A = x + 10 + 10 = x + 20$$

$$∠A + ∠B + ∠C = 180$$

$$x + 20 + x + 10 + x = 180$$

$$3x + 30 = 180$$

$$3x = 180 - 30 = 150$$

$$x = 150/3 = 50$$

$$∠C = 50$$

$$∠B =x + 10 = 50 + 10 = 60$$ . Answer (D) ...
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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 11:16
Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Define all angles in terms of ∠B (= b, ∠A = a, ∠C = c).

Quote:
∠A is 10 degrees greater than ∠B

a = b + 10
Quote:
∠B is 10 degrees greater than ∠C

b = c + 10, or

c = b - 10

a + b + c = 180, substitute values in bold type

(b + 10) + b + (b - 10) = 180

3b = 180

b = 60
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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 13:35
The sum of all the angles in a triangle is 180 Degree.
If the three angles have to differ by 10 and add to 180,
the only possibility(with a difference of 10) is when the three angles are 50,60 and 70.

Since ∠B is 10 degree greater than ∠C and 10 degree smaller than ∠A, it must be 60(Option D)
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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 18:36
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Solution:
Given:
a=b+10
b=c+10
Also, we know
a+b+c=180.----eq (1)

Therfore subsitute the value of a and c in the eq(1)
we ger
b+10+b+b-10=180.
3b=180.
b=60.

So Option D
Math Expert
Joined: 02 Aug 2009
Posts: 7765
Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 19:51
Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Hi,

Here actually difference of 10 doesn't matter. Any difference but similar would give us SAME answer as 60.. WHY?

The three angles differ by certain amount so they are in AP and central angle B would be the MEAN
So 3*B =180..... B=60
D
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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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21 Jun 2017, 21:19
If all three angles are equal: 60, 60, 60

If the three angles are in Arithmetic Progression with a difference of 10: 50, 60, 70

So the central angle is 60 only. Hence D answer
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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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22 Jun 2017, 04:50
Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Let the /_C = x
So, /_ B = x +10
/_ A = x +20

A + B + C = x+20 + x +10 +x = 180
-> 3x +30 = 180
-> x= 150/3 = 50

So, /_B = 50 +10 = 60

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In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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22 Jun 2017, 08:09
Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

Let the value of $$∠A = a°$$ , $$∠B = b°$$ and $$∠C = c°$$

We know, $$∠A + ∠B + ∠C = 180°$$ Or, $$a° + b° + c° = 180°$$
Quote:
∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C

Value of $$∠B$$ = $$( c° + 10° )$$
Value of $$∠A$$ = $$( b° + 10° )$$ => $$( c° + 10° ) + 10°$$ = $$c° + 20°$$
Value of $$∠C$$ = c°

Now, $$(c° + 20°)$$ + $$( c° + 10° )$$ + $$c°$$ = $$180°$$

Or, $$3c° + 30° = 180°$$

Or, $$3c° = 150°$$

So, c° = 50°

Now, as Value of $$∠B$$ = $$( c° + 10° )$$ = $$60°$$

Thus, answer will be (D) 60°
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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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24 Jun 2017, 08:34
Bunuel wrote:
In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degrees greater than ∠C. The value of angle B is

(A) 30
(B) 40
(C) 50
(D) 60
(E) 70

We can let angle C = x, angle B = 10 + x, and angle A = 20 + x. Since there are 180 degrees in a triangle, we have:

x + x + 10 + x + 20 = 180

3x + 30 = 180

3x = 150

x = 50

Thus, angle B is 50 + 10 = 60 degrees.

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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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24 Jun 2017, 09:25
IMO D
A+B+C=180
B+10+B+B-10=180
3B=180
B=60
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Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree  [#permalink]

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16 Dec 2017, 03:21
Here,<A+<B+<C=180 and B=C+10, A=B+10=>A=C+10+10 =>A=C+20
Hence, C+20+C+10+C=180
C=150/3=50

Re: In Triangle ABC, ∠A is 10 degrees greater than ∠B, and ∠B is 10 degree   [#permalink] 16 Dec 2017, 03:21
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