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If in the triangle shown above all three interior angles are multiples
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26 Apr 2023, 00:52
26 Apr 2023, 00:52
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47% (02:44) correct
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If in the triangle shown above all three interior angles are multiples of 10 and angle ACB is neither the largest nor the smallest angle, how many distinct values of x are possible?
0
1
2
3
More than 3
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Re: If in the triangle shown above all three interior angles are multiples
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28 May 2023, 01:24
28 May 2023, 01:24
1
Expert Reply
OE
Given that,
∠CAB = 2x
∠ACD = 3x
∠ACB = 180 – 3x ………………. [∠ACD + ∠ACB = 180]
Let ∠ABC = y
In ⊿ ACB,
∠ACB + ∠BAC + ∠ ABC = 1800
180 – 3x + 2x + y = 180
-x + y = 0
x = y
Therefore, ∠ ABC = x0
As ∠ACB is neither the largest nor the smallest angle,
So, ∠ABC < ∠ACB < ∠CAB
x < 180 – 3x < 2x
Simplifying this inequality
x < 180 – 3x
4x < 180
x < 45°
180 – 3x < 2x
5x > 180
x > 36°
36 < x < 45
We are given that all the interior angles are multiples of 10, so x being a multiple of 10
can only be 40.
So, x = 40°
Ans. (B)
Given that,
∠CAB = 2x
∠ACD = 3x
∠ACB = 180 – 3x ………………. [∠ACD + ∠ACB = 180]
Let ∠ABC = y
In ⊿ ACB,
∠ACB + ∠BAC + ∠ ABC = 1800
180 – 3x + 2x + y = 180
-x + y = 0
x = y
Therefore, ∠ ABC = x0
As ∠ACB is neither the largest nor the smallest angle,
So, ∠ABC < ∠ACB < ∠CAB
x < 180 – 3x < 2x
Simplifying this inequality
x < 180 – 3x
4x < 180
x < 45°
180 – 3x < 2x
5x > 180
x > 36°
36 < x < 45
We are given that all the interior angles are multiples of 10, so x being a multiple of 10
can only be 40.
So, x = 40°
Ans. (B)
Re: If in the triangle shown above all three interior angles are multiples
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08 Sep 2023, 08:42
08 Sep 2023, 08:42
2
Alternative to the algebra, you can plug in numbers. You know that all interior angles are multiples of 10, so x needs to be a multiple of 10. There's an upper limit of 3x to be less than 180, so the max that x can equal is 60 (180/3=60).
This leaves only a couple options for x: x = 10, 20, 30, 40 or 50.
If x = 10:
angle ACD = 30, ACB = 150 (not good, because this would make it the largest angle)
If x = 20:
angle ACD = 60, ACB = 120, CAB = 40, CBA = 30 (not good, ACB is the largest angle)
If x = 30:
ACD = 90, ACB = 90, CAB = 60, CBA = 30 (not good)
x = 40:
ACD = 120, ACB = 60, CAB = 80, CBA = 40 (good)
x = 50:
ACD = 150, ACB = 30, CAB = 100, CBA = 50 (not good, ACB is the smallest)
This means ACB is only the middle angle as a multiple of 10 one time. Choice B is correct.
This leaves only a couple options for x: x = 10, 20, 30, 40 or 50.
If x = 10:
angle ACD = 30, ACB = 150 (not good, because this would make it the largest angle)
If x = 20:
angle ACD = 60, ACB = 120, CAB = 40, CBA = 30 (not good, ACB is the largest angle)
If x = 30:
ACD = 90, ACB = 90, CAB = 60, CBA = 30 (not good)
x = 40:
ACD = 120, ACB = 60, CAB = 80, CBA = 40 (good)
x = 50:
ACD = 150, ACB = 30, CAB = 100, CBA = 50 (not good, ACB is the smallest)
This means ACB is only the middle angle as a multiple of 10 one time. Choice B is correct.
