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19 Nov 2020, 01:15
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E
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Difficulty:
55%
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Question Stats:
63% (02:15) correct
37%
(02:08)
wrong
based on 70
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In the figure shown, H is the center of the semicircle. Triangles DHF and GHF each have two vertices on the circle. The measure of ∠DHG is 120°. What is the measure of ∠FHG?
A. 50°
B. 60°
C. 75°
D. 90°
E. 100°
Attachment:
2020-10-27_14-09-58.png [ 16.83 KiB | Viewed 2649 times ]
19 Nov 2020, 01:46
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Explanation:
DHGF is Quadrilateral
So sum of inside angle will be 360
120 + 3x + 2x + x = 360
6x = 240
x = 40 deg.
So in Triangle FGH,
Angle FHG = 180 - 2x = 180 - 80
= 100
IMO-E
Posted from my mobile device
DHGF is Quadrilateral
So sum of inside angle will be 360
120 + 3x + 2x + x = 360
6x = 240
x = 40 deg.
So in Triangle FGH,
Angle FHG = 180 - 2x = 180 - 80
= 100
IMO-E
Posted from my mobile device
19 Nov 2020, 08:06
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Bunuel
both triangles are isosceles, because their sides are the radius
t1: 2x+2x+a=180
t2: x+x+(120-a)=180
equate: 4x+a=2x+120-a, x+a=60, a=60-x
subs: 2x+120-a=180, 2x-a=60, 2x-(60-x)=60, x=40
find a: a=60-x, a=20
find 120-a: 120-20=100
(E)
