50o
75o
35o
115o
30o
Correct answer is C
< WYT = \(\frac{180 - 30}{2}\)
= \(\frac{150}{2}\)
= 75o(base angles of isosceles D)
YXW = 75 - 40 = 35o(Exterior angle is equal to sum of interior angles)
Find the area of the shades segments in the figure
\(\sqrt{3}\)
4 \(\pi - \sqrt{3}\)
-\(\frac{2}{3} \pi\)
\(\frac{2\pi}{3}\) -3
Correct answer is D
Area of section = \(\frac{60^o}{360^o}\) x 11r2
= \(\frac{60}{360} \times \pi \times 2^2\)
= \(\frac{1}{6}\) x 4
= \(\frac{4\pi}{6}\)
= \(\frac{2\pi}{3}\)
Area of triangle = \(\frac{1}{2x}\)
= 2 x 28.......60
Segment Area = Area of section - Area of triangle
= \(\frac{2\pi}{3}\) -3
\(\frac{1}{2}\)
\(\frac{4}{\sqrt{3}}\)
\(\sqrt{3}\)
4\(\sqrt{3}\)
Correct answer is B
No explanation has been provided for this answer.
In the figure, PS = SR and Rx is a tangent to the circle at R, < QRX is equal to 72o. Find angle SPR
20o
36o
72o
30o
15o
Correct answer is B
XRQ = RSQ (Alternate segment)
s = 72o
(80o - 72o = 180o)
a + b + c = 190o = 2b + c = 180o
2b + 108 = 180o
2b = 72o
b = 36o
The angles marked in the figure are measured in degrees. Find x.
45o
90o
60o
15o
30o
Correct answer is E
Sum of external angles = 360o
3x + 5x + x + 2x + x = 360o
12x = 360o
x = 30o
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