In the diagram, PO and OR are radii, |PQ| = |QR| and reflex < PQR is 240o. Calculate the value x
60o
55o
50o
45o
Correct answer is A
< Q = \(\frac{240}{2}\) (angle at centre twice that at the circumference)
< Q = 120o
Also < POR = 360 - 240
= 120o
( < s at centre) since /PQ/ = /QR/, < x = < R
Byt < x + < R + O + Q = 360 (sum of interior < s of quadrilateral)
x + R + 120 = 360o
x + R = 360 - 240 = 120; Since x = R
x + x = 120
2x = 120
Since x = R
x + x = 120
2x = 120
x = \(\frac{120}{2}\)
= 60o
\(\bigtriangleup\) PQK
\(\bigtriangleup\) PKS
\(\bigtriangleup\) SKR
\(\bigtriangleup\) PSR
Correct answer is C
No explanation has been provided for this answer.
9o
10o
11o
12o
Correct answer is B
tan \(\theta\) = \(\frac{9}{50}\) = 0.18
\(\theta = tan^{-1} 0.18\)
\(\theta\) = 10.20
If 27x = 9y. Find the value of \(\frac{x}{y}\)
\(\frac{1}{3}\)
\(\frac{2}{3}\)
1\(\frac{1}{2}\)
3
Correct answer is B
27x = 9y
33x = 32y
3x = 2y
18
15
12
10
Correct answer is D
let the number of question = n
n(n + 50) = 600
n2 + 50n - 600 - 0 by quadratic formular
x = \(\frac{b \pm \sqrt{b^2 - 4ac}}{2}\)
n = \(\frac{-50 \pm \sqrt{50^2 - 4(1) (-600)}}{2(1)}\)
= \(\frac{-50 \pm \sqrt{4900}}{2}\)
= \(\frac{-50 + 70}{2}\)
= \(\frac{20}{2}\)
= 10