Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.
The sum of angles of ∆ABC is 180°, so we have
... 180° = 40° + m∠A + m∠C
Using the above expressions for m∠A and m∠C, we can write ...
... 180° = 40° + (90° -x) + (90° -3x)
... 4x = 40° . . . . . . . . . add 4x-180°
... x = 10°
From which we conclude ...
... m∠C = 90°-3x = 90° - 3·10° = 60°
The ratio of CN to CL is
... CN/CL = cos(∠C) = cos(60°)
... CN/CL = 1/2
so ...
... CN = (1/2)CL
The probability of the pointer lands on red(R) both spin =
Step-by-step explanation:
Given,
The number of parts = 6
The pointer is spun twice.
To find the probability of the pointer lands on red(R) both spin.
Formula
Probability of an event = number of required outcomes ÷ the total number of outcomes.
P(A and B) = P(A)×P(B)
Now,
For one spin,
A = RED comes.
For second spin,
B = RED comes
So,
P(A) = and P(B) =
Then,
P(A and B) = ×
=
Hence, the probability of the pointer lands on red(R) both spin =