The pythagorean theorem states that for any RIGHT-ANGLED TRIANGLE, the relationship between the side lengths are A²+B²=C², where C is the hypotenuse :P
Which means that it can only be applied on right angled triangles~
So your answer would be B, a right-angled triangle.
Ask me if you are unsure about anything <3
Answer:
Hence proved △ABE∼△CBF.
Step-by-step explanation:
Given,
ABCD is a parallelogram.
BF ⊥ CD and
BE ⊥ AD
To Prove : △ABE∼△CBF
We have drawn the diagram for your reference.
Proof:
Since ABCD is a parallelogram,
So according to the property of parallelogram opposite angles are equal in measure.
⇒1
And given that BF ⊥ CD and BE ⊥ AD.
So we can say that;
⇒2
Now In △ABE and △CBF
∠A = ∠C (from 1)
∠E = ∠F (from 2)
So by A.A. similarity postulate;
△ABE∼△CBF
Probability that a student will play both is 7/30
Step-by-step explanation:
Total students = 30
No. of students who play basketball = 18
Probability that a student will play basketball = 18/30
= 3/5
No. of students who play baseball = 9
Probability that a student will play baseball = 9/30
= 3/10
No. of students who play neither sport = 10
Probability that a student will play neither sport = 10/30
= 1/3
To find :
Probability that a student will play both = p(student will play both)
No.of students who play sport = 30 - 10
= 20
Out of 20 students 18 play basketball and 9 play baseball.
So, some students play both the sports.
No. of students who play both sports = 18 + 9 - 20
= 7
p(student will play both) = 7/30
Probability that a student will play both is 7/30
Answer:
sin (×+y)/sinx cosy =1+ cotx tany
from trig ratio Sin [x +y] = sinx Cosy + cosx siny
Sinx cosy + cosox siny./Sinx cosy
then check the attachments for further information
