Answer:
AA postulate the two triangles are similar
Answer:
should be segment BT, that's what i put and it was right. hope this helps
Step-by-step explanation:
Answer:
Step-by-step explanation:
From the given picture,
∠ABE = ∠DEF = 90° [Since, AB and DE are perpendicular to DE]
m∠ECA = m∠BFD [Given]
m∠ECA + m∠ACB = 180° [Liner pair of angles]
m∠BFD + m∠DFE = 180° [Liner pair of angles]
m∠ACB + m∠ECA = m∠BFD + m∠DFE [Transitive property]
m∠ACB = m∠DEF [Since, m∠ECA = m∠BFD]
Therefore, ΔABC ≅ ΔDEF [By AA property of similarity]
Answer:
B. subtract
Step-by-step explanation:
Subtracting is the opposite of adding which is what is originally in the equation, by subtracting you are canceling out the 14 1/6
Answer:
answer is : Cos(13pi/8) = 0.3826
Step-by-step explanation:
We have, Cos (13pi/8)
Since 13pi/8 can be shown as 3pi/2 < 13pi/8 < 2pi
Hence 13pi/8 lies on fourth quadrant.
In fourth quadrant cosine will be positive.
Cos (13pi/8) = cos(3pi/2 + pi/8)
applying formula cos(A+B) = cos A cosB - sinAsinB
i.e Cos(3pi/2 + pi/8) = cos(3pi/2)cos(pi/8) - sin(3pi/2)sin(pi/8)
∵ Remember cos(3pi/2) =0 , sin(3pi/2) = -1
Cos(3pi/2 + pi/8) = 0 cos(pi/8) - (-1)sin(pi/8)
Cos(3pi/2 + pi/8) = 0 + 0.3826
Cos(3pi/2 + pi/8) = 0.3826
Hence we got Cos(13pi/8) = 0.3826