Answer:
∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and lies between AB and KM and BK is the transversal line)
m∠MBK ≅ m∠BKM (Angles opposite to equal side of ΔBMK are equal)
Step-by-step explanation:
Given: BK is an angle bisector of Δ ABC. and line KM intersect BC such that, BM = MK
TO prove: KM ║AB
Now, As given in figure 1,
In Δ ABC, ∠ABK = ∠KBC (∵ BK is angle bisector)
Now in Δ BMK, ∠MBK = ∠BKM (∵ BM = MK and angles opposite to equal sides of a triangle are equal.)
Now ∵ ∠MBK = ∠BKM
and ∠ABK = ∠KBM
∴ ∠BKM= ∠ABK
Therefore AB ║KM (∵ ∠BKM= ∠ABK and BK is the transversal line)
Hence proved.
Is there a picture of the rectangular piece of paper
Answer:
33.85 units^2
Step-by-step explanation:
you must first draw the triangle on the plane using the equations (see
attached file), you will have a right angle triangle with a height of 192 and a base of 6.
then you calculate the angle with the tangent function = 88.21
Then you use the small triangle to find the value of a (see attached file).
Finally, you propose an equation for X to find one of the sides of the triangle, once you have x squared it, and you already have the area,
i attached procedure
Answer:
C, ON/MN
Step-by-step explanation:
Since tan is opposite/adjacent, and the opposite of angle M is NO, then NO must be the numerator. The only answer with this as the numerator is C.
