In △abc the median am is extended to ray am and point p on am is taken so that pm=am. what is the distance from point p to vertices c and b if ab=c and ac=b?

PC = c and PB = b

Step-by-step explanation:

Given: As shown in figure 1 below:

In ΔABC, AB = c and AC = b and AM is the median drawn from vertex A on BC and extended to point P e.g.,  PM = AM.

To Find: PB =? and PC = ?

Sol.  In ΔAMC and ΔBMP

BM = MC ( ∵ Median from any vertex of a triangle divides its opposite side in two equal parts)

∠BMP = ∠AMC (vertically opposite angles ∵ line BC and AP intersects each other at point M)

PM  = AM (given)

∴ ΔBMP ≅ ΔAMC (S-A-S postulate)

∴ BP = AC = b ( corresponding parts of congruent triangles)

Similarly, in  In ΔAMB and ΔCMP

AM = MP ( ∵ Median from any vertex of a triangle divides its opposite side in two equal parts)

∠AMB = ∠CMP (vertically opposite angles ∵ line BC and AP intersects each other at point M)

BM  = MC (given)

∴ ΔAMB ≅ ΔPMC (S-A-S postulate)

∴ PC = AB = c ( corresponding parts of congruent triangles)

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.

answer: idk wbu

step-by-step explanation: