Two math questions I need help on please...?

Question

Can someone please help me with these triangle problems??


1. Triangle A B C is drawn with angle bisector B H from vertex point B to point H on side A C. Angle C B H is labeled 3x + 15. Given: Segment BH is the angle bisector of angle ABC. The measure of angle ABC = 105. Find the value of x.

2. Right triangle A B C is drawn with right angle B. The altitude of the triangle is drawn and labeled B D. It goes from vertex B to side A C. If AB=10 and AC=20, find AD.

Answer 1

#1 is actually pretty easy, if I'm understanding you correctly. Angle CBH is 1/2 of angle ABC, correct? Then:

3x+15 = (1/2)(105)

3x = 52.5 - 15
3x = 37.5
x=12.5

#2 is a little more complicated. You've got right triangle similarity, which means that all three triangles are similar. As such, you can set up a proportion:

20/10 = 10/AD, and AD = 5

That should do it! :)

Answer 2

1. Since angle ABC was bisected, 3x + 15 is half of 105. Half of 105 is 52.5. So
3x + 15 = 52.5
Subtract 15 from both sides then divide by 3 to get x.

@ has to do with similar triangles. 10 is the hypotenuse of ΔADB but a leg of ΔABC, and if AD, a corresponding leg of ΔACB is x, then AC (hyp. of ΔABC) is 20

So the proportion to use is x/10 = 10/20

[leg over hyp. = leg over hyp.]

cross multiply then divide to get x

Answer 3

2)By similaryty
AC/AB=BC/BD so
20/10=sqrt(300)/BD so BD =5*sqrt(3)