triangles

[Cmorano]

in triangle ABC, the measure of angle A = x, measure of angle B = 2x + 2, and measure of angle C = 3x + 4
i'm looking for the value of x

since all 3 angles should add up to 180 degrees, do i add x + 2x + 2 + 3x + 4 ?
and then solve for x?

 

[Deleted member 4993]

in triangle ABC, the measure of angle A = x, measure of angle B = 2x + 2, and measure of angle C = 3x + 4
i'm looking for the value of x

since all 3 angles should add up to 180 degrees, do i add x + 2x + 2 + 3x + 4 ?
and then solve for x?

yes

 

[Cmorano]

in triangle ADB, measure triangle BDA = 90, AD=5 sq.root 2, and AB= 2 sq.root 15
do i use pythagorum theorem? when trying to figure out what is the length of line BD ?

 

[Deleted member 4993]

in triangle ADB, measure triangle BDA = 90, AD=5 sq.root 2, and AB= 2 sq.root 15
do i use pythagorum theorem? when trying to figure out what is the length of line BD ?

So AB is the hypoteneuse - and use Pythagorian theorem to calculate BD ? BD[sup:zsy2ej1w]2[/sup:zsy2ej1w] = AB[sup:zsy2ej1w]2[/sup:zsy2ej1w] - AD[sup:zsy2ej1w]2[/sup:zsy2ej1w]

 

[Cmorano]

o.kay, so,
yes, AB is the hypotheneuse,
and,
the equation so far looks like this:
(BD)squared = (2sq.root 15)squared - (5 sq.root 2) squared

i'm changing BD into "x"

x squared = 4 sq.root 30 - 50
x squared = 46 sq.root 30
i'm not sure about this at all ?!?

 

[Mrspi]

o.kay, so,
yes, AB is the hypotheneuse,
and,
the equation so far looks like this:
(BD)squared = (2sq.root 15)squared - (5 sq.root 2) squared

i'm changing BD into "x"

x squared = 4 sq.root 30 - 50
x squared = 46 sq.root 30
i'm not sure about this at all ?!?

This was suggested in an earlier post.....

(BD)[sup:16t0gnxq]2[/sup:16t0gnxq] = (AB)[sup:16t0gnxq]2[/sup:16t0gnxq] - (AD)[sup:16t0gnxq]2[/sup:16t0gnxq]

You're using x for BD, I guess.....
x[sup:16t0gnxq]2[/sup:16t0gnxq] = (2 sqrt 15)[sup:16t0gnxq]2[/sup:16t0gnxq] - (5 sqrt 2)[sup:16t0gnxq]2[/sup:16t0gnxq]

Looks like you might need a REVIEW of how to square something like 2 sqrt 15......

(2 sqrt 15)[sup:16t0gnxq]2[/sup:16t0gnxq] means
(2 sqrt 15)*(2 sqrt 15)
If you are multiplying more than two factors together, you can rearrange them in any order, and group them any way you want to do the multiplication. So, I'll rearrange and regroup like this:

(2*2)*(sqrt 15)*(sqrt 15)
4 * sqrt 15[sup:16t0gnxq]2[/sup:16t0gnxq]
4 * 15
60


Use the same procedure to calculate ( 5 sqrt 2)[sup:16t0gnxq]2[/sup:16t0gnxq]

Then continue.....

 

[Cmorano]

o.kay, so,
(BD)squared = 60 - 50 = 10
if BD = x, then
x squared = 10
x = sq.root of 10,
then,
BD = sq.rt.10

by the way,
how do i type "squared" on this web site ?
thanks

 

[lookagain]

by the way,
how do i type "squared" on this web site ?
thanks

x^2 or

\(\displaystyle x^2\) <--------- The word "tex" inside of square brackets, then "x^2,"
and finally "/tex" inside of square brackets

 

[Denis]

...please can anyone suggest me easy tricks to solve math

NO.
Attend your math classes.
Listen to teacher.
Do your homework.

 

[mmm4444bot]

lisa25 is a SPAMMER from down under.