## Trending News

# I need help solving a trigonometry math question.?

Use your imagination that this is a triangle side AB is longer than CA and the dots are a line from A to P

A

/.\ CA=6.862

/ . \ AB=40.815

/ : \ CAB= 89*31'10"

C/__ _ :_______\B

:

P

I need this answer to find gps coordinates

Find side CB and distance AP

The angle APB is a right angle (90*)

Note: this is way above my math level. If you could find the answer and provide an email so that I can double check the answer (cant do until I have an answer).

### 1 Answer

- sCoRpIoTaRiUsLv 69 years agoFavorite Answer
side b=CA=6.862

side c=AB=40.815

<A= CAB= 89*31'10

<A=89° 31' 10"= 89.594444

side a=CB=?

.

apply cosine law to get a or CB

.

2bc cos A=b^2 plus c^2-a^2

.

[2*6.862*40.815]*cos 89.594444= 6.862^2 plus 40.815^2 - a^2

.

(560..14506*0.0070782)=47.087044 plus 1665.864225 -a^2

.

3.9648346=1712.951269 -a^2

.

a^2=1712.951269-3.9648346

.

a=sqrt (1708.9864344)

.

a=41.3398891

CB=41.3398891 answer (@_@)

.

.

find height h or AP

x=CP

y=PB

h=AP

a=CB

b=AC

c=AB

.

x plus y =a

x=a-y

.

use pythagorean:

eq 1: h^2 =b^2-x^2

.

eq 2: y^2 plus h^2=c^2

.

apply the value of h^2 to eq 2:

y^2 plus [b^2-x^2]=c^2

.

apply the value of x:

y^2 plus b^2 - (a-y)^2=c^2

.

expand:

y^2 plus b^2-(a^2- 2ay plus y^2)=c^2

.

y^2 is cancel when bracket is open, apply all equivalent values:

.

6.862^2- 41.3398891^2 plus (2*41.3398891y)=40.815^2

.

47.087044 -1708.9864308 plus 82.6797782y=1665.864225

.

-1661.8993868 plus 82.6797782y =1665.864225

.

82.6797782y=1665.864225 plus 1661.8993868

.

y=3327.7636118/82.6797782

y=40.2488212

.

x=41.3398891-40.2488212

x=1.0910678

.

h^2=b^2-x^2

h=sqrt (6.862^2 - 1.0910678^2)

h=sqrt (45.8966151)

h=6.774704

AP=6.774704 answer (@_@)

.

.

check:

y^2 plus h^2=c^2

40.2488212^2 plus 6.774704^2=c^2

.

sqrt (1665.8642221)=c

40.8149999=c

.

given from above:

c=AB=40.815

.

.

Edit:

if you want the angles apply the value of x and y to hypotenuse b and c, you know what to do now, right?

.

.

hope this helps (@_@)