1. ## dot product

Think im being stupid with this one but ill ask in anycase:

I have a descent down a hill measured at -2. I need to find the angle of the descent so i use the following formula:

/a//b/cos(theta)

so rearranging :

arccos a . b / /a//b/

where a = -2 and b is the unit vector i. Right when Iam working this out i get pi which is ridiculous as this means the descent is vertical!!! clearly im working something out wrong but not sure what it is.

(Note : I am learning this stuff in my spare time - i am firefighter full time - so if this seems really basic to some i do apolgise)

2. Calculate $arccos\left(\frac{[-2,1]\cdot[1,0]}{\sqrt{5}}\right)$. It should give you the angle. Remember that the cosine is symmetrical along the x-axis (in the unit circle).

3. 153.4 degrees or 2.67 rad s-1 ????
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 1, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9738531708717346, "perplexity": 953.1606342279077}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-04/segments/1484560281202.94/warc/CC-MAIN-20170116095121-00279-ip-10-171-10-70.ec2.internal.warc.gz"}