{"id":358,"date":"2022-04-13T10:46:41","date_gmt":"2022-04-13T10:46:41","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=358"},"modified":"2024-11-28T09:50:30","modified_gmt":"2024-11-28T09:50:30","slug":"s-2-trigonometric-ratios-csc-sec-and-cot","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/s-2-trigonometric-ratios-csc-sec-and-cot\/","title":{"rendered":"S.2 Trigonometric ratios: csc, sec and cot"},"content":{"rendered":"\n

Trigonometric ratios: csc, sec and cot<\/strong><\/h2>\n\n\n\n

key notes :<\/strong><\/p>\n\n\n\n

The cotangent (cot) of an acute angle in a right triangle is a ratio. It is the length of the adjacent leg (adj) divided by the length of the opposite leg (opp). The adjacent leg is the leg next to the specified angle and the opposite leg is the side across from the specified angle.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Learn with an example<\/strong><\/p>\n\n\n\n

\n
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Find the cosecant of \u2220G.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Simplify your answer and write it as a proper fraction, improper fraction, or whole number.<\/p>\n\n\n\n

csc (G) =________<\/p>\n<\/div><\/div>\n\n\n\n

Take the formula for the cosecant of \u2220G and plug in the relevant side lengths.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

csc ( G ) = hyp \/ opp Definition of cosecant<\/p>\n\n\n\n

= GI \/ HI Substitute hyp=GI and opp=HI<\/p>\n\n\n\n

= 13 \/ 12 Plug in GI=13 and HI=12<\/p>\n\n\n\n

So, csc (G) = 13\/12 .<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

Find the cosecant of \u2220G.<\/strong><\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

Simplify your answer and write it as a proper fraction, improper fraction, or whole number.<\/p>\n\n\n\n

csc (G) = ________<\/p>\n<\/div><\/div>\n\n\n\n

Take the formula for the cosecant of \u2220G and plug in the relevant side lengths.<\/p>\n\n\n\n

\"\"<\/figure>\n\n\n\n

csc (G) = hyp \/ opp Definition of cosecant<\/p>\n\n\n\n

= FG \/ EF Substitute hyp=FG and opp=EF<\/p>\n\n\n\n

= 53 \/ 45 Plug in FG=53 and EF=45<\/p>\n\n\n\n

So, csc(G)= = 53 \/ 45 .<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

Find the cotangent of \u2220U.<\/strong><\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

Simplify your answer and write it as a proper fraction, improper fraction, or whole number. <\/p>\n\n\n\n

cot(U)= _______<\/p>\n<\/div><\/div>\n\n\n\n

Take the formula for the cotangent of \u2220U and plug in the relevant side lengths.<\/p>\n\n\n

\n
\"\"<\/figure><\/div>\n\n\n

cot(U) = adj \/ opp<\/p>\n\n\n\n

= SU \/ ST Substitute adj=SU and opp=ST<\/p>\n\n\n\n

= 28 \/ 45 Plug in SU=28 and ST=45<\/p>\n\n\n\n

So, cot(U) = 28 \/ 45 .<\/p>\n<\/div><\/div>\n\n\n\n

let’s practice!<\/strong><\/p>\n\n\n\n

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\"\"<\/a><\/figure>\n<\/div>\n\n\n\n
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\"\"<\/a><\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Trigonometric ratios: csc, sec and cot key notes : The cotangent (cot) of an acute angle in a right triangle is a ratio. It is the length of the adjacent leg (adj) divided by the length of the opposite leg (opp). The adjacent leg is the leg next to the specified angle and the oppositeContinue reading “S.2 Trigonometric ratios: csc, sec and cot”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"footnotes":""},"class_list":["post-358","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/358","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=358"}],"version-history":[{"count":18,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/358\/revisions"}],"predecessor-version":[{"id":14938,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/358\/revisions\/14938"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=358"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}