{"id":350,"date":"2022-04-13T10:45:22","date_gmt":"2022-04-13T10:45:22","guid":{"rendered":"http:\/\/10thclass.deltapublications.in\/?page_id=350"},"modified":"2025-01-10T09:50:05","modified_gmt":"2025-01-10T09:50:05","slug":"r-10-angles-in-inscribed-right-triangles","status":"publish","type":"page","link":"https:\/\/10thclass.deltapublications.in\/index.php\/r-10-angles-in-inscribed-right-triangles\/","title":{"rendered":"R.10 Angles in inscribed right triangles"},"content":{"rendered":"\n

Angles in inscribed right triangles<\/strong><\/h2>\n\n\n\n

Key Notes :<\/strong><\/p>\n\n\n\n

If AC is a diameter of a circle, then the inscribed angle \u2220ABC is a right angle.<\/p>\n\n\n

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

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

\n
\n

\ud83d\udd14 What is \u2220K?<\/strong><\/p>\n\n\n

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

\u2220K=______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n

Since IK is a diameter of the circle, \u2220J is a right angle.<\/p>\n\n\n

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

So, \u25b3IJK is a right triangle and \u2220I and \u2220K are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220K.<\/p>\n\n\n\n

\u2220I+\u2220K=90\u00b0<\/p>\n\n\n\n

36\u00b0<\/strong>+\u2220K=90\u00b0 Plug in \u2220I=36\u00b0<\/p>\n\n\n\n

\u2220K=54\u00b0 Subtract 36\u00b0 from both sides<\/p>\n\n\n\n

\u2220K is 54\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

\ud83d\udd14 What is \u2220F?<\/strong><\/p>\n\n\n

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

\u2220F=_______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n

Since FG is a diameter of the circle, \u2220H is a right angle.<\/p>\n\n\n

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

So, \u25b3FGH is a right triangle and \u2220G and \u2220F are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220F.<\/p>\n\n\n\n

\u2220G+\u2220F=90\u00b0<\/p>\n\n\n\n

52\u00b0<\/strong>+\u2220F=90\u00b0 Plug in \u2220G=52\u00b0<\/p>\n\n\n\n

\u2220F=38\u00b0 Subtract 52\u00b0 from both sides<\/p>\n\n\n\n

\u2220F is 38\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n

\n
\n

\ud83d\udd14 What is \u2220H?<\/strong><\/p>\n\n\n

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

\u2220H=______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n

Since HI is a diameter of the circle, \u2220J is a right angle.<\/p>\n\n\n

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

So, \u25b3HIJ is a right triangle and \u2220I and \u2220H are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220H.<\/p>\n\n\n\n

\u2220I+\u2220H=90\u00b0<\/p>\n\n\n\n

30\u00b0<\/strong>+\u2220H=90\u00b0 Plug in \u2220I=30\u00b0<\/p>\n\n\n\n

\u2220H=60\u00b0 Subtract 30\u00b0 from both sides<\/p>\n\n\n\n

\u2220H is 60\u00b0.<\/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":"

Angles in inscribed right triangles Key Notes : If AC is a diameter of a circle, then the inscribed angle \u2220ABC is a right angle. Learn with an example \ud83d\udd14 What is \u2220K? \u2220K=______ \u00b0 Since IK is a diameter of the circle, \u2220J is a right angle. So, \u25b3IJK is a right triangle and \u2220I and \u2220K are complementary. Write an equation setting the sum of their measures equal to 90\u00b0, and solve for \u2220K.Continue reading “R.10 Angles in inscribed right triangles”<\/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-350","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/350","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=350"}],"version-history":[{"count":11,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/350\/revisions"}],"predecessor-version":[{"id":17376,"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/350\/revisions\/17376"}],"wp:attachment":[{"href":"https:\/\/10thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=350"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}