{"id":212,"date":"2022-04-13T08:43:44","date_gmt":"2022-04-13T08:43:44","guid":{"rendered":"http:\/\/9thclass.deltapublications.in\/?page_id=212"},"modified":"2025-12-04T12:22:17","modified_gmt":"2025-12-04T12:22:17","slug":"l-10-angles-in-inscribed-right-triangles","status":"publish","type":"page","link":"https:\/\/9thclass.deltapublications.in\/index.php\/l-10-angles-in-inscribed-right-triangles\/","title":{"rendered":"L.10 Angles in inscribed right triangles"},"content":{"rendered":"\n<h2 class=\"wp-block-heading has-text-align-center has-text-color\" style=\"color:#00056d;text-transform:uppercase\"><strong>Angles in inscribed right triangles<\/strong><\/h2>\n\n\n\n<p class=\"has-text-color has-link-color has-huge-font-size wp-elements-aae0d72b0df05b1a7d05750bd97517d2\" style=\"color:#74008b\"><strong>Key Notes :<\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Inscribed Triangle Definition<\/strong><\/h3>\n\n\n\n<p>An <strong>inscribed triangle<\/strong> is a triangle whose <strong>all vertices lie on a circle<\/strong>.<br>\ud83d\udc49 This circle is called the <strong>circumcircle<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Right Triangle in a Circle<\/strong><\/h3>\n\n\n\n<p>A <strong>right triangle can always be inscribed in a circle<\/strong>.<br>\u2728 <strong>The hypotenuse becomes the diameter<\/strong> of the circle.<\/p>\n\n\n\n<p>\ud83d\udfe2 <strong>Important Rule:<\/strong><br>\u27a1\ufe0f <em>If a triangle is inscribed in a circle and one side is the diameter,<\/em><br>then the angle opposite the diameter is <strong>a right angle (90\u00b0)<\/strong>.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Thales&#8217; Theorem<\/strong> \ud83c\udfaf<\/h3>\n\n\n\n<p>This amazing theorem states:<br>\ud83d\udc49 <em>Any angle inscribed in a semicircle is a <strong>right angle (90\u00b0)<\/strong>.<\/em><br>So, if you see a triangle inside a circle with one side as diameter \u2192 <strong>90\u00b0 for sure!<\/strong> \u2714\ufe0f<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Angle Relationships<\/strong><\/h3>\n\n\n\n<p>In an inscribed right triangle:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>The <strong>right angle<\/strong> is opposite the <strong>diameter<\/strong><\/li>\n\n\n\n<li>The other two angles are <strong>acute angles<\/strong><\/li>\n\n\n\n<li>These two acute angles add up to <strong>90\u00b0<\/strong><\/li>\n<\/ul>\n\n\n\n<p>\ud83e\udde9 Example: If one angle is 30\u00b0, the other must be 60\u00b0.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Central Angle vs Inscribed Angle<\/strong> \ud83d\udd35\ud83d\udcd0<\/h3>\n\n\n\n<p>An <strong>inscribed angle<\/strong> is half of the <strong>central angle<\/strong> that subtends the same arc.<br>\ud83d\udc49 In a right triangle, the arc corresponding to the right angle is a semicircle (180\u00b0).<br>So, inscribed angle = 180\u00b0 \u00f7 2 = <strong>90\u00b0<\/strong> \ud83d\udc4d<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Identifying Right Angles in Diagrams<\/strong> \ud83d\udc40<\/h3>\n\n\n\n<p>If you spot:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>A triangle inside a circle<\/li>\n\n\n\n<li>One side as the <strong>diameter<\/strong><\/li>\n<\/ul>\n\n\n\n<p>\ud83c\udf89 Then you instantly know the angle opposite is <strong>90\u00b0<\/strong>!<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83d\udd39 <strong>Real-World Uses<\/strong> \ud83c\udf0d<\/h3>\n\n\n\n<p>Inscribed right triangles help in:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Finding missing angles<\/li>\n\n\n\n<li>Solving circle geometry problems<\/li>\n\n\n\n<li>Understanding trigonometry basics<\/li>\n\n\n\n<li>Proving theorems in coordinate geometry<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading has-large-font-size\">\ud83c\udf08 <strong>Summary<\/strong><\/h3>\n\n\n\n<p>\ud83d\udccc The hypotenuse of an inscribed right triangle is always the <strong>diameter<\/strong>.<br>\ud83d\udccc The angle opposite that diameter is always <strong>90\u00b0<\/strong>.<br>\ud83d\udccc The other two angles are acute and complementary.<br>\ud83d\udccc This is based on <strong>Thales\u2019 Theorem<\/strong>.<\/p>\n\n\n\n<p class=\"has-text-align-center has-text-color has-large-font-size\" style=\"color:#105000\"><strong>Learn with an example<\/strong><\/p>\n\n\n\n<div class=\"wp-block-group has-primary-color has-text-color has-background has-link-color has-large-font-size wp-elements-5e5cdf94a8c4eb9d702daef0ad8f0614\" style=\"background-color:#e7deff\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<div class=\"wp-block-group has-primary-color has-background-background-color has-text-color has-background has-link-color wp-elements-998edbaf6b3d45938a1dd91666466b4f\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<p class=\"has-text-color has-link-color wp-elements-669cf4cac7d804867fb2ea194ae5ea4a\" style=\"color:#b00012\"><strong>\ud83d\udd14 What&nbsp;is&nbsp;\u2220K?<\/strong><\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/d-5.png\" alt=\"\" class=\"wp-image-11837\" style=\"width:374px;height:auto\"\/><\/figure><\/div>\n\n\n<p>\u2220K=______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n<p>Since&nbsp;IK&nbsp;is a diameter of the circle,&nbsp;\u2220J&nbsp;is a right&nbsp;angle.<\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/image-removebg-preview-19.png\" alt=\"\" class=\"wp-image-11839\" style=\"width:371px;height:auto\"\/><\/figure><\/div>\n\n\n<p>So,&nbsp;\u25b3IJK&nbsp;is a right triangle and&nbsp;\u2220I&nbsp;and&nbsp;\u2220K&nbsp;are complementary. Write an equation setting the sum of their measures equal to&nbsp;90\u00b0,&nbsp;and solve for&nbsp;\u2220K.<\/p>\n\n\n\n<p>\u2220I+\u2220K=90\u00b0<\/p>\n\n\n\n<p><strong>36\u00b0<\/strong>+\u2220K=90\u00b0                Plug&nbsp;in&nbsp;\u2220I=36\u00b0<\/p>\n\n\n\n<p>\u2220K=54\u00b0                    Subtract&nbsp;36\u00b0&nbsp;from both&nbsp;sides<\/p>\n\n\n\n<p>\u2220K is 54\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n<div class=\"wp-block-group has-primary-color has-text-color has-background has-link-color has-large-font-size wp-elements-d0c50fc20e4462fb57f7271d4d9a1b3e\" style=\"background-color:#f9dcfc\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<div class=\"wp-block-group has-primary-color has-background-background-color has-text-color has-background has-link-color wp-elements-74151d9c1737ec5f141873f986b28724\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<p class=\"has-text-color has-link-color wp-elements-0219176e6636562692c567da88e4e77d\" style=\"color:#b00012\"><strong>\ud83d\udd14 What&nbsp;is&nbsp;\u2220F?<\/strong><\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/d-6.png\" alt=\"\" class=\"wp-image-11844\" style=\"width:380px;height:auto\"\/><\/figure><\/div>\n\n\n<p>\u2220F=_______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n<p>Since&nbsp;FG&nbsp;is a diameter of the circle,&nbsp;\u2220H&nbsp;is a right&nbsp;angle.<\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/image-removebg-preview-20.png\" alt=\"\" class=\"wp-image-11846\" style=\"width:387px;height:auto\"\/><\/figure><\/div>\n\n\n<p>So,&nbsp;\u25b3FGH&nbsp;is a right triangle and&nbsp;\u2220G&nbsp;and&nbsp;\u2220F&nbsp;are complementary. Write an equation setting the sum of their measures equal to&nbsp;90\u00b0,&nbsp;and solve for&nbsp;\u2220F.<\/p>\n\n\n\n<p>\u2220G+\u2220F=90\u00b0<\/p>\n\n\n\n<p><strong>52\u00b0<\/strong>+\u2220F=90\u00b0                   Plug&nbsp;in&nbsp;\u2220G=52\u00b0<\/p>\n\n\n\n<p>\u2220F=38\u00b0                      Subtract&nbsp;52\u00b0&nbsp;from both&nbsp;sides<\/p>\n\n\n\n<p>\u2220F is 38\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n<div class=\"wp-block-group has-primary-color has-text-color has-background has-link-color has-large-font-size wp-elements-9f61ef325bcb5a292f222a32791db78a\" style=\"background-color:#f9f2af\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<div class=\"wp-block-group has-primary-color has-background-background-color has-text-color has-background has-link-color wp-elements-f683d993fff118a5cee806a6dac3e9c1\"><div class=\"wp-block-group__inner-container is-layout-flow wp-block-group-is-layout-flow\">\n<p class=\"has-text-color has-link-color wp-elements-5d0f894d7fd7c0515f3ea5390aa700dd\" style=\"color:#b00012\"><strong>\ud83d\udd14 What&nbsp;is&nbsp;\u2220H?<\/strong><\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/d-7.png\" alt=\"\" class=\"wp-image-11849\" style=\"width:409px;height:auto\"\/><\/figure><\/div>\n\n\n<p>\u2220H=______ \u00b0<\/p>\n<\/div><\/div>\n\n\n\n<p>Since&nbsp;HI&nbsp;is a diameter of the circle,&nbsp;\u2220J&nbsp;is a right&nbsp;angle.<\/p>\n\n\n<div class=\"wp-block-image size-full\">\n<figure class=\"aligncenter is-resized\"><img decoding=\"async\" src=\"https:\/\/10thclass.deltapublications.in\/wp-content\/uploads\/2024\/01\/image-removebg-preview-21.png\" alt=\"\" class=\"wp-image-11850\" style=\"width:355px;height:auto\"\/><\/figure><\/div>\n\n\n<p>So,&nbsp;\u25b3HIJ&nbsp;is a right triangle and&nbsp;\u2220I&nbsp;and&nbsp;\u2220H&nbsp;are complementary. Write an equation setting the sum of their measures equal to&nbsp;90\u00b0,&nbsp;and solve for&nbsp;\u2220H.<\/p>\n\n\n\n<p>\u2220I+\u2220H=90\u00b0<\/p>\n\n\n\n<p><strong>30\u00b0<\/strong>+\u2220H=90\u00b0                Plug&nbsp;in&nbsp;\u2220I=30\u00b0<\/p>\n\n\n\n<p>\u2220H=60\u00b0                   Subtract&nbsp;30\u00b0&nbsp;from both&nbsp;sides<\/p>\n\n\n\n<p>\u2220H is 60\u00b0.<\/p>\n<\/div><\/div>\n\n\n\n<p class=\"has-text-color has-huge-font-size\" style=\"color:#d90000\"><strong>let&#8217;s practice!<\/strong><\/p>\n\n\n\n<div class=\"wp-block-columns is-layout-flex wp-container-core-columns-is-layout-9d6595d7 wp-block-columns-is-layout-flex\">\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<figure class=\"wp-block-image size-full\"><a href=\"https:\/\/wordwall.net\/play\/89043\/261\/109\"><img loading=\"lazy\" decoding=\"async\" width=\"500\" height=\"500\" src=\"https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-3-74.png\" alt=\"\" class=\"wp-image-7910\" srcset=\"https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-3-74.png 500w, https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-3-74-300x300.png 300w, https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-3-74-150x150.png 150w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><\/figure>\n<\/div>\n\n\n\n<div class=\"wp-block-column is-layout-flow wp-block-column-is-layout-flow\">\n<figure class=\"wp-block-image size-full\"><a href=\"https:\/\/wordwall.net\/play\/89042\/465\/858\"><img loading=\"lazy\" decoding=\"async\" width=\"500\" height=\"500\" src=\"https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-1-2-83.png\" alt=\"\" class=\"wp-image-7911\" srcset=\"https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-1-2-83.png 500w, https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-1-2-83-300x300.png 300w, https:\/\/9thclass.deltapublications.in\/wp-content\/uploads\/2023\/05\/Worksheet-1-1-2-83-150x150.png 150w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a><\/figure>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Angles in inscribed right triangles Key Notes : \ud83d\udd39 Inscribed Triangle Definition An inscribed triangle is a triangle whose all vertices lie on a circle.\ud83d\udc49 This circle is called the circumcircle. \ud83d\udd39 Right Triangle in a Circle A right triangle can always be inscribed in a circle.\u2728 The hypotenuse becomes the diameter of the circle.<a class=\"more-link\" href=\"https:\/\/9thclass.deltapublications.in\/index.php\/l-10-angles-in-inscribed-right-triangles\/\">Continue reading <span class=\"screen-reader-text\">&#8220;L.10 Angles in inscribed right triangles&#8221;<\/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-212","page","type-page","status-publish","hentry","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/212","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/comments?post=212"}],"version-history":[{"count":12,"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/212\/revisions"}],"predecessor-version":[{"id":18297,"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/pages\/212\/revisions\/18297"}],"wp:attachment":[{"href":"https:\/\/9thclass.deltapublications.in\/index.php\/wp-json\/wp\/v2\/media?parent=212"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}