{"id":1754,"date":"2020-04-16T10:24:12","date_gmt":"2020-04-16T10:24:12","guid":{"rendered":"http:\/\/ibalmaths.com\/?post_type=knowledgebase&#038;p=1754"},"modified":"2020-06-12T08:02:16","modified_gmt":"2020-06-12T08:02:16","slug":"practice-non-right-angled-triangle-trigonometry","status":"publish","type":"knowledgebase","link":"https:\/\/ibalmaths.com\/index.php\/ibdp-math-hl-2\/non-right-angled-triangle-trigonometry\/practice-non-right-angled-triangle-trigonometry\/","title":{"rendered":"Practice &#8211; Non Right-Angled Triangle Trigonometry"},"content":{"rendered":"<p>1. The sides of a triangle are in arithmetic sequence and the greatest angle is double the smallest angle. Prove that the ratio of its sides is 4:5:6.<\/p>\r\n<div id=\"link1-link-1754\" class=\"sh-link link1-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link1', 1754, 'Solution', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link1-toggle-1754\">Solution<\/span><\/a><\/div><div id=\"link1-content-1754\" class=\"sh-content link1-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>\u00a0<\/div>\r\n<p>2. If\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>a<\/mi><mi>cos<\/mi><mi>A<\/mi><mo>=<\/mo><mi>b<\/mi><mi>cos<\/mi><mi>B<\/mi><\/math>\r\n , prove that\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mo>&#8710;<\/mo><mi>A<\/mi><mi>B<\/mi><mi>C<\/mi><mo>&#160;<\/mo><\/math>\r\n is either isosceles, or right angled.\u00a0<\/p>\r\n<div id=\"link2-link-1754\" class=\"sh-link link2-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link2', 1754, 'Solution', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link2-toggle-1754\">Solution<\/span><\/a><\/div><div id=\"link2-content-1754\" class=\"sh-content link2-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>\u00a0<\/div>\r\n<p>3. The angles of a triangle are in the ratio 1:2:7, then show that the ratio of the greatest side to the least side is\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><msqrt><mn>5<\/mn><\/msqrt><mo>+<\/mo><mn>1<\/mn><mo>&#160;<\/mo><mo>:<\/mo><mo>&#160;<\/mo><msqrt><mn>5<\/mn><\/msqrt><mo>&#8211;<\/mo><mn>1<\/mn><\/math>\r\n .<\/p>\r\n<div id=\"link3-link-1754\" class=\"sh-link link3-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link3', 1754, 'Solution', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link3-toggle-1754\">Solution<\/span><\/a><\/div><div id=\"link3-content-1754\" class=\"sh-content link3-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>\u00a0<\/div>\r\n<p>4. If the angles of a triangle are\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mn>30<\/mn><mo>&#176;<\/mo><\/math>\r\n and\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mn>45<\/mn><mo>&#176;<\/mo><\/math>\r\n and the included side is\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mfenced><mrow><msqrt><mn>3<\/mn><\/msqrt><mo>+<\/mo><mn>1<\/mn><\/mrow><\/mfenced><\/math>\r\n cm, the area of the triangle is\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mfrac><mn>1<\/mn><mn>2<\/mn><\/mfrac><mfenced><mrow><msqrt><mn>3<\/mn><\/msqrt><mo>+<\/mo><mn>1<\/mn><\/mrow><\/mfenced><\/math>\r\n cm<sup>2<\/sup>.<\/p>\r\n<div id=\"link4-link-1754\" class=\"sh-link link4-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link4', 1754, 'Solution', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link4-toggle-1754\">Solution<\/span><\/a><\/div><div id=\"link4-content-1754\" class=\"sh-content link4-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>\u00a0<\/div>\r\n<p>5. The sides of a triangle are\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mn>2<\/mn><mo>,<\/mo><mo>&#160;<\/mo><mn>3<\/mn><mo>,<\/mo><mo>&#160;<\/mo><msqrt><mn>19<\/mn><\/msqrt><\/math>\r\n then find the value of the greatest angle of the triangle.<\/p>\r\n<div id=\"link5-link-1754\" class=\"sh-link link5-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link5', 1754, 'Solution', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link5-toggle-1754\">Solution<\/span><\/a><\/div><div id=\"link5-content-1754\" class=\"sh-content link5-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>\r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mn>120<\/mn><mo>&#176;<\/mo><\/math>\r\n<\/p>\r\n<p>\u00a0<\/div>\r\n<p>6. The diagram shows a sector of a circle OAB. C is the midpoint of OB. Angle AOB is\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>&#952;<\/mi><\/math>\r\n radians.<\/p>\r\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1906\" src=\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/04\/1-50-300x215.jpg\" alt=\"\" width=\"300\" height=\"215\" srcset=\"https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/04\/1-50-300x215.jpg 300w, https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/04\/1-50.jpg 657w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\r\n<p>Given that the area of the triangle OAC is equal to one quarter of the area of the sector OAB, show that\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>&#952;<\/mi><mo>=<\/mo><mn>2<\/mn><mi>sin<\/mi><mi>&#952;<\/mi><\/math>\r\n .\u00a0 \u00a0 \u00a0[4 marks]<\/p>\r\n<div id=\"link6-link-1754\" class=\"sh-link link6-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link6', 1754, 'Solution6', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link6-toggle-1754\">Solution6<\/span><\/a><\/div><div id=\"link6-content-1754\" class=\"sh-content link6-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p>Answer Given<\/p>\r\n<p>\u00a0<\/div>\r\n<p>7.<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1922\" src=\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1-2-300x155.png\" alt=\"\" width=\"300\" height=\"155\" srcset=\"https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1-2-300x155.png 300w, https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1-2.png 642w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\r\n<p>The diagram shows a circle with centre O and radius \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>r<\/mi><\/math>\r\n cm. The points A and B lie on the circle and AT is a tangent to the circle. Angle AOB = 1 radians and OBT is a straight line.<br \/>\r\n(i) Express the area of the shaded region in terms of\u00a0 \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>r<\/mi><\/math>\r\n and \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>l<\/mi><\/math>\r\n . [3 marks]<\/p>\r\n<p>(ii) In the case where \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>r<\/mi><mo>=<\/mo><mn>3<\/mn><\/math>\r\n and \r\n<math xmlns=\"http:\/\/www.w3.org\/1998\/Math\/MathML\"><mi>&#952;<\/mi><mo>=<\/mo><mn>1<\/mn><mo>.<\/mo><mn>2<\/mn><\/math>\r\n , find the perimeter of the shaded region. [4 marks]<\/p>\r\n<div id=\"link-link-1754\" class=\"sh-link link-link sh-hide\"><a href=\"#\" onclick=\"showhide_toggle('link', 1754, 'Solution7', 'Hide Solution'); return false;\" aria-expanded=\"false\"><span id=\"link-toggle-1754\">Solution7<\/span><\/a><\/div><div id=\"link-content-1754\" class=\"sh-content link-content sh-hide\" style=\"display: none;\"><\/p>\r\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-medium wp-image-1924\" src=\"http:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1-300x246.jpg\" alt=\"\" width=\"300\" height=\"246\" srcset=\"https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1-300x246.jpg 300w, https:\/\/ibalmaths.com\/wp-content\/uploads\/2020\/06\/1.jpg 394w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\r\n<p>\u00a0<\/div><!-- AddThis Advanced Settings generic via filter on the_content --><!-- AddThis Share Buttons generic via filter on the_content -->","protected":false},"excerpt":{"rendered":"1. The sides of a triangle are in arithmetic sequence and the greatest angle is double the smallest angle. Prove that the ratio of its sides is 4:5:6. 2. If\u00a0 acosA=bcosB , prove that\u00a0 &#8710;ABC&#160; is either isosceles, or right angled.\u00a0 3. The angles of a triangle are in the ratio 1:2:7, then show that [&hellip;]<!-- AddThis Advanced Settings generic via filter on get_the_excerpt --><!-- AddThis Share Buttons generic via filter on get_the_excerpt -->","protected":false},"author":4,"featured_media":0,"comment_status":"closed","ping_status":"closed","template":"","class_list":["post-1754","knowledgebase","type-knowledgebase","status-publish","hentry","knowledgebase_cat-non-right-angled-triangle-trigonometry","no-wpautop"],"_links":{"self":[{"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/knowledgebase\/1754","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/knowledgebase"}],"about":[{"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/types\/knowledgebase"}],"author":[{"embeddable":true,"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/users\/4"}],"replies":[{"embeddable":true,"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/comments?post=1754"}],"version-history":[{"count":9,"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/knowledgebase\/1754\/revisions"}],"predecessor-version":[{"id":1927,"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/knowledgebase\/1754\/revisions\/1927"}],"wp:attachment":[{"href":"https:\/\/ibalmaths.com\/index.php\/wp-json\/wp\/v2\/media?parent=1754"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}