{"id":8241,"date":"2024-04-15T02:57:15","date_gmt":"2024-04-15T02:57:15","guid":{"rendered":"https:\/\/exam.pscnotes.com\/mcq\/?p=8241"},"modified":"2024-04-15T02:57:15","modified_gmt":"2024-04-15T02:57:15","slug":"according-to-napiers-rules-of-circular-parts-for-a-right-angled-triangle-sine-of-middle-part-equals-the-product-of-a-tangents-of-two-adjacent-parts-b-sines-of-two-adjacent-parts-c-cosines-of-two","status":"publish","type":"post","link":"https:\/\/exam.pscnotes.com\/mcq\/according-to-napiers-rules-of-circular-parts-for-a-right-angled-triangle-sine-of-middle-part-equals-the-product-of-a-tangents-of-two-adjacent-parts-b-sines-of-two-adjacent-parts-c-cosines-of-two\/","title":{"rendered":"According to Napier&#8217;s Rules of circular parts for a right angled triangle, sine of middle part equals the product of A. Tangents of two adjacent parts B. Sines of two adjacent parts C. Cosines of two adjacent parts D. Both (A) and (B) above"},"content":{"rendered":"<p>\r\n    <!-- Check if it's an AMP page -->\r\n            <!-- Non-AMP version -->\r\n        <div class=\"mcq-container\" data-quiz-id=\"quizState_6a9ed9642d5a1\">\r\n                                            <div class=\"option\" data-option-key=\"option1\" data-is-correct=\"false\">\r\n                    Tangents of two adjacent parts                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option2\" data-is-correct=\"false\">\r\n                    Sines of two adjacent parts                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option3\" data-is-correct=\"false\">\r\n                    Cosines of two adjacent parts                <\/div>\r\n                                            <div class=\"option\" data-option-key=\"option4\" data-is-correct=\"true\">\r\n                    Both (A) and (B) above                <\/div>\r\n                            \r\n            <!-- Feedback messages for non-AMP -->\r\n            <div class=\"feedback\" data-feedback=\"wrong\">Answer is Right!<\/div>\r\n            <div class=\"feedback\" data-feedback=\"right\">Answer is Wrong!<\/div>\r\n        <\/div>\r\n\r\n        <script>\r\n        document.addEventListener('DOMContentLoaded', function () {\r\n            var containers = document.querySelectorAll('.mcq-container');\r\n\r\n            containers.forEach(function(container) {\r\n                var options = container.querySelectorAll('.option');\r\n                var feedbackSelect = container.querySelector('[data-feedback=\"select\"]');\r\n                var feedbackWrong = container.querySelector('[data-feedback=\"wrong\"]');\r\n                var feedbackRight = container.querySelector('[data-feedback=\"right\"]');\r\n\r\n                options.forEach(function(option) {\r\n                    option.addEventListener('click', function() {\r\n                        var selectedOption = option.getAttribute('data-option-key');\r\n                        var isCorrect = option.getAttribute('data-is-correct') === 'true';\r\n\r\n                        \/\/ Remove previous selections\r\n                        options.forEach(function(opt) {\r\n                            opt.classList.remove('correct', 'incorrect');\r\n                        });\r\n\r\n                        \/\/ Add the correct\/incorrect class\r\n                        if (isCorrect) {\r\n                            option.classList.add('correct');\r\n                            feedbackRight.hidden = false;\r\n                            feedbackWrong.hidden = true;\r\n                        } else {\r\n                            option.classList.add('incorrect');\r\n                            feedbackRight.hidden = true;\r\n                            feedbackWrong.hidden = false;\r\n                        }\r\n\r\n                        \/\/ Hide select feedback\r\n                        feedbackSelect.hidden = true;\r\n                    });\r\n                });\r\n            });\r\n        });\r\n        <\/script>\r\n    \r\n    <!--more--><\/p>\n<p>The correct answer is: <strong>D. Both (A) and (B) above<\/strong><\/p>\n<p>Napier&#8217;s Rules of Circular Parts are a set of rules that can be used to find the <div class=\"youtube-subscribe-container\">\r\n        <a href=\"https:\/\/www.youtube.com\/channel\/UCNHT8lW-JmLC68rjBfZhdkg?sub_confirmation=1\" target=\"_blank\" class=\"youtube-subscribe-button\">\r\n            <span class=\"youtube-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 576 512\">\r\n                    <path d=\"M549.7 124.1c-6.3-23.7-24.8-42.3-48.3-48.6C458.8 64 288 64 288 64S117.2 64 74.6 75.5c-23.5 6.3-42 24.9-48.3 48.6-11.4 42.9-11.4 132.3-11.4 132.3s0 89.4 11.4 132.3c6.3 23.7 24.8 41.5 48.3 47.8C117.2 448 288 448 288 448s170.8 0 213.4-11.5c23.5-6.3 42-24.2 48.3-47.8 11.4-42.9 11.4-132.3 11.4-132.3s0-89.4-11.4-132.3zm-317.5 213.5V175.2l142.7 81.2-142.7 81.2z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Subscribe on YouTube\r\n        <\/a>\r\n  <div class=\"telegram-channel-container\">\r\n        <a href=\"https:\/\/t.me\/pscnotes2025\" target=\"_blank\" class=\"telegram-channel-button\">\r\n            <span class=\"telegram-icon\">\r\n                <svg xmlns=\"http:\/\/www.w3.org\/2000\/svg\" viewBox=\"0 0 496 512\">\r\n                    <path fill=\"white\" d=\"M248,8C111,8,0,119,0,256s111,248,248,248s248-111,248-248S385,8,248,8z M362,177L320,367c-3,14-10,18-20,14l-56-41l-27,26 c-3,3-5,5-10,5l4-63L323,196c5-5-1-7-8-3l-98,62l-42-13c-9-3-10-9,2-14l162-63C351,160,365,164,362,177z\"\/>\r\n                <\/svg>\r\n            <\/span>\r\n            Join Our Telegram Channel\r\n        <\/a>\r\n    <\/div>   <\/div> trigonometric ratios of any angle in a right triangle. The rules state that the sine of the middle part of a right triangle is equal to the product of the tangents of the two adjacent parts, and the cosine of the middle part is equal to the product of the sines of the two adjacent parts.<\/p>\n<p>For example, in the right triangle below, the sine of the angle $\\theta$ is equal to the product of the tangents of the angles $A$ and $B$, and the cosine of the angle $\\theta$ is equal to the product of the sines of the angles $A$ and $B$.<\/p>\n<p>[asy]<br \/>\nunitsize(1 cm);<\/p>\n<p>draw((0,0)&#8211;(4,0)&#8211;(2,2.82843)&#8211;cycle);<\/p>\n<p>label(&#8220;$A$&#8221;, (0,0), S);<br \/>\nlabel(&#8220;$B$&#8221;, (4,0), N);<br \/>\nlabel(&#8220;$\\theta$&#8221;, (2,2.82843), E);<\/p>\n<p>draw((0,0)&#8211;(2,0));<br \/>\ndraw((2,0)&#8211;(2,2.82843));<\/p>\n<p>label(&#8220;$1$&#8221;, (1,0), S);<br \/>\nlabel(&#8220;$1$&#8221;, (2,1), E);<br \/>\n[\/asy]<\/p>\n<p>Therefore, the answer to the question is: <strong>D. Both (A) and (B) above<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Join Our Telegram Channel Subscribe on YouTube<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[654],"tags":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.2 (Yoast SEO v23.3) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>According to Napier&#039;s Rules of circular parts for a right angled triangle, sine of middle part equals the product of A. Tangents of two adjacent parts B. Sines of two adjacent parts C. Cosines of two adjacent parts D. 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