{"id":11546,"date":"2022-07-18T14:43:02","date_gmt":"2022-07-18T09:13:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11546"},"modified":"2022-07-18T14:43:07","modified_gmt":"2022-07-18T09:13:07","slug":"in-an-acute-triangle-the-square-of-the-side-opposite-to-an-acute-angle-is-equal-to-the-sum-of-the-squares-of-other-two-sides-minus-twice-the-product-of-one-side-and-the-projection-of-the-other-on-fir","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-an-acute-triangle-the-square-of-the-side-opposite-to-an-acute-angle-is-equal-to-the-sum-of-the-squares-of-other-two-sides-minus-twice-the-product-of-one-side-and-the-projection-of-the-other-on-fir\/","title":{"rendered":"In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first."},"content":{"rendered":"

Solution :<\/h2>\n

Given<\/strong> : In triangle ABC angle B is an acute angle and AD \\(\\perp\\) BC\"triangle\"<\/p>\n

To Prove<\/strong> : \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) – 2BC.BD<\/p>\n

Proof<\/strong> : Since \\(\\triangle\\) ADC is a right triangle, angled at D. Therefore, by Pythagoras theorem,<\/p>\n

\\({AC}^2\\) = \\({AD}^2\\) + \\({CD}^2\\)<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({BC – BD}^2\\)\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0(because DC = BC – BD)<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({BC}^2\\) + \\({BD}^2\\) – 2BC.BD<\/p>\n

Since, In triangle ADB, \\({AB}^2\\) = \\({AD}^2\\) + \\({BD}^2\\)<\/p>\n

\\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) – 2BC.BD<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Given : In triangle ABC angle B is an acute angle and AD \\(\\perp\\) BC To Prove : \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) – 2BC.BD Proof : Since \\(\\triangle\\) ADC is a right triangle, angled at D. Therefore, by Pythagoras theorem, \\({AC}^2\\) = \\({AD}^2\\) + \\({CD}^2\\) \\(\\implies\\)\u00a0 \\({AC}^2\\) = \\({AD}^2\\) + \\({BC – …<\/p>\n

In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first.<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[43,912],"tags":[],"yoast_head":"\nIn an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first. - Mathemerize<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/in-an-acute-triangle-the-square-of-the-side-opposite-to-an-acute-angle-is-equal-to-the-sum-of-the-squares-of-other-two-sides-minus-twice-the-product-of-one-side-and-the-projection-of-the-other-on-fir\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first. - Mathemerize\" \/>\n<meta property=\"og:description\" content=\"Solution : Given : In triangle ABC angle B is an acute angle and AD (perp) BC To Prove : ({AC}^2) = ({AB}^2) + ({BC}^2) – 2BC.BD Proof : Since (triangle) ADC is a right triangle, angled at D. Therefore, by Pythagoras theorem, ({AC}^2) = ({AD}^2) + ({CD}^2) (implies)\u00a0 ({AC}^2) = ({AD}^2) + ({BC – … In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first. 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