{"id":11613,"date":"2022-08-02T14:22:19","date_gmt":"2022-08-02T08:52:19","guid":{"rendered":"https:\/\/mathemerize.com\/?p=11613"},"modified":"2022-08-02T14:22:45","modified_gmt":"2022-08-02T08:52:45","slug":"in-triangle-abc-right-angled-at-b-ab-24-cm-bc-7-cm-determine-i-sin-a-cos-a-ii-sin-c-cos-c","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/in-triangle-abc-right-angled-at-b-ab-24-cm-bc-7-cm-determine-i-sin-a-cos-a-ii-sin-c-cos-c\/","title":{"rendered":"In triangle ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) Sin A, Cos A (ii) Sin C, Cos C"},"content":{"rendered":"
Let us draw a right \\(\\triangle\\) ABC,<\/p>\n
By using Pythagoras Theorem, we have :<\/p>\n
\\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) = \\((24)^2\\) + \\((7)^2\\)<\/p>\n
= 576 + 49 = 625<\/p>\n
So,\u00a0 AC = \\(\\sqrt{625}\\) = 25 cm<\/p>\n
(i)<\/strong>\u00a0 Sin A = \\(BC\\over AB\\) = \\(7\\over 25\\)<\/strong>,\u00a0 Cos A = \\(AB\\over AC\\) = \\(24\\over 25\\)<\/strong><\/p>\n (ii)<\/strong>\u00a0 Sin C = \\(AB\\over AC\\) = \\(24\\over 25\\)<\/strong>,\u00a0 Cos C = \\(BC\\over AC\\) = \\(7\\over 25\\)<\/strong><\/p>\n","protected":false},"excerpt":{"rendered":" Solution : Let us draw a right \\(\\triangle\\) ABC, By using Pythagoras Theorem, we have : \\({AC}^2\\) = \\({AB}^2\\) + \\({BC}^2\\) = \\((24)^2\\) + \\((7)^2\\) = 576 + 49 = 625 So,\u00a0 AC = \\(\\sqrt{625}\\) = 25 cm (i)\u00a0 Sin A = \\(BC\\over AB\\) = \\(7\\over 25\\),\u00a0 Cos A = \\(AB\\over AC\\) = \\(24\\over 25\\) …<\/p>\n