Solution : (i) In \(\triangle\) ABC, \(cot \theta\) = \(7\over 8\) = \(AB\over BC\) Let AB = 7k and BC = 8k Now, AC = \(\sqrt{{AB}^2 + {BC}^2}\) = \(\sqrt{113k^2}\) So, AC = \(\sqrt{113}k\) Thus, \(sin \theta\) = \(8k\over \sqrt{113}k\) = \(8\over \sqrt{113}\) \(cos \theta\) = \(7k\over \sqrt{113}k\) = \(7\over \sqrt{113}\) Now, \({(1 + sin\theta)(1 …
Solution : Let us consider two right triangles PQA and RSB in which cos A = cos B (see figure). We have : cos A = \(QA\over PA\) and cos B = \(SB\over RB\) Thus, it is given that \(QA\over PA\) = \(SB\over RB\) So, \(QA\over SB\) = \(PA\over RB\) …
Solution : Consider a triangle ABC in which \(\angle\) A = \(\theta\) and \(\angle\) B = 90 Then, Base = AB, perp = BC and Hypo. = AC, \(\therefore\) \(sec \theta\) = \(perp\over hypo\) = \(BC\over AC\) = \(3\over 4\) Let AC = 13k and AB = 12k. Then, By using Pythagoras Theorem, \({BC}^2\) = …
Given \(sec \theta\) = \(13\over 12\), calculate all other trigonometric ratios. Read More »
Solution : We have, 15 cot A = 8 \(\implies\) cot A = \(8\over 15\) Draw a right triangle ABC, cot A = \(AB\over BC\) = \(8\over 15\) If BC = 15k, then AB = 8k, where k is any positive number. By using Pythagoras Theorem, \({AC}^2\) = \({AB}^2\) + \({BC}^2\) = \(64k^2\) + \(225k^2\) …
Solution : Consider a triangle ABC in which \(\angle\) B = 90 For \(\angle\) A, we have : Base = AB, Perpendicular = BC and Hypotenuse = AC, \(\therefore\) Sin A = \(perpendicular\over hypotenuse\) = \(BC\over AC\) = \(3\over 4\) Let BC = 3k and AC = 4k, Then, \({AB}^2\) = \({AC}^2 – {BC}^2\) = …
If Sin A = \(3\over 4\), Calculate Cos A and Tan A. Read More »
Solution : In triangle PQR, we have \(\angle\) Q = 90, PQ = 12 cm and PR = 13 cm Using Pythagoras Theorem, \({PR}^2\) = \({PQ}^2\) + \({QR}^2\) \(\implies\) \({QR}^2\) = \({PR}^2\) – \({PQ}^2\) = \((13)^2\) – \((12)^2\) = 25 \(\implies\) QR = 5 cm We know that, tan P = \(QR\over PQ\) = tan …
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, AC = \(\sqrt{625}\) = 25 cm (i) Sin A = \(BC\over AB\) = \(7\over 25\), Cos A = \(AB\over AC\) = \(24\over 25\) …
Solution : The value of cos 54 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Proof : We know that value of sin 36 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Since 54 degree is the complement of 36 degree. \(\therefore\) cos 54 = sin(90 – 36) = sin 36 = \(\sqrt{10 – 2\sqrt{5}}\over 4\) Hence, cos 54 …
Solution : The value of sin 54 degrees is \(\sqrt{5} + 1\over 4\). Proof : We know that value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Since 54 degree is the complement of 36 degree. \(\therefore\) sin 54 = sin(90 – 36) = cos 36 = \(\sqrt{5} + 1\over 4\) Hence, sin 54 …
Solution : The value of sin 36 degrees is \(\sqrt{10 – 2\sqrt{5}}\over 4\). Proof : We know that the value of cos 36 degrees is \(\sqrt{5} + 1\over 4\). Putting \(\theta\) = 36 in \(cos \theta\) = \(\sqrt{1 – sin^2 \theta}\), we get cos 36 = \(\sqrt{1 – sin^2 36}\) = \(\sqrt{1 – ({\sqrt{5} + …
