mathemerize

Solution : Consider a \(\triangle\) ABC, in which \(\angle\) B = 90 For \(\angle\) A, we have : Base = AB, Perp. = BC,  and   Hyp. = AC, tan A = \(\perp\over base\) = \(BC\over AB\) = \(1\over \sqrt{3}\) Let BC = k and AB = \(\sqrt{3} k, AC = \(\sqrt{{AB}^2 + {BC}^2}\) = 2k …

In \(\triangle\) ABC right angled at B, it tan A = \(1\over \sqrt{3}\), find the value of (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C Read More »

Solution : We have,   3 cot A = 4    \(\implies\)  cot A = \(4\over 3\) = \(AB\over BC\)  Let  AB = 4k, then BC = 3k By Pythagoras Theorem, \({AC}^2\) = \({AB}^2\) + \({BC}^2\) \(\implies\)  \({AC}^2\) = \(25k^2\) \(\implies\)  AC = 5k Thus,  tan A = \(BC\over AB\) = \(3k\over 4k\) = \(3\over 4\) …

If 3 cot A = 4, check whether \(1 – tan^2A\over 1 + tan^2A\) = \(cos^2A – sin^2A\) or not. Read More »

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 …

If cot \(\theta\) = \(7\over 8\), evaluate : (i) \({(1 + sin\theta)(1 – sin\theta)}\over {(1 + cos\theta)(1 – cos\theta)}\) (ii) \(cot^2 \theta\) Read More »

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, \(QA\over SB\) = \(PA\over RB\) …

If \(\angle\) A and \(\angle\) B are acute angles such that cos A = cos B, then show that \(\angle\) A = \(\angle\) B. Read More »

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, \({AC}^2\) = \({AB}^2\) + \({BC}^2\) = \(64k^2\) + \(225k^2\) …

Given 15 cot A = 8, find sin A and sec A. Read More »

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 …

In the figure, find tan P – cot R ? Read More »

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 = \(AB\over AC\) = \(24\over 25\) …

In triangle ABC, right angled at B, AB = 24 cm, BC = 7 cm. Determine : (i) Sin A, Cos A (ii) Sin C, Cos C Read More »