Not the exact question you were looking for?

Paste your question to our Mathbee AI Mentor below to get an instant step-by-step solution.

Introduction To Trigonometry
EXERCISE 8.1
CBSE_NCERT_TEXTBOOK
Grade 10

Question:

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

Step-by-Step Solution

Key Concept: Use the definition of cotangent \(\cot A = \frac{\text{adjacent}}{\text{opposite}}\) and the Pythagorean relation \(\sin^2 A + \cos^2 A = 1\). From \(\cot A\) we obtain \(\tan A\), construct a right‑angled triangle, compute the hypotenuse, and then evaluate \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}}\) and \(\sec A = \frac{\text{hypotenuse}}{\text{adjacent}}\).
1. Given condition
\[15\cot A = 8 \quad\Rightarrow\quad \cot A = \frac{8}{15}.\]
2. Relation between cot and tan
\[\cot A = \frac{1}{\tan A} \;\Rightarrow\; \tan A = \frac{1}{\cot A}=\frac{15}{8}.\]
3. Form a right‑angled triangle
Let the side opposite \(A\) be \(15\) units and the side adjacent to \(A\) be \(8\) units (consistent with \(\tan A = \frac{\text{opposite}}{\text{adjacent}} = \frac{15}{8}\)).
4. Find the hypotenuse using Pythagoras theorem:
\[\text{hypotenuse} = \sqrt{8^{2}+15^{2}} = \sqrt{64+225}=\sqrt{289}=17.\]
5. Compute \(\sin A\)
\[\sin A = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{15}{17}.\]
6. Compute \(\sec A\)
\[\sec A = \frac{\text{hypotenuse}}{\text{adjacent}} = \frac{17}{8}.\]
7. Answer
\[\sin A = \frac{15}{17}, \qquad \sec A = \frac{17}{8}.\]

Correct Answer: sin A = 15/17, sec A = 17/8
Mathbee AI Mentor (Free Demo)

Confused by the solution? Ask the AI to explain a specific step, tell you where you went wrong, or break down the key trap in this question.

Master Introduction To Trigonometry with Mathbee

Practice this topic under real exam conditions with strict timers, or ask our AI Mentor to explain the concepts step-by-step.

Start Practicing for Free