Jump to
- Rational and Irrational Numbers
- Compound Interest
- Expansions
- Factorization
- Simultaneous Linear Equations
- Problems on Simultaneous Linear Equations
- Quadratic Equations
- Indices
- Logarithms
- Triangles
- Mid Point Theorem
- Pythagoras Theorem
- Rectilinear Figures
- Theorems on Area
- Circle
- Mensuration
- Trigonometric Ratios
- Trigonometric Ratios and Standard Angles
- Coordinate Geometry
- Statistics
ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
Prev
If O is an acute angle and tan = 8/15 then find sec + cosec
Answer:
Given tan θ = 8/15
θ is an acute angle
in the figure triangle, OMP is a right-angled triangle,
\angle \mathrm{M}=90^{\circ} \text { and } \angle \mathrm{Q}=\theta
Tan θ = PM/OL = 8/15
Therefore, PM = 8, OM = 15
\begin{aligned} &\text { But } \mathrm{OP}^{2}=\mathrm{OM}^{2}+\mathrm{PM}^{2} \text { using Pythagoras theorem, }\\ &=15^{2}+8^{2}\\ &=225+64\\ &=289\\ &=17^{2} \end{aligned}
Therefore, OP = 17
Sec θ = OP/OM
= 17/15
Cosec θ = OP/PM
= 17/8
Now,
Sec θ + cosec θ = (17/15) + (17/8)
= (136 + 255)/ 120
= 391/120
Related Questions
Facebook
Whatsapp
Copy Link
Exercises
Chapters
Lido
Courses
Quick Links
Terms & Policies
Terms & Policies
Connect with us on social media!
2022 © Quality Tutorials Pvt Ltd All rights reserved