Engineering
Question : If $\tan \theta=\frac{8}{15}$, then the value of $\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$ is:
Option 1: $\frac{1}{5}$
Option 2: $\frac{3}{5}$
Option 3: $\frac{2}{5}$
Option 4: $\frac{4}{5}$
Correct Answer: $\frac{3}{5}$
Solution : $\tan \theta=\frac{8}{15}=\frac{\text{Perpendicular}}{\text{Base}}$ $\therefore$ Hypotenuse = $\sqrt{8^2+15^2}=\sqrt{289}= 17$ So, $\sin \theta=\frac{8}{17}$ $\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sqrt{\frac{1-\frac{8}{17}}{1+\frac{8}{17}}}=\sqrt{\frac{\frac{9}{17}}{\frac{25}{17}}}=\sqrt{\frac{9}{25}}=\frac{3}{5}$ Hence, the correct answer is $\frac{3}{5}$.
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Engineering is one of the top fields with wide scope and vast number of opportunities. Engineering colleges in Bangalore are given below:
1) IIIT Bangalore
2) Ramaiah Institute of Technology
3) JAIN (Deemed-to-be-university), Faculty of Engineering and Technology
4) Dayananda Sagar College of Engineering
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Question : Directions: If a mirror is placed on the line AB, then out of the option figures which figure will be the right image of the question figure?
Option 1:
Option 2:
Option 3:
Option 4:
Correct Answer:
Solution : As per the mirror image properties, closer things appear closer to the mirror in the reflection. Here, according to the information provided, the mirror is placed on the right side of the figure (on line AB). So, the left side of the reflected image will appear
Hello Aman
You can get lateral entry to b tech In civil engineering course after completing diploma in civil engineering.
After completing diploma , you will have to appear for entrance exam .
Eligibility for b tech course - you must have passed 10+2 exam with physics, chemistry and mathematics
If you are looking forward to pursue B.tech in Computer Science from a reputed or from a top college with great facilities and high placements, then you will have to give entrance exams. To apply to colleges or universities in India, you need to score highly on entrance exams
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