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
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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
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