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A right angle triangle has a base of \(25m\) and its area is \(375m^{2}\) find its height

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Question : A triangle has a base of  \(25m\)  and its area is  \(375m^{2}\)  find its height?  Answer : Height of the right angle triangle is  \(30m\)  . Explanation : Given base is  \(25m\)  and area is  \(375m^{2}\) Figure-1 (right angle triangle) Now we know area of right angle triangle=  \(\frac{1}{2}\times base \times height\) So according to question,  \(\frac{1}{2}\times base \times height\) =375 \[\Rightarrow base \times height= 375 \times 2\] \[\Rightarrow 25 \times height= 750\] \[\Rightarrow height= \frac{750}{25}\] \[\Rightarrow height= 30\] Hence, height of the right angle triangle is \(30m\)  . (Answer)
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Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle

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Question : Express \(Sine(sin)\), \(Cosine(cos)\) and \(tangent(tan)\) in terms of sides of triangle  Answer : For a right angle triangle  we can find sine or sin, cosine or cos, tangent or tan in terms of their side.  To know more accurately about sin, cos and tan we must have to know about (1)base, (2)height (perpendicular) and (3)hypotenuse. Figure-1 (right angle triangle)    In general we use base, hypotenuse and height(perpendicular) as shown in figure.  But this method is not   always right.    Let's know what is right (1) Definition of hypotenuse:       The side opposite to the right angle or \(90^{o}\) is called hypotenuse, it is the largest side of right angle triangle.  (2) Definition of perpendicular:   It depends on the angle which we are going to consider. If we consider angle C the  perpendicular  is AB.  If we consider angle A then  perpendicular  is  BC. (See be...
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Four point charges qA=2μC, qB=−5μC, qC=2μC, and qD=−5μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square?

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Question Id:  PH0000007 Question :   Four point charges qA = 2 µC, qB = –5 µC, qC = 2 µC, and qD = –5 µC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1 µC placed at the centre of the square? Answer : zero  Explanation :  According to the given question you can draw the figure look like as given below.   Fig-1: Coulomb's Force at the centre of square Let, side of the square is a unit. Then length of the corner is \(a\sqrt{2}\) . So, the distance froom centre to corner is \(\frac{a}{\sqrt{2}}\) . Given \(2\mu C\) charge will apply foce towards the centre. For both the \(2\mu C\) the applied forces on \(1\mu C\) chrge are equal and opposite to each other. And value of the force is  \(F=\frac{1}{4 \pi \epsilon_0} \times \frac{1 \times 2}{(\frac{a}{\sqrt 2})^{2}} \mu N\) \[\Rightarrow F=\frac{1}{4 \pi \epsilon_0} \times \frac{4}{a^{2}} \mu N\] As the force are opposite and equal so  the resultant...
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