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# Tutor profile: Murali Krishna Varma N.

Murali Krishna Varma N.
Have been tutoring for more than 15 years, working as a Tutor Quality Manager.

## Questions

### Subject:Pre-Calculus

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Question:

Solve the triangle ABC if a=2, b=6, and A = 30 degrees.

Murali Krishna Varma N.

Every triangle ABC has six elements: 1) Three angles: <A, <B, and <C 2) The sides opposite to them: a, b, and c. By solving a triangle, we mean finding the unknown elements of the triangle. It is possible only in the following cases: 1) The three sides are given. 2) Two sides and the included angle are given. 3) Two sides and the angle opposite to one of them given. 4) One side and two angles are given. In our problem, two sides 'a' and 'b' and the angle 'A' which is opposite to the side 'a' are given. So, this is the case (3) and in which we use the law of sines. That is, sin(A)/a = sin(B)/b = sin(C)/c. From, sin(A)/a = sin(B)/b, we get sin(B)= b*sin(A)/a. Now, we might have three possibilities: 1) sin(B)>1 2) sin(B)=1 3) 0<sin(B)<1 Note that 0<=sin(B)<=1 for any angle B, and sin(B) is not equal to zero as b>0. Also, sin(B)>0 when B is in the first two quadrants. So, 1) sin(B)>1 implies existence of no such triangle, 2) sin(B)=1 results in a unique right-triangle, right angled at B, and 3) 0<sin(B)<1 results two triangles. If we can find the angle B, we can easily find the values of C and c. m <C = 180-(m<A + m <B). Now, sin(B) = b*sin(A)/a = 6*sin(30)/2 = 6*(1/2)/2 = 3/2 = 1.5>1. This result shows that there is no such triangle.

### Subject:Geometry

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Question:

A regular hexagon with a perimeter of 24 inches is inscribed in a circle. How far from the center is each side?

Murali Krishna Varma N.

A regular hexagon has six sides with equal length. Let the length of each side be x inches. Then, its perimeter would be 6x inches. So, we have 6x= 24 so that x= 24/6 = 4 inches. That is, each side of the hexagon is of length 4 inches. Each side of the hexagon subtends an angle of 360/6 = 60 degrees at the center. By choosing one side of the hexagon and joining its endpoints (vertices) with the center, we form an isosceles triangle with vertex angle 60 degrees. Let the altitude of this triangle be h inches. This altitude bisects the vertex angle and the base of the isosceles triangle and hence bisects the triangle into two 30-60-90 special triangles. Since the base is 4 inches, half of it would be 2 inches. So, in a 30-60-90 triangle, the side opposite to 30 degrees is of length 2 inches, and the side opposite to 60 degrees is of length h inches. The ratio of the sides of a 30-60-90 triangle is 1: sqrt(3):2. So, 2/h= 1/sqrt(3) and hence h= 2sqrt(3) inches. That is, the center is located at a distance of 2*sqrt(3) inches from each side of the given regular hexagon.

### Subject:Calculus

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Question:

Does the series 1+ (1/2) +(1/3)+(1/4)+......converge?

Murali Krishna Varma N.

This is one of the basic results in this concept that we can use to check the convergence of other series. Here, we use the Monotone Convergence Theorem. It states: The sequence {s(n)} is convergent if it is EITHER 1. Increasing and 2. Bounded above OR 1. Decreasing and 2. Bounded below. Now, let us consider nth partial sums s(n) of the terms of the given series: 1+(1/2)+(1/3)+(1/4)+..... s(1) = 1 s(2) = 1+(1/2) s(4) = 1+ 1/2 + (1/3 + 1/4) = 1+ 1/2 +7/12> 1+ 1/2 + 1/2=1+(1/2)(2) (Note: 2^2 = 4) s(8) = 1+1/2 + (1/3 + 1`/4) + (1/5+1/6+1/7+1/8)> 1+ 1/2 + 1/2 + 1/2 = 1 + (1/2)(3) (Note: 2^3=8) ------- s(2^k) > 1 + (1/2 + 1/2 + .....(k times)) = 1+ (1/2)k From this, we can see that the sequence {s(n)} of partial sums is increasing and not bounded above. Therefore, it cannot be convergent by the Monotone Convergence Theorem and hence it is divergent. Since the sequence of partial sums is divergent, the given series is also divergent.

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