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# Tutor profile: Raffa M.

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Raffa M.
Tutor for more than 10 years (private and at University)
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## Questions

### Subject:Trigonometry

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Question:

Consider a triangle with the sides $$a$$, $$b$$ and $$c$$ and the angles $$\alpha$$, $$\beta$$ and $$\gamma$$ where $$\alpha$$ is the angle opposite of $$a$$ etc. We know that $$b = 5$$ $$cm$$, $$\beta = 90°$$ and $$\alpha = 36,87°$$ Calculate $$a$$, $$c$$, $$\gamma$$ and the area $$A$$ of the triangle.

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Raffa M.

To calculate $$\gamma$$ we use that for each triangle the sum of all angles equals $$180°$$. Therefore we know that $$\alpha + \beta + \gamma = 180°$$, so that $$\gamma = 90° - 36,87° = 53,13°$$ Since $$\beta = 90°$$ we have a right triangle. To calculate $$c$$ we use that $$\cos(\alpha) = \frac{c}{b}$$. Since we know $$\alpha$$ and $$b$$ we get $$c = \cos(\alpha)b = 0,8 * b = 4$$ $$cm$$ To calculate $$a$$ we can either use the angles again or the Pythagorean theorem (because we have a right triangle). Since $$b$$ is the side opposite to the right angle it is the hypotenuse in our triangle, so $$b^2 = a^2 + c^2$$. That means $$a^2 = \sqrt{b^2 - c^2} = \sqrt{25 - 16} = 3$$ $$cm$$. The formula for the area of a triangle is $$A = \frac{1}{2}gh$$ where $$g$$ is a side of the triangle and $$h$$ is the height of the triangle, perpendicular to $$g$$. In our example we can use $$a$$ as the height since it is perpendicular to $$c$$. We get for the area $$A = \frac{1}{2}ca = \frac{1}{2}4*3 = 6$$ $$cm^2$$

### Subject:Basic Math

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Question:

Let $$a$$, $$b$$ and $$c$$ all be different from $$0$$. Solve for $$a$$: $$4a + 12,5b^2 = \frac{14}{3}ac + abc$$ What condition must be fulfilled for $$a$$ to have a solution?

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Raffa M.

At first we use equivalent transformations to bring all the terms with $$a$$ on one side: $$12,5b^2 = \frac{14}{3}ac + abc - 4a$$ Ultimately to solve the equation for a we need to have a result that looks like $$a = ...$$. All the terms on the right side of the equation contain $$a$$ but $$a$$ is not standing by itself. To get that we need to bracket $$a$$ out of the terms on the right side, so the right side will look like $$a(...+...+...)$$. In general to bracket $$x$$ out looks like that: $$xy + xy = x(y + z)$$. Doing that on the right side of our equation we get: $$12,5b^2 = a(\frac{14}{3}c + bc - 4)$$ Now the right side simply is $$a$$ times a number depending on $$b$$ and $$c$$. To have a standing alone we divide both sides by $$(\frac{14}{3}c + bc - 4)$$. Doing that we get $$a = \frac{12,5b^2}{(\frac{14}{3}c + bc - 4)}$$. For $$a$$ to have a solution the denominator of the fraction $$\frac{12,5b^2}{(\frac{14}{3}c + bc - 4)}$$ must be different from zero. If we look at $$(\frac{14}{3}c + bc - 4) = 0$$ and add $$4$$ on both sides to have an expression for $$b$$ and $$c$$ we get $$\frac{14}{3}c + bc) = 4$$. That means the denominator is different from zero if $$\frac{14}{3}c + bc) \neq 4$$.

### Subject:Calculus

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Question:

What is the derivative of $$f(x) = \cos(x)\sin(x) + e^{\sin(x)}$$ ?

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Raffa M.

To determine $$f'(x)$$ we use the sum rule for $$f(x) = g(x) + h(x)$$, so $$f'(x) = g'(x) + h'(x)$$ and part the problem into two: At first we take a look at $$g(x) = \cos(x)\sin(x)$$. Here we have a product of two functions which means that we need to use the product rule to derive $$g(x)$$. As a reminder, the product rule says $$(u(x)v(x))' = u'(x)v(x) + u(x)v'(x)$$. We know that $$\cos'(x) = -\sin(x)$$ and $$\sin'(x) = \cos(x)$$. Therefore using the product rule we get $$g'(x) = \cos^2(x) - \sin^2(x)$$. Now we take a look at $$h(x) = e^{\sin(x)}$$. Here we have a composite function which means that we need to use the chain rule to derive it. The chain rule says: $$(u(v(x))' = v'(x)u'(v(x))$$. That means $$v(x)$$ is the inner function and $$u(x)$$ is the outer function. In our example $$e^x$$ is the outer function and $$\sin(x)$$ is the inner function. We know that $${e^x}' = e^x$$ and $$\sin'(x) = \cos(x)$$. Using the chain rule we get $$h'(x) = \cos(x)e^{\sin(x)}$$. Now using $$f'(x) = g'(x) + h'(x)$$ we get $$f'(x) = \cos^2(x) - \sin^2(x) + \cos(x)e^{\sin(x)}$$.

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