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# Tutor profile: Harry D.

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Harry D.
GCSE and A-Level Maths Tutor, 5 years experience and 800+ lessons taught.
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## Questions

### Subject:Trigonometry

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Question:

Given a right-angled triangle $$ABC$$, where angle $$B = 90^{\circ}$$, angle $$C = 30^{\circ}$$ and side length $$AC = 10$$cm. Find the length of side $$AB.$$

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Harry D.

The first step of any trigonometry question is to label the sides of the triangle $$H$$, $$O$$ and $$A$$ - standing for for hypotenuse, opposite and adjacent. $$AC$$ is the opposite side to the right-angle, hence is the hypotenuse. $$AB$$ is the opposite side to the known angle and hence is the opposite side, leaving $$AC$$ as the adjacent side. Since we know the length of the hypotenuse, $$H = 10$$cm, and want to know the length of the opposite side, $$O = AB$$, we need the trigonometric ratio from SOHCAHTOA that contains $$H$$ and $$O$$. Hence we use the formula corresponding to SOH, which is $(sin(x) = \frac{O}{H}$) where $$x$$ is the known angle. Substituting in the known values, this becomes $(sin(30^{\circ}) = \frac{AB}{10},$) making $$AB$$ the subject, we get $(AB = 10sin(30^{\circ}) = 5.$) Hence the side length $$AB = 5$$cm.

### Subject:Differential Equations

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Question:

Find the general solution of the differential equation $(\frac{dy}{dx} = \frac{1-x}{y+5}.$)

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Harry D.

First, we begin by separating the variables, giving $((y + 5)\frac{dy}{dx} = 1 - x.$)Integrating with respect to $$x$$ on both sides gives $(\int(y+5)\frac{dy}{dx}dx = \int(1-x) dx.$) The left integral becomes $(\int(y+5)dy = \int(1-x) dx$) and integrating both sides gives $(\frac{y^2}{2} + 5y = x - \frac{x^2}{2} + c$) where $$c$$ is the constant of integration. Multiplying by $$2$$ on both sides gives $(y^2 + 10y = 2x - x^2 + k$) where $$k = 2c.$$ Hence the general solution is $(y^2 + 10y + x^2 -2x= k.$)

### Subject:Algebra

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Question:

Solve the quadratic equation $$x^2 + 6x - 10 = 0$$ by completing the square.

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Harry D.

To complete the square, we need to write the equation in the form $$(x + a)^2 + b = 0$$. Expanding this gives $$x^2 + 2ax + a^2 + b = 0$$, by comparing the coefficients of $$x$$ in this equation and the quadratic given in the question, we see that $$2a = 6$$ and hence $$a = 3$$. Comparing the constant terms gives $$a^2 + b = -10$$, giving $$b = -10 - 3^2 = -19$$. So we have that the quadratic equation is equivalent to $$(x+ 3)^2 - 19 = 0$$. Adding $$19$$ on both sides gives $$(x+3)^2 = 19$$. We can now take the square root on both sides. Remembering both the positive and negative root of $$19$$ gives the following solutions: $$x+ 3 = \sqrt{19} \implies x = - 3 + \sqrt{19}$$ or $$x+ 3 = -\sqrt{19} \implies x = - 3 - \sqrt{19}.$$ So the two solutions are $$x = -3 \pm \sqrt{19}.$$

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