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# Tutor profile: Mary Z.

Mary Z.
Masters in Applied Math with teaching & tutoring experience

## Questions

### Subject:Trigonometry

TutorMe
Question:

A man measures the angle of elevation from a point on the ground to the top of a building and finds it to be 15$$^{\circ}$$. He walks 100ft. closer to the building and finds the angle of elevation from this new point to be 25$$^{\circ}$$. How tall is the building?

Mary Z.

If we were to sketch this problem, we would see that the we have two triangles: Note: not to scale Starting point: Walks 100ft closer: | \ |\ | \ | \ $$h$$ | \ $$h$$ | \ | \ | \ |____$$15^{\circ}$$_\ |_$$25^{\circ}$$ \ $$x$$ $$x-100$$ Height of the building: $$h$$ Distance from the starting point to the building: $$x$$ Distance from the second point to the building: $$x-100$$ We get this from taking the starting point distance, $$x$$, and subtracting the distance walked toward the building. With the information given, we can use tangent for each triangle to give us the equations needed to find the height of the building. (The tangent of an angle is the opposite side over the adjacent side) So we have: $$\tan(15^{\circ})=\frac{h}{x}$$ AND $$\tan(25^{\circ})=\frac{h}{x-100}$$ Now, we solve both of these equations for $$x$$: $$\tan(15^{\circ})=\frac{h}{x}$$ $$x\tan(15^{\circ})=h$$ $$x=\frac{h}{\tan(15^{\circ})}$$ AND $$\tan(25^{\circ})=\frac{h}{x-100}$$ $$(x-100)\tan(25^{\circ})=h$$ $$x-100=\frac{h}{\tan(25^{\circ})}$$ $$x=100+\frac{h}{\tan(25^{\circ})}$$ So we have: $$x=\frac{h}{\tan(15^{\circ})}$$ AND $$x=100+\frac{h}{\tan(25^{\circ})}$$ Since both of these equations are equal to $$x$$, they must also be equal to each other. So we set the two equations equal to each other: $$\frac{h}{\tan(15^{\circ})}=100+\frac{h}{\tan(25^{\circ})}$$ Now, solve for $$h$$: $$\frac{h}{\tan(15^{\circ})}-\frac{h}{\tan(25^{\circ})}=100$$ $$\frac{h\tan(25^{\circ})-h\tan(15^{\circ})}{\tan(15^{\circ})\tan(25^{\circ})}=100$$ $$h\frac{\tan(25^{\circ})-\tan(15^{\circ})}{\tan(15^{\circ})\tan(25^{\circ})}=100$$ $$h=100\frac{\tan(15^{\circ})\tan(25^{\circ})}{\tan(25^{\circ})-\tan(15^{\circ})}$$ $$h=100(0.6299304)=62.993$$ Which means the building is 62.993 ft. tall.

### Subject:Calculus

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Question:

A 12ft. ladder is resting on a wall at a point 10ft. high. The ladder is sliding down the wall at a rate of $$\frac{1}{5}$$ ft./sec. How fast is the bottom of the ladder sliding across the floor 25 seconds after it begins to fall?

Mary Z.

If we were to sketch this problem, we would see that the ladder forms a triangle with the wall and floor with the following parameters: |\ | \ $$y$$ | \ $$12ft$$ | \ |_____ \ $$x$$ Length of the ladder: $$12 ft.$$ Height of the ladder relative to time: $$y(t)=y$$ Distance of the bottom of the ladder from the floor relative to time: $$x(t)=x$$ **NOTE: We are simplifying both functions to make the calculations easier moving forward** We know that the rate that the height of the ladder, $$y$$, is changing is $$\frac{1}{5}$$ ft./sec. And since the derivative of a function is the rate at which it's changing, we know that: $$y'=-\frac{1}{5}$$ Note: Our derivative is negative because the height of the ladder is decreasing. So, we know the rate at which $$y$$ is changing, but we want to determine how fast the position of the bottom of the ladder, $$x$$, is changing. And again, since the rate of change is the derivative, we are looking for $$x'$$. To do this, first we need to find the relationship between our two functions $$x$$ and $$y$$. Since our ladder forms a triangle with the wall and floor, we can use the Pythagorean Theorem. $$x^{2}+y^{2}=12^{2}$$ or $$x^{2}+y^{2}=144$$ Again, since we are solving for the rate of change, we need differentiate both sides, which gives us: $$2xx'+2yy'=0$$ Now we isolate $$x'$$, which gives us: $$x'=\frac{-2yy'}{2x}=\frac{-yy'}{x}$$ We know $$y'$$ but to solve this equation we also need to determine the values $$x$$ and $$y$$ at 25 seconds. Since we know that the initial height of the ladder is 10ft., we can determine how far it moved using the distance formula: $$change$$ $$in$$ $$distance = (rate$$ $$of$$ $$change)*(time)$$ So, $$y=10+(-\frac{1}{5})(25)=10-5=5$$ And plugging this back into our original equation allows us to solve for $$x$$. $$x^{2}+y^{2}=144$$ $$x^{2}+5^{2}=144$$ $$x^{2}+25=144$$ $$x=\sqrt{144-25}=\sqrt{119}$$ Now we have all the information we need to solve for $$x'$$. $$x'=\frac{-yy'}{x}$$ $$x'=\frac{-(5)(-\frac{1}{5})}{\sqrt{119}}=\frac{1}{\sqrt{119}}=0.09167 ft./sec$$ So at 25 seconds after the ladder began falling, the bottom of the ladder is sliding across the floor at a rate of $$0.09167 ft./sec$$.

### Subject:Algebra

TutorMe
Question:

Find the x and y intercepts of the following equation: $$(x-2)^{2}+y^{2}=4$$

Mary Z.

To find the x intercepts, set $$y=0$$ and solve for x: $$(x-2)^{2}+(0)^{2}=4$$ $$(x-2)^{2}=4$$ $$x^{2}-4x+4=4$$ $$x^{2}-4x=0$$ $$x(x-4)=0$$ $$x=0$$ and $$x-4=0$$...so $$x=0$$ and $$x=4$$ The x intercepts are $$(4,0)$$ and $$(0,0)$$ To find the y intercepts, set $$x=0$$ and solve for y: $$(0-2)^{2}+y^{2}=4$$ $$(-2)^{2}+y^{2}=4$$ $$4+y^{2}=4$$ $$y^{2}=0$$ so $$y=0$$ The y intercept is $$(0,0)$$

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