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  1. #1
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    Triangle (Show)

    Anyone can show me the method?

    1. In a triangle PQR, PQ=a, QR=a+d, RP=a+2d and $\displaystyle \angle PQR > 120^o$, where a and d are positive constant, show that $\displaystyle \frac{2}{3}a<d<a$

    2. A man observed the top of a tower from the points A, B and C along a straight, horizontal road, where AB=a and BC=b. The angle of elevation of the top of the tower from A is $\displaystyle \alpha$ while the angle of elevation from B and C are both equal to $\displaystyle \beta$. If $\displaystyle \vartheta$ is the angle of elevation of the top of the tower from the point E wich is the mid-point of BC, show that

    i)$\displaystyle \vartheta$ is the largest angle of elevation,
    ii) $\displaystyle 4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha$
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  2. #2
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    Hello cloud5
    Quote Originally Posted by cloud5 View Post
    Anyone can show me the method?

    1. In a triangle PQR, PQ=a, QR=a+d, RP=a+2d and $\displaystyle \angle PQR > 120^o$, where a and d are positive constant, show that $\displaystyle \frac{2}{3}a<d<a$

    2. A man observed the top of a tower from the points A, B and C along a straight, horizontal road, where AB=a and BC=b. The angle of elevation of the top of the tower from A is $\displaystyle \alpha$ while the angle of elevation from B and C are both equal to $\displaystyle \beta$. If $\displaystyle \vartheta$ is the angle of elevation of the top of the tower from the point E wich is the mid-point of BC, show that

    i)$\displaystyle \vartheta$ is the largest angle of elevation,
    ii) $\displaystyle 4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha$
    Here's the outline for number 1.

    $\displaystyle 120^o< \angle Q <180^o \Rightarrow -\tfrac12>\cos Q >-1$
    $\displaystyle \Rightarrow -\tfrac12>\frac{a^2+(a+d)^2-(a+2d)^2}{2a(a+d)}>-1$, using the Cosine Rule.
    Expand and simplify:
    $\displaystyle \Rightarrow ...$
    Factorise the numerator, and cancel.
    $\displaystyle ...$
    Reverse the direction of the inequality (multiply through by -1).
    $\displaystyle ...$
    $\displaystyle \Rightarrow \tfrac23a<d<a$
    Can you complete it?

    Grandad
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  3. #3
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    Hello again cloud5
    Quote Originally Posted by cloud5 View Post
    ...2. A man observed the top of a tower from the points A, B and C along a straight, horizontal road, where AB=a and BC=b. The angle of elevation of the top of the tower from A is $\displaystyle \alpha$ while the angle of elevation from B and C are both equal to $\displaystyle \beta$. If $\displaystyle \vartheta$ is the angle of elevation of the top of the tower from the point E wich is the mid-point of BC, show that

    i)$\displaystyle \vartheta$ is the largest angle of elevation,
    ii) $\displaystyle 4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha$
    ... and here's the outline for number 2.

    Suppose that $\displaystyle P$ is the foot and $\displaystyle T$ the top of the tower, and that $\displaystyle TP = h$. Then we have:
    $\displaystyle \angle TAP = \alpha$

    $\displaystyle \angle TBP = \angle TCP = \beta$

    $\displaystyle \angle TEP = \theta$

    $\displaystyle AB = a, BE = EC = \tfrac12b$
    For part (i): $\displaystyle PB = h\cot\beta = PC$
    $\displaystyle \Rightarrow \triangle BPC$ is isosceles

    $\displaystyle \Rightarrow PE \perp BC$

    $\displaystyle \Rightarrow PE$ is the shortest distance from the foot of the tower to $\displaystyle BC$.

    So?
    (ii) $\displaystyle PE^2 = h^2\cot^2\theta = PB^2-BE^2$
    $\displaystyle \Rightarrow ...$

    $\displaystyle \Rightarrow h^2 = \frac{b^2}{4(\tan^2\beta-\cot^2\theta)}$ (1)
    $\displaystyle AP^2 = h^2\cot^2\alpha = PE^2+AE^2$
    $\displaystyle \Rightarrow ...$

    $\displaystyle \Rightarrow h^2 = \frac{(2a+b)^2}{4(\cot^2\alpha - \cot^2\theta)}$ (2)
    Eliminate $\displaystyle h$ from (1) and (2) and the result follows.

    Can you fill in the gaps?

    Grandad
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  4. #4
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    Quote Originally Posted by Grandad View Post
    Can you complete it?

    Grandad
    Can you show the rest of the method, I just wanna double check mine. Thanks a lot!
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  5. #5
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    Hello cloud5

    Question 1

    $\displaystyle -\tfrac12>\frac{a^2+(a+d)^2-(a+2d)^2}{2a(a+d)}>-1
    $

    $\displaystyle \Rightarrow-\tfrac12>\frac{a^2+(a^2+2ad+d^2)-(a^2+4ad+4d^2)}{2a(a+d)}>-1$

    $\displaystyle \Rightarrow-\tfrac12>\frac{a^2-2ad-3d^2}{2a(a+d)}>-1$

    $\displaystyle \Rightarrow-\tfrac12>\frac{(a-3d)(a+d)}{2a(a+d)}>-1$

    $\displaystyle \Rightarrow-\tfrac12>\frac{a-3d}{2a}>-1$

    $\displaystyle \Rightarrow\tfrac12<\frac{3d-a}{2a}<1$

    $\displaystyle \Rightarrow a<3d-a<2a$

    $\displaystyle \Rightarrow 2a<3d<3a$

    $\displaystyle \Rightarrow \tfrac23a<d<a$

    Question 2

    (ii) $\displaystyle PE^2 = h^2\cot^2\theta =PB^2-BE^2$
    $\displaystyle =h^2\cot^2\beta-\tfrac14b^2$
    $\displaystyle \Rightarrow h^2(\cot^2\beta-\cot^2\theta) = \tfrac14b^2$

    $\displaystyle \Rightarrow h^2=\frac{b^2}{4(\cot^2\beta-\cot^2\theta)}$ (1) (Note the typo in my previous post: not $\displaystyle \tan^2\beta$.)

    Also $\displaystyle AP^2 = h^2\cot^2\alpha = PE^2+AE^2$
    $\displaystyle =h^2\cot^2\theta+(a+\tfrac12b)^2$
    $\displaystyle \Rightarrow h^2(\cot^2\alpha-\cot^2\theta) = (a+\tfrac12b)^2$

    $\displaystyle \Rightarrow h^2=\frac{(2a+b)^2}{4(\cot^2\alpha-\cot^2\theta)}$
    $\displaystyle =\frac{b^2}{4(\cot^2\beta-\cot^2\theta)}$ from (1)
    $\displaystyle \Rightarrow (2a+b)^2(\cot^2\beta-\cot^2\theta)=b^2(\cot^2\alpha-\cot^2\theta)$

    $\displaystyle \Rightarrow (2a+b)^2\cot^2\beta-b^2\cot^2\alpha = (4a^2+4ab+b^2)\cot^2\theta - b^2\cot^2\theta$
    $\displaystyle =4a(a+b)\cot^2\theta$
    Grandad
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  6. #6
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    Quote Originally Posted by Grandad View Post


    (ii) $\displaystyle PE^2 = h^2\cot^2\theta =PB^2-BE^2$
    $\displaystyle =h^2\cot^2\beta-\tfrac14b^2$
    $\displaystyle \Rightarrow h^2(\cot^2\beta-\cot^2\theta) = \tfrac14b^2$

    $\displaystyle \Rightarrow h^2=\frac{b^2}{4(\cot^2\beta-\cot^2\theta)}$ (1) (Note the typo in my previous post: not $\displaystyle \tan^2\beta$.)

    Also $\displaystyle AP^2 = h^2\cot^2\alpha = PE^2+AE^2$
    $\displaystyle =h^2\cot^2\theta+(a+\tfrac12b)^2$
    $\displaystyle \Rightarrow h^2(\cot^2\alpha-\cot^2\theta) = (a+\tfrac12b)^2$

    $\displaystyle \Rightarrow h^2=\frac{(2a+b)^2}{4(\cot^2\alpha-\cot^2\theta)}$
    $\displaystyle =\frac{b^2}{4(\cot^2\beta-\cot^2\theta)}$ from (1)
    $\displaystyle \Rightarrow (2a+b)^2(\cot^2\beta-\cot^2\theta)=b^2(\cot^2\alpha-\cot^2\theta)$

    $\displaystyle \Rightarrow (2a+b)^2\cot^2\beta-b^2\cot^2\alpha = (4a^2+4ab+b^2)\cot^2\theta - b^2\cot^2\theta$
    $\displaystyle =4a(a+b)\cot^2\theta$
    Grandad
    Would you mind showing the sketch of the triangle for Q2? Can you explain why you're using $\displaystyle \cot$ for Q2i ?
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  7. #7
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    Hello cloud5
    Quote Originally Posted by cloud5 View Post
    Would you mind showing the sketch of the triangle for Q2? Can you explain why you're using $\displaystyle \cot$ for Q2i ?
    Look at the attached diagram.

    $\displaystyle AP, BP, EP, CP$ are all horizontal lines, and therefore are perpendicular to $\displaystyle PT$. So from the right-angled triangles, we get:
    $\displaystyle AP = h\cot\alpha$

    $\displaystyle BP = h\cot\beta$

    $\displaystyle EP = h\cot\theta$

    $\displaystyle CP = h \cot\beta$
    Thus $\displaystyle BP = PC$, and $\displaystyle \triangle BPC$ is therefore isosceles, with $\displaystyle E$ the mid-point of $\displaystyle BC$. Hence:
    $\displaystyle \angle PEB = 90^o$

    $\displaystyle E$ is therefore the foot of the perpendicular from $\displaystyle P$ to $\displaystyle BC$

    Therefore $\displaystyle PE$ is the shortest distance from $\displaystyle P$ to $\displaystyle BC$, and so:

    $\displaystyle \theta$ is therefore the largest angle of elevation of $\displaystyle T$ from the line $\displaystyle BC$
    The remaining working uses Pythagoras' Theorem on the right-angled triangles $\displaystyle APE$ and $\displaystyle BPE$.

    Grandad
    Attached Thumbnails Attached Thumbnails Triangle (Show)-untitled.jpg  
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