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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 \angle PQR > 120^o, where a and d are positive constant, show that  \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 \alpha while the angle of elevation from B and C are both equal to \beta. If \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) \vartheta is the largest angle of elevation,
    ii) 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 \angle PQR > 120^o, where a and d are positive constant, show that  \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 \alpha while the angle of elevation from B and C are both equal to \beta. If \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) \vartheta is the largest angle of elevation,
    ii) 4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha
    Here's the outline for number 1.

    120^o< \angle Q <180^o \Rightarrow -\tfrac12>\cos Q >-1
    \Rightarrow -\tfrac12>\frac{a^2+(a+d)^2-(a+2d)^2}{2a(a+d)}>-1, using the Cosine Rule.
    Expand and simplify:
    \Rightarrow ...
    Factorise the numerator, and cancel.
    ...
    Reverse the direction of the inequality (multiply through by -1).
    ...
    \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 \alpha while the angle of elevation from B and C are both equal to \beta. If \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) \vartheta is the largest angle of elevation,
    ii) 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 P is the foot and T the top of the tower, and that TP = h. Then we have:
    \angle TAP = \alpha

    \angle TBP = \angle TCP = \beta

    \angle TEP = \theta

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

    \Rightarrow PE \perp BC

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

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

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

    \Rightarrow h^2 = \frac{(2a+b)^2}{4(\cot^2\alpha - \cot^2\theta)} (2)
    Eliminate 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

     -\tfrac12>\frac{a^2+(a+d)^2-(a+2d)^2}{2a(a+d)}>-1<br />

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

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

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

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

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

    \Rightarrow a<3d-a<2a

    \Rightarrow 2a<3d<3a

    \Rightarrow \tfrac23a<d<a

    Question 2

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

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

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

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

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

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

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

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

     \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
    =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 \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 \cot for Q2i ?
    Look at the attached diagram.

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

    BP = h\cot\beta

    EP = h\cot\theta

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

    E is therefore the foot of the perpendicular from P to BC

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

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

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