Triangle (Show)

**cloud5** · January 11th 2010, 12:02 AM

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) $4a(a+b)cot^2 \vartheta = (2a+b)^2 cot^2 \beta - b^2 cot^2 \alpha$

**Grandad** · January 11th 2010, 04:35 AM

Hello cloud5

Originally Posted by cloud5

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

**Grandad** · January 11th 2010, 05:00 AM

Hello again cloud5

Originally Posted by cloud5

...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

**cloud5** · January 12th 2010, 02:55 AM

Originally Posted by Grandad

Can you complete it?

Grandad

Can you show the rest of the method, I just wanna double check mine. Thanks a lot!

**Grandad** · January 12th 2010, 08:02 AM

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

**cloud5** · January 14th 2010, 01:30 AM

Originally Posted by Grandad

(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 ?

**Grandad** · January 14th 2010, 05:03 AM

Hello cloud5

Originally Posted by cloud5

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

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