three trigo problems

**afeasfaerw23231233** · January 17th 2008, 05:15 AM

**Soroban** · January 17th 2008, 10:57 AM

Hello, afeasfaerw232312331

22) Prove that in $\Delta ABC$ :

(a) $\cot\frac{A}{2} \:=\:\tan\left(\frac{B}{2}+\frac{C}{2}\right)\text {, by using the identity: }\cot x \:=\:\tan(90^o-x)$

(b) Hence show that: . $\cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} \;=\;\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\ frac{C}{2}$

You solved part (a) . . . Nice work!

(b) We have: . $\cot\frac{A}{2} \;\;=\:\;\tan\left(\frac{B}{2} \:+ \:\frac{C}{2}\right) \quad\Rightarrow\quad\frac{1}{\tan\frac{A}{2}} \;\;=\;\;\frac{\tan\frac{B}{2} \:+ \:\tan\frac{C}{2}}{1-\tan\frac{B}{2}\cdot\tan\frac{C}{2}}$

. . . . $1 \:- \:\tan\frac{B}{2}\cdot\tan\frac{C}{2} \;\;=\;\;\tan\frac{A}{2}\cdot\tan\frac{B}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{C}{2}$

. . . . $\tan\frac{B}{2}\cdot\tan\frac{C}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{C}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{B}{2} \;\;=\;\;1$

. . . . $\frac{1}{\cot\frac{B}{2}\cdot\cot\frac{C}{2}} \:+ \:\frac{1}{\cot\frac{A}{2}\cdot\cot\frac{C}{2}} \:+ \:\frac{1}{\cot\frac{A}{2}\cdot\cot\frac{B}{2}} \;\;=\;\;1$

Multiply through by: . $\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{ C}{2}$

. . . . $\cot\frac{A}{2} \:+ \:\cot\frac{B}{2} \:+ \:\cot\frac{C}{2} \;\;=\;\;\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\ cot\frac{C}{2}$

**Soroban** · January 17th 2008, 01:33 PM

Hello again, afeasfaerw23231233!

26. $\Delta ABC$ is equilateral with $AB = a.$

$M$ is a point outside the triangle such that: . $\angle AMB = 20^o,\;\angle AMC = 30^o,\;\angle BAM = \alpha.$

(a)(i) By considering $\Delta ABM$ , express $AM$ in terms of $a\text{ and }\alpha.$

Your answer is correct, but I would write it this way:

. $AM \:=\:\frac{a\cdot\sin[180^o -(20^o+\alpha)]}{\sin20^o} \quad\Rightarrow\quad AM\;=\; \frac{a\cdot\sin(20+\alpha)}{\sin20^o}$ .[1]

(a)(ii) By considering $\Delta ACM$ , express $AM$ in terms of $a\text{ and }\alpha.$

Since $\angle = 60^o,\;\angle MAC \:=\:60^o-\alpha$
.Then: . $\angle MCA \:=\:180^o - 30^o - (60^o-\alpha) \;=\;90^o + \alpha$

We have: . $\frac{AM}{\sin(90+\alpha)} \:=\:\frac{a}{\sin30^o} \quad\Rightarrow\quad AM \;=\;\frac{a\cdot\sin(90^o+\alpha)}{\frac{1}{2}}$

. . $AM \;=\;2a\cdot\sin(90^o+\alpha)\quad\Rightarrow\quad AM \;=\;2a\cdot\cos\alpha$ .[2]

(b) Hence, find $\alpha.$

Equate [1] and [2]: . $\frac{a\cdot\sin(20^o + \alpha)}{\sin20^o} \;=\;2a\cdot\cos\alpha$

. . . . . . $\sin(20^o+\alpha) \;=\;2\cdot\sin20^o\cos\alpha$

. . $\overbrace{\sin20^o\cos\alpha + \sin\alpha\cos20^o} \;=\;2\cdot\sin20^o\cos\alpha$

. . $\underbrace{\sin\alpha\cos20^o - \sin20^o\cos\alpha} \;=\;0$

. . . . . . $\sin(\alpha-20^o) \;=\;0$

. . . . . . . $\alpha - 20^o \;=\;0^o$

. . . . . . . . $\alpha \:=\:20^o$

**afeasfaerw23231233** · January 27th 2008, 11:13 PM

i thought that the the diagram was a 3D pyramid-like solid! so i spent a great deal of time and could not do [a][ii]. now i know it is a 2D diagram. thanks

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