# Thread: A question based on the sine & cosine formulas

1. ## A question based on the sine & cosine formulas

The acute triangle $\triangle ABC$ inscribes a circle. D, E, and F are the tangent points. It is given that: $\angle EFD=\alpha$, $\angle FED=\beta$, and AF=a (see picture).
Using $\alpha ,\beta$ and a, express the following:
(a) the length of EF.
(b) the length of DF.

Well, first thing I thought about is that the angle between a tangent and chord is equal to the subtended angle on the opposite side of the chord, hence $\angle BFD=\beta ,$ and $\angle AFE=180-\alpha -\beta$, and since the triangle $\triangle AFE$ is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), $\angle AEF=180-\alpha -\beta$ as well. In addition, $\angle D=180-\alpha -\beta$ and $\angle A=180-2*(180-\alpha- \beta )= -180+2\alpha +2\beta$ (is that correct?)

Does what I did help in solving the question? And what am I supposed to do next? (I must use the sine/cosine formulas)

Thanks.

2. Hello, loui1410!

Everything you did was splendid!

Since $\Delta AFE$ isosceles: . $\angle AEF \:=\:\angle AFE \:=\:180 - \alpha - \beta$

We case use the Law of Sines on $\Delta AFE$

. . $\frac{EF}{\sin A} \:=\:\frac{a}{\sin(\angle AEF)}\quad\Rightarrow\quad EF \:=\:\frac{a\cdot\sin(2\alpha + 2\beta - 180)}{\sin(180 - \alpha - \beta)}$

3. Hmm, then how can I reach the textbook's answer:
$EF=-2 * \alpha * \cos ( \alpha + \beta )$
?

4. Originally Posted by loui1410
The acute triangle $\triangle ABC$ inscribes a circle. D, E, and F are the tangent points. It is given that: $\angle EFD=\alpha$, $\angle FED=\beta$, and AF=a (see picture).
Using $\alpha ,\beta$ and a, express the following:
(a) the length of EF.
(b) the length of DF.

Well, first thing I thought about is that the angle between a tangent and chord is equal to the subtended angle on the opposite side of the chord, hence $\angle BFD=\beta ,$ and $\angle AFE=180-\alpha -\beta$, and since the triangle $\triangle AFE$ is an isosceles triangle (tangents drawn from a point outside the circle are equal in length), $\angle AEF=180-\alpha -\beta$ as well. In addition, $\angle D=180-\alpha -\beta$ and $\angle A=180-2*(180-\alpha- \beta )= -180+2\alpha +2\beta$ (is that correct?)

Does what I did help in solving the question? And what am I supposed to do next? (I must use the sine/cosine formulas)

Thanks.

a)

$\frac {EF}{sin\angle EAF}=\frac {a}{sin\angle AEF}$
$\frac {EF}{sin(2\alpha +2\beta -180^o)}=\frac {a}{sin(180^o -\alpha -\beta)}$
$EF=\frac {asin(2\alpha +2\beta -180^o)}{sin(180^o -\alpha -\beta)}$
$EF=\frac {a[sin2(\alpha+\beta)cos180^o -sin180^ocos2(\alpha+\beta)]}{sin180^o cos(\alpha +\beta)-sin(\alpha +\beta)cos180^o}$
$EF=\frac {a[-sin2(\alpha+\beta)]}{sin(\alpha +\beta)}$
$EF=\frac {-a[sin2(\alpha+\beta)]}{sin(\alpha +\beta)}$
$EF=\frac {-a[2sin(\alpha+\beta)cos(\alpha+\beta)]}{sin(\alpha +\beta)}$
$EF=-2acos(\alpha+\beta)$

b)

$\frac {DF}{sin \beta}=\frac {EF}{sin(2\alpha +2\beta -180^o)}$
$DF=\frac {-2acos(\alpha+\beta)sin\beta}{sin(2\alpha +2\beta -180^o)}$
$DF=\frac {-2acos(\alpha+\beta)sin\beta}{-2sin(\alpha +\beta)cos(\alpha +\beta)}$
$DF=\frac {asin\beta}{sin(\alpha +\beta)}$

The end.

5. Thanks a lot SengNee. The teacher said today that there is still a phase of abstraction that we haven't learnt yet and therefore we don't have to get the same answer as the textbook's, so this is why I didn't know how to proceed.

6. Hello, loui1410!

Then how can I reach the textbook's answer: . $EF\:=\:-2a\cdot\cos(\alpha + \beta)$ ?
We need a few identities . . .

. . $\sin(-\theta)\:=\:-\sin\theta$

. . $\sin(180^o - \theta) \:=\:\sin\theta$

. . $\sin2\theta \:=\:2\sin\theta\cos\theta$

We had: . $EF \;=\;\frac{a\cdot\sin(2\alpha + 2\beta - 180)}{\sin(180-\alpha-\beta)}$ .[1]

The numerator is: . $a\cdot\sin\bigg(-[180 - 2(\alpha + \beta)]\bigg) \;=\;-a\cdot\sin[180 - 2(\alpha + \beta)]$

. . $= \;-a\cdot\sin[2(\alpha + \beta)] \;=\;-2a\cdot\sin(\alpha+\beta)\cdot\cos(\alpha+\beta)$

The denominator is: . $\sin[180 - (\alpha+\beta)] \;=\;\sin(\alpha+\beta)$

Then [1] becomes: . $EF \;=\;\frac{-2a\cdot\sin(\alpha+\beta)\cdot\cos(\alpha+\beta)}{ \sin(\alpha+\beta)} \;=\;-2a\cdot\cos(\alpha+\beta)$