0
$\begingroup$

In the adjoining figure(below)

Figure of the problem

CZ is perpendicular to XY and the ratio of AZ to ZB is $1 : 2$. The angle $ACX$ is $\alpha$ and the angle $BCY$ is $\beta$. Find an expression for the angle $AZC$ in terms of $\alpha$ and $\beta$.

By using the sine rule (in my opinion, which can be used here):

$\dfrac{AZ}{sin(90^{\circ}-\alpha)}=\dfrac{ZC}{sin\angle{ACB}}=\dfrac{AC}{sin\angle{AZC}}$.........(1)

$\dfrac{BZ}{sin(90^{\circ}-\beta)}=\dfrac{ZC}{sin\angle{ABC}}=\dfrac{BC}{sin({\pi-AZC})=sin\angle{AZC}}$..........(2)

Now $\dfrac{AZ}{ZB}=\dfrac{1}{2}$(Given)

$\dfrac{\dfrac{AZ}{ZC}}{\dfrac{ZB}{ZC}}=\dfrac{1}{2}$........(3)

Using $(1),(2) \space and \space (3)$, we get,

$\dfrac{AC}{BC}=\dfrac{sin\angle{ABC}}{sin\angle{CAB}}$= $\dfrac{1}{2}\cdot\dfrac{cos\beta}{cos\alpha}$

Again $\angle{CAB}+\angle{ABC}=\alpha+\beta$(by careful observation)

Again using cosine rule on $\Delta{ACZ}$, we get

$\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}= AZ$

Putting value of AZ above in $(1)$, we get,

$\dfrac{\sqrt{AC^2+ZC^2-2AC\cdot ZC \cdot sin \alpha}}{cos\alpha}=\dfrac{AC}{sin\angle{AZC}}$

But here I am stuck as I think I am going somewhere else. Please give me any suggestion, idea or directly, the answer(if you'd actually do that I'll be obliged)

The probable solution is below:

enter image description here

$\endgroup$
  • $\begingroup$ In the future, please post your attempted solution that you would like checked in the main post, and use images from software or mathjax instead of pictures - these will significantly improve the response you get! $\endgroup$ – B. Mehta Apr 17 '18 at 18:33
0
$\begingroup$

Your solution looks good to me.

$\endgroup$
0
$\begingroup$

Here is the solution for the above question. Please see if there is some error.

enter image description here

In the 3rd line it should be angle $ZCA=90-\alpha$, wrongly written as $ZAC=90-\alpha$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.