1
$\begingroup$

right_triangle_with_altitude

Can't really figure out this problem where I have to find side $RS$, this is covered in my high-school trig curriculum, and it is in the section which deals with all concepts before sine and cosine law, so likely they want as solution using none of cosine or sine law. Nonetheless if nothing works... cosine or sine law solutions would also be accepted :).

I just can't really get any info from here other than the fact that since the right triangles are right, and we know the angles, we can get info about the other triangle angle measures. That's all...

|cite|improve this question
$\endgroup$
  • $\begingroup$ Can you show us some work? $\endgroup$ – Julian Rachman Jan 19 '15 at 3:11
  • $\begingroup$ I can't really show any work. All I know is that interior angles add up to 180, 180 - 90 - 32 = 58, etc. I can't figure out anything else. $\endgroup$ – user164403 Jan 19 '15 at 3:13
  • $\begingroup$ Ok. Give me a second. $\endgroup$ – Julian Rachman Jan 19 '15 at 3:13
  • $\begingroup$ This diagram looks very familiar. I used it in grade 12 (along with the law of sines) to show that (for acute angles) that $$\sin(A+B) = \sin A\cos B + \cos A\sin B$$ $\endgroup$ – John Joy Jan 19 '15 at 17:17
3
$\begingroup$

From what we can see, we get an equation for finding the $50$ to solve for segment $RS$:

$$RS(\tan (26^\circ)+\tan (32^\circ))=50$$ $$RS=\frac{50}{\tan (26^\circ)+\tan (32^\circ)}.$$

Thus, we can then solve for $RS$:

$$RS=\frac{50}{1.11260194...}=\boxed{44.939 \ \text{m}}.$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ don't you need the degree symbol \^circ for the arguments of $\tan?$ $\endgroup$ – abel Jan 19 '15 at 8:14
  • $\begingroup$ @abel OK. I fixed it. $\endgroup$ – Julian Rachman Jan 19 '15 at 8:15
3
$\begingroup$

You have $RS(\tan 26^\circ+\tan 32^\circ)=50$ from the right triangles.

|cite|improve this answer
$\endgroup$
0
$\begingroup$

My approach is a little bit more geometrical:

I'll name the remaining vertex: the upper vertex (the one in the right triangle whose other angle is 32) is U and the only vertex left is T.

Please note that the angle in U is $ 180-90-32= 58$. Which is the same as the angle in S ($32+26= 90$). Therefore the triangle we are dealing with is isosceles with $UT=US$. If we draw the altitude from U it would be the same length as the altitude from S (which is RS), because the triangle is isosceles. If we call X the point where the altitude from U meets the side TS of the triangle, we can simply say that $UX=RS$. But since, $UT=50$ and the angle in T can be calculated by doing the following operation $180-90-26 = 64$, therefore $UX= \sin(64) UT = 50\sin (64)$.

Let me know if I wasn't clear in any way (English isn't my native language).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ The perpendicular segment from a vertex to the opposite side of a triangle is called an altitude. Its length is called the height. $\endgroup$ – N. F. Taussig Jan 19 '15 at 12:42
  • $\begingroup$ It is true that the measure of angle $U$ is $58^\circ$. Thus, $\angle TUS \cong \angle UST$. The isosceles triangle theorem states that the sides of the triangle opposite congruent angles are congruent. Therefore, $UT \cong ST$, so $ST = 50~\text{m}$. It is also true that the measure of angle $T$ is $64^\circ$. From there, you should be able to complete the problem easily. $\endgroup$ – N. F. Taussig Jan 19 '15 at 13:05

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.