Trigonometry problem

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chijioke

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Good morning 🌅 everyone.IMG-20260301-WA0008.jpgIMG_20260301_063732.jpg
I am trying to find the marked sides. Please help me confirm if my solutions are true also if solved problem well.
 
That looks correct. There are, of course, several other ways you could have solved it.

Is there a reason you were not sure?
 
1 think I used the cosine ratios wrongly. It should be CAH and not CHA when I was trying to compute for m in the first work I posted. But by main looking m should be more than y, not the same as y because that would be impossible in real life.
 
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chijioke said:
1 think I used the cosine ratios wrongly. It should be CAH and not CHA when I was trying to compute for m in the first work I posted. But by main looking m should be more than y, not the same as y because that would be impossible in real life.
Click to expand...

The first work still looks right to me. In calculating m, you used cos(60) = d/m = 6/m, which is correct; that is adjacent over hypotenuse, as you say. And you did find that m was greater than y, namely 12 as compared to 3. Why do you say you did that wrong?

As for the new work, there you say cos(60) = m/d = m/6, which is wrong. Obviously y = 0 is wrong. Are your comments really about this rather than the original?
 
Dr.Peterson said:
The first work still looks right to me. In calculating m, you used cos(60) = d/m = 6/m, which is correct; that is adjacent over hypotenuse, as you say. And you did find that m was greater than y, namely 12 as compared to 3. Why do you say you did that wrong?

As for the new work, there you say cos(60) = m/d = m/6, which is wrong. Obviously y = 0 is wrong. Are your comments really about this rather than the original?
Click to expand...
triangle.jpg
I said it is wrong because when I am making reference to triangle ABC, I am seeing the hypotenuse as d because the right angle the altitude |AR| makes with |RC| at point R. I am then seeing the adjacent as |BC| =m and the opposite as |BA| because angle 60 degrees faces |BA|.
Please could you correct my thinking as to what is the hypotenuse, adjacent and opposite sides respectively, when making reference to triangle ABC?
 
chijioke said:
I said it is wrong because when I am making reference to triangle ABC, I am seeing the hypotenuse as d because the right angle the altitude |AR| makes with |RC| at point R. I am then seeing the adjacent as |BC| =m and the opposite as |BA| because angle 60 degrees faces |BA|.
Please could you correct my thinking as to what is the hypotenuse, adjacent and opposite sides respectively, when making reference to triangle ABC?
Click to expand...

Surely you know that the hypotenuse is opposite the right angle, and that in triangle ABC, the right angle is at A. Point R is irrelevant to that triangle, because that is not a vertex of ABC.

It may help if you redraw the figure to isolate one triangle at a time. To find x, you used only triangle ARC, with right angle at R:

1772456419323.png

But to find m, you are using triangle ABC, with right angle at A:

1772456597295.png

Do you see now that m is the hypotenuse?
 
Dr.Peterson said:
Surely you know that the hypotenuse is opposite the right angle, and that in triangle ABC, the right angle is at A. Point R is irrelevant to that triangle, because that is not a vertex of ABC.

It may help if you redraw the figure to isolate one triangle at a time. To find x, you used only triangle ARC, with right angle at R:

View attachment 39967

But to find m, you are using triangle ABC, with right angle at A:

View attachment 39968

Do you see now that m is the hypotenuse?
Click to expand...
Yes I fully understand now? But I still want to ask. How will one know that angle BAC is 90 degree? I am asking this question because I know that the hypotenuse m should face M that is side BC.
 
chijioke said:
Yes I fully understand now? But I still want to ask. How will one know that angle BAC is 90 degree? I am asking this question because I know that the hypotenuse m should face M that is side BC.
Click to expand...
You know that ∠BAC is 90° because it is marked as such (in the original drawing).
 
chijioke said:
You mean that sign I indicated with green arrow is a sign of a right angle?
Click to expand...
Yes, that is the symbol for a right angle1772540874284.png

and, although the altitude appears to bisect the apex, it clearly does not (as seen by the angles I have now added in).

Furthermore, although this looks like an isosceles triangle, again it is not; you have to interpret the diagram by the given parameters, not by how it looks (as with any "sketch").
 
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The Highlander said:
Yes, that is the symbol for a right angleView attachment 39979

and, although the altitude appears to bisect the apex, it clearly does not (as seen by the angles I have now added in).

Furthermore, although this looks like an isosceles triangle, again it is not; you have to interpret the diagram by the given parameters, not by how it looks (as with any "sketch").
Click to expand...
I am now very cleared. Thank you so much.
 
With the sketch marked up the way I have done above, you might then see a more direct (& elegant?) way to determine the lengths of x & y. For example....

Using exact values for the trig ratios: \(\displaystyle tan\, 60°=\sqrt{3} \text{ and } tan\,30°=\frac{1}{\sqrt{3}}\)

From the diagram...

\(\displaystyle tan\,60°=\frac{x}{3}\implies x=3\times tan\,60°=\underline{\underline{3\sqrt{3}}}\quad(\approx 5.196)\)

\(\displaystyle tan\,30°=\frac{x}{y}\implies y=\frac{x}{tan\,30°}=\frac{3\sqrt{3}}{\frac{1}{\sqrt{3}}}=3\times\sqrt{3}\times\sqrt{3}=3\times3=\underline{\underline{9}}\)

Therefore: x = 3\(\displaystyle \sqrt{3}\) cm (≈ 5.196 cm) and y = 9 cm.

Hope that helps. 😊
 
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