Perimeter of Isoceles Triangle without Base
- Thread starter jceriotti
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mmm4444bot
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Two ways that come to mind are (1) break the triangle into two 30-60-90 right triangles, and (2) the Law of Cosines.
But I do not know the altitude to use the 30-60-90 triangles...we thought of that earlier but are stuck.
This is from class and math counts website. I want to get this but do not know about cosines yet
if i do not have the altitude/height then how can i get to the base
This is from class and math counts website. I want to get this but do not know about cosines yet
if i do not have the altitude/height then how can i get to the base
mmm4444bot
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- Oct 6, 2005
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The right-triangle definition of the cosine function is the ratio of an angle's adjacent side to the hypotenuse.
I was thinking that you're learning trigonometry because you posted on the trigonometry board.
If you want to use an algebraic approach, then how about the concept of similar right-triangles?
Take an equilateral triangle with unit sides, and break it into two 30-60-90 right triangles. The side opposite the 30-degree angle will be one-half unit in length, and the hypotenuse will be one unit. Use the Pythagorean Theorem to find the length of the side opposite the 60-gree angle.
If you enlarge this 30-60-90 triangle by a factor of 10, it will match the 30-60-90 triangle in your exercise.
I was thinking that you're learning trigonometry because you posted on the trigonometry board.
If you want to use an algebraic approach, then how about the concept of similar right-triangles?
Take an equilateral triangle with unit sides, and break it into two 30-60-90 right triangles. The side opposite the 30-degree angle will be one-half unit in length, and the hypotenuse will be one unit. Use the Pythagorean Theorem to find the length of the side opposite the 60-gree angle.
If you enlarge this 30-60-90 triangle by a factor of 10, it will match the 30-60-90 triangle in your exercise.
jceriotti said:
You have an isosceles triangle, with the two sides each equal to 10. The angle formed by these two equal sides is 120, right?
No, you DON'T have the altitude. BUT...you can DRAW the perpendicular from the vertex angle (where sides you called "a" and "c" meet).
That perpendicular bisects the vertex angle...making two 60-degree angles. And that perpendicular is the altitude to the base of the isosceles triangle.
I suspect you are studying 30-60-90 triangles...that perpendicular you drew makes TWO of those.
If you've learned the relationships between the legs and the hypotenuse of a 30-60-90 triangle, you don't NEED to know about cosines.
If you haven't learned any trig yet, and you don't know about 30-60-90 triangles, you need to talk to your teacher.
Divide the isosceles triangle into two right triangles. You obtain two triangles with angles 30, 60, 90. I those triangles the relation between their sides is
always k*5, k*3, k*4. check them up in the formula and figure out which k is for your triangle.
always k*5, k*3, k*4. check them up in the formula and figure out which k is for your triangle.
Hello, jceriotti!
Another approach . . .
\(\displaystyle \text{Divide the isosceles triangle into two congruent right triangles}\)
. . \(\displaystyle \text{with altitude }BD.\)
\(\displaystyle \text{With a 30-60 right triangle, we know the ratio of ths sides.}\)
. . . \(\displaystyle \underbrace{1}_{\text{opp.30}^o} : \underbrace{\sqrt{3}}_{\text{opp.60}^o} : \underbrace{2}_{\text{hyp}}\)
\(\displaystyle \text{Since }hyp = 19\text{, multiply through by 5: }\quad5\::\;5\sqrt{3}\::\:10\)
\(\displaystyle \text{Hence: }AD\text{ (side opp. }60^o) \;=\;5\sqrt{3}\)
\(\displaystyle \text{Therefore: }\;AC \;=\;10\sqrt{3}\)
Another approach . . .
\(\displaystyle \text{Divide the isosceles triangle into two congruent right triangles}\)
. . \(\displaystyle \text{with altitude }BD.\)
Code:
B
o
* : *
10 * 60d : * 10
* : *
* : *
* 30d : *
A o - - - - - - - - o - - - - - - - - o C
D\(\displaystyle \text{With a 30-60 right triangle, we know the ratio of ths sides.}\)
. . . \(\displaystyle \underbrace{1}_{\text{opp.30}^o} : \underbrace{\sqrt{3}}_{\text{opp.60}^o} : \underbrace{2}_{\text{hyp}}\)
\(\displaystyle \text{Since }hyp = 19\text{, multiply through by 5: }\quad5\::\;5\sqrt{3}\::\:10\)
\(\displaystyle \text{Hence: }AD\text{ (side opp. }60^o) \;=\;5\sqrt{3}\)
\(\displaystyle \text{Therefore: }\;AC \;=\;10\sqrt{3}\)
garf said:
I don't think what you've said is correct. Suppose you have a triangle with sides of 5k, 3k and 4k.
Is this a right triangle? You can check that using the converse of the Pythagorean Theorem...
If the sides a, b, and c of a triangle where a and b are both smaller than c satisfy the relationship a[sup:kv0mp788]2[/sup:kv0mp788] + b[sup:kv0mp788]2[/sup:kv0mp788] = c[sup:kv0mp788]2[/sup:kv0mp788], then the triangle is a right triangle.
Ok...you've suggested a triangle with sides 5k, 3k and 4k.
3k and 4k are the two smaller sides. Is it true that
(3k)[sup:kv0mp788]2[/sup:kv0mp788] + (4k)[sup:kv0mp788]2[/sup:kv0mp788] = (5k)[sup:kv0mp788]2[/sup:kv0mp788]?
9k[sup:kv0mp788]2[/sup:kv0mp788] + 16k[sup:kv0mp788]2[/sup:kv0mp788] = 25k[sup:kv0mp788]2[/sup:kv0mp788]
Yep, that's true....a triangle with sides 5k, 3k and 4k IS a right triangle.
BUT...is it a right triangle with angles of 30, 60, and 90?
Pick the angle opposite the leg of length 3k...let's call that A. And let's use B to represent the angle opposite the leg of length 4k. Please draw and label a diagram...the hypotenuse of this triangle should have length 5k.
Now, tan A = opposite leg/adjacent leg
tan A = 3k / 4k
tan A = 3/4
Use your calculator to find the measure of angle A...it ISN'T 30 OR 60.
We appreciate your attempts to help here, and we all have made mistakes from time to time. BUT...it is a good idea to check your approach before posting.
tan A = 3k / 4k