Up next in 10
Geometry Olympiad Problem l Solve for the Missing Side of a Right Triangle
0 views
Jun 28, 2025
In this video, I tackle an exciting Geometry Olympiad problem. We are given a right triangle with a hypotenuse of four units and one angle measuring fifty-four degrees. Watch as I solve for the missing side using trigonometric methods like cosine. Try solving it on your own before watching the full solution! Don’t forget to like, comment, and subscribe for more challenging math problems.
View Video Transcript
0:00
the hypotenuse of a right triangle is
0:02
equal to 4 and one of its angles is
0:06
54° find the leg opposite this angle as
0:10
usual before watching further try to
0:12
solve this problem on your own the sign
0:15
of the acute angle in a right triangle
0:17
in our case the sign of 54° is equal to
0:20
the ratio of the opposite leg which is
0:22
what we need to find to the hypotenuse
0:25
which according to the problem is four
0:28
next we can conveniently switch to the
0:30
coine of
0:31
36° as 54° is what remains to reach 90°
0:36
from
0:36
36° According to the reduction formula
0:39
the S of 54° can be replaced by the cine
0:42
of
0:43
36° thus the cine of 36° will be equal
0:46
to x / 4 from which the leg we need will
0:50
be equal to 4 * the cine of
0:53
36° now let's try to find the coine of
0:57
36° separately notice that if we take
1:00
Take 36° 5 times we get
1:02
180° let's split the left side of the
1:05
resulting equation into two terms
1:07
meaning we take 36° three times and add
1:11
36° two times and as a result we get the
1:15
same
1:17
180° Express 3 * 36° as 180° - 2 * 36°
1:23
now let's find the coine of the triple
1:26
angle the coine of 3 * 36° will be equal
1:30
to the cine of 180° - 2 * 36° by the
1:35
reduction formula the cine of 180° -
1:39
Alpha is equal to netive cine of alpha
1:42
thus we get Negative coine of 2 * 36° so
1:45
the cine of the triple angle 3 * 36°
1:49
should be equal to negative coine of the
1:52
double angle 2 *
1:54
36° but remember we need to find the
1:57
coine of 36° on the left side of the
2:00
equation we have the cine of the triple
2:02
angle which by the standard formula
2:05
equals 4 * the cube of the cine of 36°
2:08
minus 3 * the cine of
2:11
36° on the right side we have negative
2:14
coine of the double angle the cine of
2:17
the double angle is equal to 2 * the
2:19
square of the cine of 36° - 1 with the
2:24
negative sign we switch the signs giving
2:28
us 1 - 2 * the square of the coine of
2:32
36° we move everything to the left side
2:35
for convenience and replace the cine of
2:38
36° with t in the result we get 4 * T
2:42
Cub + 2 * T ^ 2 - 3 t - 1 = 0 one of the
2:48
roots can be easily found among the
2:50
divisors of the constant term indeed t
2:53
equal to -1 is a root of our equation
2:56
which means we can divide the polinomial
2:58
by t + 1
3:00
or factor it
3:01
out to do this we add 2 t^ 2 and through
3:06
2 T ^2 we factor out four * T
3:10
^2 in the parentheses we are left with t
3:14
+ 1 then we write -3 t as -2
3:21
tus1 from the next two terms we factor
3:24
out -1 as a result we get -2 T * t + 1
3:30
and we still have
3:33
t-1 we factor out -1 getting t + 1 and
3:37
the entire equation equals 0 factoring
3:41
out t + 1 we are left with 4 t^2 - 2 t -
3:46
1 which equal
3:48
0 therefore either t + 1 = 0 or 4 T ^2 -
3:54
2 t - 1 = 0 from the first equation t
3:59
must be equal to1 but T is the coine of
4:04
36° which cannot be -1 this leaves us
4:07
with the second equation to solve we
4:09
find the discriminant but it's more
4:11
convenient to find D1 which is the
4:14
quarter discriminant solving gives us 1
4:16
+ 4 which equals 5 this means the roots
4:20
of the quadratic equation will be of the
4:23
Form B / 2 or 1 + or minus the square <
4:28
TK of D1
4:30
which is the square < TK of 5 All divid
4:34
4 let's recall that t is the coine of
4:39
36° since the coine of an angle in the
4:41
first quadrant is positive we take only
4:44
the positive root therefore T will be
4:48
equal to 1 + the < TK of 5 / 4 thus the
4:54
coine of 36° = 1 + the < TK 5 / 4
5:00
returning to our formula for the leg X
5:02
will be equal to 4 * 1 + < TK 5 /
5:07
4 simplifying we get that the leg equal
5:11
1 + the < TK of 5 the problem is solved
5:15
if you understood the solution give it a
5:18
like leave a comment and don't forget to
5:21
subscribe to the channel
#Science