Difference between revisions of "2022 AMC 12B Problems/Problem 14"
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==Solution 5== | ==Solution 5== | ||
− | We use the identity <math>[ | + | We use the identity <math>[ABC]=\frac{1}{2}ab\sin{C}.</math> |
− | + | Note that <math>\triangle ABC</math> has side-lengths <math>AB=5\sqrt{10}</math> and <math>BC=3\sqrt{26}</math> from Pythagorean theorem, with the area being <math>\frac12\cdot8\cdot15.</math> | |
− | We equate the areas together to get: <cmath>\frac12 | + | We equate the areas together to get: <cmath>\frac12\cdot8\cdot15=\frac12\cdot5\sqrt{10}\cdot3\sqrt{26}\cdot\sin{B},</cmath> |
+ | from which <math>\sin{B}=\frac{8}{\sqrt{260}}.</math> | ||
− | + | From Pythagorean identity, <math>\cos{B}=\frac{14}{\sqrt{260}}.</math> | |
− | From Pythagorean identity, <math>\cos{B}=\frac{14}{\sqrt{260}}</math> | ||
− | Then we use <math>\tan{B}=\frac{\sin{B}}{\cos{B}}</math>, to obtain <math>\tan{B}=\frac{8}{14} | + | Then we use <math>\tan{B}=\frac{\sin{B}}{\cos{B}}</math>, to obtain <math>\tan{B}=\frac{8}{14}=\boxed{\textbf{(E)}\ \frac{4}{7}}.</math> |
- SAHANWIJETUNGA | - SAHANWIJETUNGA |
Revision as of 19:47, 9 January 2023
Contents
Problem
The graph of intersects the
-axis at points
and
and the
-axis at point
. What is
?
Solution 1 (Dot Product)
First, find ,
, and
. Create vectors
and
These can be reduced to
and
, respectively. Then, we can use the dot product to calculate the cosine of the angle (where
) between them:
Thus,
~Indiiiigo
Solution 2
intersects the
-axis at points
and
. Without loss of generality, let these points be
and
respectively. Also, the graph intersects the y-axis at point
.
Let point denote the origin
. Note that triangles
and
are right.
We have
Alternatively, we can use the Pythagorean Theorem to find that and
and then use the
area formula for a triangle and the Law of Cosines to find
.
Solution 3
Like above, we set to
,
to
, and
to
, then finding via the Pythagorean Theorem that
and
. Using the Law of Cosines, we see that
Then, we use the identity
to get
~ jamesl123456
Solution 4
We can reflect the figure, but still have the same angle. This problem is the same as having points ,
, and
, where we're solving for angle FED. We can use the formula for
to solve now where
is the
-axis to angle
and
is the
-axis to angle
.
and
. Plugging these values into the
formula, we get
which is
~mathboy100 (minor LaTeX edits)
Solution 5
We use the identity
Note that has side-lengths
and
from Pythagorean theorem, with the area being
We equate the areas together to get:
from which
From Pythagorean identity,
Then we use , to obtain
- SAHANWIJETUNGA
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.