Geometry Questions Involving Trigonometric Ratios -Korean math Question
Hello, I’m Teacher JOJO, teaching mathematics in South Korea.
Let’s solve an interesting geometry Question today!
This problem is covered in the curriculum for third-year middle school students in Korea.
If you’re familiar with trigonometric ratios, give this Question a try.
[Question – Geometry Problems Involving Trigonometric Ratios]
Geometry Question Involving Trigonometric Ratios – Korean math Question
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_0D79EE5F1CC0-1-1024x702.jpeg)
The radius length is 6, and the semicircle O.
\angle AOB=135^\circ,\; \angle ACB=90^\circ
Find the value of tan\;x.
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Let’s take our time and give it a try.
Once you’ve solved it, please check the explanation below.
There are different ways to approach this Question.
If your solution differs from mine, please leave a comment.
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[Solution]
First, to find the tangent of x, let’s consider \triangle ABC.
tan x=\cfrac{\overline{BC}}{\overline{AC}}
Therefore, we need to know the lengths of segments AC and BC.
The length of segment OB is 6 since it’s the radius of the circle. The angle BOC is 45 degrees.
{\color{blue}\overline{OB}=6, \angle BOC=45^\circ}
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_42B4B1967457-1-1024x615.jpeg)
Let’s look at triangle BOC.
Of course, we could use the Pythagorean theorem here to find the lengths of segments BC and OC.
However, I’ll use trigonometric ratios.
Let’s try to find them using sine and cosine.
Let’s find the length of segment BC.
sin45^\circ = \cfrac{\overline{BC}}{\overline{OB}}
\cfrac{\sqrt{2}}{2} = \cfrac{BC}{\overline{OB}}
\overline{BC}=3\sqrt{2}
Let’s fine the length of segment OC.
cos45^\circ = \cfrac{\overline{OC}}{\overline{OB}}
\cfrac{\sqrt{2}}{2} = \cfrac{OC}{\overline{OB}}
\overline{OC}=3\sqrt{2}
{\color{blue}\overline{BC}=3\sqrt{2}}, \;{\color{blue}\overline{OC}=3\sqrt{2}}
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_1BE668361D0E-1.jpeg)
Finally, let’s find the value of tangent x.
tan x = \cfrac{\overline{BC}}{\overline{AC}}
tan x = \cfrac{3\sqrt{2}}{6+3\sqrt{2}} = \sqrt{2}-1
{\color{red}tan x = \sqrt{2}-1}
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Can you solve Korean SAT Math Question? -korean math Question #18
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[Korean ver]
안녕하세요. 저는 한국에서 수학을 가르치고있는 조조쌤입니다.
오늘은 재미있는 도형문제를 가지고 왔습니다.
이 문제는 중학교 3학년 학생들이 푸는 문제입니다.
도형문제를 좋아한다면 한번 풀어보세요!
[문제]
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_0D79EE5F1CC0-1-1024x702.jpeg)
다음 반원의 반지름은 6입니다.
\angle AOB=135^\circ,\; \angle ACB=90^\circ 일때,
tan\;x의 값은 얼마인가요?
천천히 생각하며 풀어보고 아래의 풀이와 비교해 보세요.
[풀이]
우선, tan x 를 찾기 위해서는 \triangle ABC를 봐야합니다.
tan x=\cfrac{\overline{BC}}{\overline{AC}} 이므로,
우리는 선분AC와 선분BC의 길이를 알아내야합니다.
선분OB는 원의 반지름이기 때문에 길이가 6입니다.
각BOC는 45 ^\circ 라는 것을 알수있습니다.
{\color{blue}\overline{OB}=6, \angle BOC=45^\circ}
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_42B4B1967457-1-1024x615.jpeg)
삼각형BOC를 봅시다.
다양한 방법으로 선분BC, 선분OC의 길이를 구할수 있을 것입니다.
저는 삼각비를 이용하여 길이를 구해보도록 하겠습니다.
선분BC를 구하기위해서 sin45^\circ를 이용하겠습니다.
sin45^\circ = \cfrac{\overline{BC}}{\overline{OB}}
\cfrac{\sqrt{2}}{2} = \cfrac{BC}{\overline{OB}}
\overline{BC}=3\sqrt{2}
이번에는 cos45^\circ을 이용하여 선분OC를 구해보겠습니다.
cos45^\circ = \cfrac{\overline{OC}}{\overline{OB}}
\cfrac{\sqrt{2}}{2} = \cfrac{OC}{\overline{OB}}
\overline{OC}=3\sqrt{2}
{\color{blue}\overline{BC}=3\sqrt{2}}, \;{\color{blue}\overline{OC}=3\sqrt{2}}
![](https://www.jojoteacher.com/wp-content/uploads/2023/09/IMG_1BE668361D0E-1.jpeg)
이제 tan x의 값을 찾으면 됩니다.
tan x = \cfrac{\overline{BC}}{\overline{AC}}
tan x = \cfrac{3\sqrt{2}}{6+3\sqrt{2}} = \sqrt{2}-1
{\color{red}tan x = \sqrt{2}-1}
끝났습니다.
오늘 문제는 어땠나요?
너무 쉬웠나요?
사실, 삼각비를 안쓰고도 충분히 풀수있습니다.
하지만 이 문제는 삼각비 단원에서 나오는 문제여서, 그냥 삼각비를 이용하여 풀어보았습니다.
여러분들은 다른 방법으로 풀었겠죠?
잘하셨습니다.
그럼 다음에 재미있는 문제로 또 찾아오도록하겠습니다.
다음에 또봐요.
빠잉.
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