Does anyone have a non-trigonometry solution for this problem:

Suppose we have a triangle \(ABC\).It is known that \(C=30\) degrees.From \(A\) draw a segment \(AD\) such that \(D\) is on \(BC\) and \(ADC\) is \(40\) degrees.Also \(AB=CD\).Find \(B\).

Note by Lawrence Bush
3 years, 8 months ago

No vote yet
1 vote

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

Comments

Sort by:

Top Newest

My thinking is as follows: Locate the circum-centre of triangle ADC . Let's call it O. /AOD=2*/ACD=2*30 or 60°. This makes triangle AOD equilateral or AD=CO = R, the circumradius of Tr. ADC. Now in triangles BAD & DCO, we've AB=CD (Given), AD=CO & /_ BDA=/DOC both =140° . This isn't the included angle but since it is clearly the largest angle of each triangle, this is a special case of congruence. This means /ABD or /_ B = /_CDO = 20°

One Top - 3 years, 7 months ago

Log in to reply

Thanks

Lawrence Bush - 3 years, 7 months ago

Log in to reply

Amazing! You're awesome @One Top How did you think of it? I though of drawing the circumcircle of \(\Delta ABC\)...

Satvik Golechha - 3 years, 7 months ago

Log in to reply

From sine law in triangle ADC we have AD/2R=sin30°=1/2,so AD=R. That is hiding behind his result.That is why we locate circum-centre of triangle ADC.Circumcentre of tringle ADC is outside of the triangle,A and O are on the oposite sides of line CD.
When we apply sine law in triangle ADC to line CD we have that CD/2R=sin110°,so CD=2Rsin110°=2Rsin70°=2Rcos20° ,but AB=CD,so AB=2Rcos20°,from sine law aplied to the triangle ABD(which has circumcentre r) and line AD we get AD=2rsinB, so R=2rsinB,so sinB=R/(2r).But when apply sin law to the triangle ABC and line AB we have AB=2rsin140°=2rsin40° so 2rsin40°=2Rcos20°,and we get R/r=sin40°/cos20°.Now we replace it and get sinB=0,5(sin40°/cos20)=0,5(2sin20°cos20°/cos20°)=sin20°,so b=20° or b=160°,but B<90° so answer is B=20°.

Nikola Djuric - 3 years, 6 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...