Here is my effort:
[math]D\widehat{B}C=35\degree~(\text{base >s of isosceles triangle})[/math]
[math]B\widehat{D}C=180-2(D\widehat{B}C)=2\times 35=110\degree ~(\text{sum of>s in a triangle})[/math]
[math]A\widehat{B}D=90\degree~(\text{ > in a semi circle})[/math]
[math]B\widehat{A}O=180-(90+110)=-20\degree~\text{or perhaps}~ 20\degree[/math]
But I am thinking something is wrong somewhere.
[math]B\widehat{A}O ~\text{cannot be}~ =20\degree~\text{because if}~B\widehat{A}O =20\degree \\
\text{then}~ B\widehat{A}O+ A\widehat{B}D + B\widehat{D}A=180\degree[/math]
That is [math]\underset{20 \degree}{B\widehat{A}O} + \underset{90\degree} {A \widehat{B}D } + \underset{110\degree }{B\widehat{D}A} \underset {220 \degree} {\ne180\degree}[/math]
So what do you think?
View attachment 35926
Here is my effort: View attachment 35927
[math]D\widehat{B}C=35\degree~(\text{base >s of isosceles triangle})[/math]
[math]B\widehat{D}C=180-2(D\widehat{B}C)=2\times 35=110\degree ~(\text{sum of>s in a triangle})[/math]
[math]A\widehat{B}D=90\degree~(\text{ > in a semi circle})[/math]
[math]B\widehat{A}O=180-(90+110)=-20\degree~\text{or perhaps}~ 20\degree[/math]
But I am thinking something is wrong somewhere.
[math]B\widehat{A}O ~\text{cannot be}~ =20\degree~\text{because if}~B\widehat{A}O =20\degree \\
\text{then}~ B\widehat{A}O+ A\widehat{B}D + B\widehat{D}A=180\degree[/math]
That is [math]\underset{20 \degree}{B\widehat{A}O} + \underset{90\degree} {A \widehat{B}D } + \underset{110\degree }{B\widehat{D}A} \underset {220 \degree} {\ne180\degree}[/math]
So what do you think?
You correctly found [imath]m(\angle BDC)=110^o[/imath].
Using supplementary angles w get [imath]m(\angle BDA)=70^o[/imath].
Therefore, [imath]m(\widehat{AB})=140^o[/imath] HOW and/or WHY?
Then what is [imath]m(\widehat{BD})=?^o[/imath]
So [imath]x=?^o[/imath]
[imath][/imath][imath][/imath][imath][/imath]
I think that, for some (considerable) time now, I have been encouraging you to use sketches but you now seem to have adopted an approach whereby you say: "Here is my effort:" and promptly reproduce a(n exact) copy of the diagram that comes with the question!
That is not the same as drawing your own sketch or, at the very least, modifying the question's diagram to show relevant information that may be deduced from it (as I have done in the picture below, q.v.).
Drawing your own sketch is useful because it means you have to think about the relationships involved in the figures as you construct it and allows you to include other information that comes from those relationships as you go along.
I believe I tried to demonstrate this to you in your last problem (see here), where a sketch simplified the problem down to just three lines.
I trust you will see that the diagram below similarly simplifies this problem (using much of what you already stated in your attempt that went awry).
Following the 'logic' from redto green to blue, thus determining the two blue angles, can you now simply see what x must be?
[math]D\widehat{B}C=35\degree~(\text{base >s of isosceles triangle})[/math]
[math]B\widehat{D}C=180-2(D\widehat{B}C)=2\times 35=110\degree ~(\text{sum of>s in a triangle})[/math][math]A\widehat{B}D=90\degree~(\text{ > in a semi circle})[/math][math]\angle ABD= 180\degree -110\degree=70\degree~(\text{ sum >s on a straight line})[/math][math]B\widehat{A}O=180\degree - (70+90)\degree = 20\degree ~(\text{sum of >s in a}~ \lhd)[/math]
Yes, that is the correct answer, but don't you see how much simpler it is with a sketch (or, as in this case, the modified diagram from the question).
I would suggest that if you had made a sketch (or even modified the diagram the way I did) then you would have been much less likely to make the mistake you did!
Drawing your own sketch has many advantages and will stand you in good stead especially if you can make an attempt to do it with a reasonable degree of accuracy.
I bought this little "Maths Set" in 'Poundland' for just £1 (admittedly, that was back when (almost) everything in 'Poundland' still cost £1); I expect it may be dearer now. And, of course, there may not be a 'Poundland' where you live (I suspect you are not in the UK). I have also seen it on eBay (for £2.95, including delivery) but, again, I have no idea if that's "available" to you either. (eBay item number:191973939634)
However, it would be worth your while investing in a set of instruments like this (cheap ones are fine for sketching if you can get them) as the practice you would get in constructing your own diagrams would definitely not be time wasted!
I used that little set to construct the sketch below. I started by constructing the triangles. I started with C at the end of a horizontal line, then a line from C at 35° to my horizontal (using protractor), then D four boxes along the line to the left of C. I could have used the pair of compasses to get DB the same length as DC but (1 step ahead ?) I just drew a line at 70° from D and where it crossed my 35° line was Point B, then a line at 90° from B gave me A (where it crossed my horizontal). I then just added the circle by centering it (approximately) half way along AD and 'adjusting' the compasses to 'catch' A, B & D.
That allowed me to actually measure x (as 20°) to confirm my geometrical analysis. ?
You can also now see how different a "proper" diagram of the situation described in the question actually looks compared to their (rather poor) sketch!
Little exercises like that would help your understanding of these problems and also help you to avoid the kind of errors we have been seeing you fall foul of.
PS: you can use these characters: "sum of ∠s in a Δ"
instead of: sum of ">s in a ⊲"
I also keep this lot (below) in a text file I named "Special Characters.txt".
Feel free to copy them into one of your own.
(In the time you spend creating LaTex expressions you could probably construct your own sketch and write your working out in pencil; you could then just submit a (good, clear) picture of that. ?)
My list of "handy" characters:-
| ± ×{\times in LaTex} ÷ + - ˉ ∑ ∙ ⋮ √ ∞ ∩ ≈ ≅ ≠ ≡ ≤ ≥⇐⇒σ ΩωΦϕα β θ º π Ψ ∠ ∴(therefore) ∵(because)
ABB̈ΓΔEZHΘIKΛMNΞOΠPΣTΥΦXΨΩ α β γ δ ϵ ζ η θ ι κ λ μ ν ξ o π ρ σ τ υ ϕ χ ψ ω ε ϑ ϖ ϱ ς φ ϝαβγδϵζηθικλμνξoπ∂
\rightarrow (→) & \Rightarrow (=>) \[b or x]cancel{text to strike thru with / [or \ or X]}
Sup/scrpt: ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ plus in LaTex: ^{sup/scrpt text} & _{sub/scrpt text}
Fractions: ¼ ½ ¾ ⅓ ⅔ ⅛ ⅜ ⅝ ⅞
Arrows: ← ↑ → ↓ ↔ ↕ ↨
∆ (Difference) Δ (Gk Delta)
Unit vectors: Inline: \(\displaystyle \hat{i} \hat{j} \hat{k} \)
Centred: [math] \hat{i} \hat{j} \hat{k} [/math] (NB: Spaces between are ignored!)
See also: List of Logic Symbols (https://en.wikipedia.org/wiki/List_of_logic_symbols)
En Quad (Character that looks like a Space) " "
Also: \; - a thick space; \: - a medium space; \, - a thin space & \! - a negative thin space
Colon & Question Mark in Filenames [：？]
└ ─ ┬ ─ ┘ └───┬───┘
or, for better 'bracketting' (in LaTex):-
\underbrace{things to be bracketted}_{\text{words under bracket}}
\overbrace{things to be bracketted}^{\text{words over bracket}}
Colon & Question Mark in Filenames [：？]
I would suggest that if you had made a sketch (or even modified the diagram the way I did) then you would have been much less likely to make the mistake you did!
No, I don't think so. Making/modifying a diagram may not necessarily have affected my work in anyway since one have only been provided and I do understand it. It is just a mistake that I made and needed an assistance. I only try make a diagram when one is not provided. Though I appreciate your concerns and contributions.
However, it would be worth your while investing in a set of instruments like this (cheap ones are fine for sketching if you can get them) as the practice you would get in constructing your own diagrams would definitely not be time wasted!
PS: you can use these characters: "sum of ∠s in a Δ"
instead of: sum of ">s in a ⊲"
I also keep this lot (below) in a text file I named "Special Characters.txt".
Feel free to copy them into one of your own.
(In the time you spend creating LaTex expressions you could probably construct your own sketch and write your working out in pencil; you could then just submit a (good, clear) picture of that. ?)
My list of "handy" characters:-
| ± ×{\times in LaTex} ÷ + - ˉ ∑ ∙ ⋮ √ ∞ ∩ ≈ ≅ ≠ ≡ ≤ ≥⇐⇒σ ΩωΦϕα β θ º π Ψ ∠ ∴(therefore) ∵(because)
ABB̈ΓΔEZHΘIKΛMNΞOΠPΣTΥΦXΨΩ α β γ δ ϵ ζ η θ ι κ λ μ ν ξ o π ρ σ τ υ ϕ χ ψ ω ε ϑ ϖ ϱ ς φ ϝαβγδϵζηθικλμνξoπ∂
\rightarrow (→) & \Rightarrow (=>) \[b or x]cancel{text to strike thru with / [or \ or X]}
Sup/scrpt: ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ plus in LaTex: ^{sup/scrpt text} & _{sub/scrpt text}
Fractions: ¼ ½ ¾ ⅓ ⅔ ⅛ ⅜ ⅝ ⅞
Arrows: ← ↑ → ↓ ↔ ↕ ↨
∆ (Difference) Δ (Gk Delta)
Unit vectors: Inline: \(\displaystyle \hat{i} \hat{j} \hat{k} \)
Centred: [math] \hat{i} \hat{j} \hat{k} [/math] (NB: Spaces between are ignored!)
See also: List of Logic Symbols (https://en.wikipedia.org/wiki/List_of_logic_symbols)
En Quad (Character that looks like a Space) " "
Also: \; - a thick space; \: - a medium space; \, - a thin space & \! - a negative thin space
Colon & Question Mark in Filenames [：？]
└ ─ ┬ ─ ┘ └───┬───┘
or, for better 'bracketting' (in LaTex):-
\underbrace{things to be bracketted}_{\text{words under bracket}}
\overbrace{things to be bracketted}^{\text{words over bracket}}
Colon & Question Mark in Filenames [：？]
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