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Image of the triangle form the question I am struggling with this question. I am unsure if it can be solved or if there is a mistake in it. Specifically, $\angle ADC$ looks to be $90^\circ$ but is not marked as such. So, given that $\angle ADC$ is unknown, can this be solved using the sine or cosine rule? This is meant to be a GCSE level question and I am really struggling to see a simple solution. Thank you.

Aiden Chow
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88_matsy
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  • There is probably a mistake. Without the $90^\circ$ it is impossible to solve, since the ratio $DC:CB$ can then be changed at will. – player3236 Sep 16 '20 at 20:12

2 Answers2

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The only features of the diagram that are fixed is the triangle $\triangle ABC$. You can solve for all angle measures and all side lengths for this triangle. However, point $D$ relative to this triangle can be positioned at any location along the ray $BC$ such that $BC < BD$. Therefore, the length of $CD$ is not uniquely determined. An additional assumption is required but not stated.

heropup
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Suppose AD is not joined as given and it is required to solve only the acute angled triangle part ABC:

$$\angle CBA = 47^{\circ};\;\angle CAB = 59-47=12^{\circ};AB=49\;; $$

This way it can be solved.

Narasimham
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