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Question

Consider the following figure:
$$\sin B$$ is $$\displaystyle \frac{12}{m}$$, m is 

185867_227b8b0385d64896bd5393746d641fe9.png


Solution

In $$\triangle ABC$$, $$AD \perp BC$$ $$AB = 13$$, $$BD = 5$$, $$DC =16$$
Now, In $$\triangle ABD$$,
$$AB^2 = AD^2 + BD^2$$
$$13^2 = AD^2 + 5^2$$
$$AD^2 = 144$$
$$AD = 12$$ 
Now, in $$\triangle ADC$$,
$$AC^2 = AD^2 + CD^2$$
$$AC^2 = 12^2 + 16^2$$
$$AC^2 = 144 + 256$$
$$AC = 20$$ cm
Now, $$\sin B = \frac{P}{H} = \frac{AD}{AB} = \frac{12}{13}$$

Mathematics

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