Question

(a) Evaluate value of tanx
cos y from
the figure
where AC=13
AD=5
BD = 16​

Answer 1

Refer to the above attachment.

Answer 2

Answer:

$\huge{\boxed{\mathfrak{Answer=1)\frac{5}{12}}}}$

$\huge{\boxed{\mathfrak{Answer=2)\frac{4}{5}}}}$

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From ∆ACD,

$\tt {AC}^{2} = {AD}^{2} + {CD}^{2} \\ \leadsto\tt {13}^{2} ={ 5}^{2} + {CD}^{2} \\ \leadsto\tt 169 = 25 + {CD}^{2} \\ \leadsto\tt {CD}^{2} = 169 - 25 \\ \leadsto\tt CD =\sqrt{144}\:unit \\ \leadsto\tt CD = 12 \:unit.$

Now from ∆BCD,

$\tt {BC}^{2} = {BD}^{2} + {CD}^{2} \\ \leadsto\tt {BC}^{2} = {16}^{2} + {12}^{2} \\ \leadsto\tt {BC}^{2} = 256 + 144 \\ \leadsto\tt {BC}^{2} =400 \\ \leadsto\tt BC =\sqrt{400}\:unit \\ \leadsto\tt BC = 20 \:unit$

Now, at first,

$\tt tanx =\frac{AD}{CD} \\ \leadsto\tt tanx = \frac{5}{12}$

Secondly,

$\tt Cosy =\frac{BD}{BC} \\ \leadsto\tt Cosy =\frac{16}{20} \\ \leadsto\tt Cosy =\frac{4}{5}$