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Consider a triangle with sides ${5}'',{8}''\ and\ {10}''.$ Find the $\angle A$ , $\angle B,$ and $\angle C$ .

Option: 1
$44.5^{\circ},90^{\circ},45.4^{\circ}$

Option: 2
$90^{\circ},44.5^{\circ},45.4^{\circ}$

Option: 3
$44.5^{\circ},90^{\circ},44.5^{\circ}$

Option: 4
$44.5^{\circ},44.5^{\circ},90^{\circ}$

Answers (1)

best_answer

Given that,

The triangle with sides ${5}'',{8}''\ and\ {10}''.$

Using the law of cosine,

$cos\ A=\frac{\left ( b^{2}+c^{2}-a^{2} \right )}{2bc}$

$cos\ A=\frac{\left ( 64+100-25 \right )}{160}$

$cos\ A=0.7125$

$A=44.5^{\circ}$

Using the law of sine to find the $\angle B,$

$\frac{sin\ B}{b}=\frac{sin\ A}{a}$

$sin\ B=\frac{b}{a}sin\ A$

$sin\ B=\frac{8}{5}sin\ 44.5^{\circ}$

$sin\ B=\frac{8}{5}\times 0.7^{\circ}$

$sin\ B=1.12$

$B\approx 90^{\circ}$

The third angle is,

$C=180^{\circ}-A-B$

$C=45.4^{\circ}$

Posted by

Suraj Bhandari

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