How do you solve the $triangle NPQ$ given $P=109^circ, Q=57^circ, n=22$ ?

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Answer 1 · 2017-01-28T07:19:22

$m/_N=44^@$ , $m/_P=109^@$ , $m/_Q=27^@$
and $n=22$ , $p=29.94$ and $q=14.38$

Let the triangle be as shown below:
enter image source here

Observe that as $m/_P=109^@$ and $m/_Q=27^@$ , we have $m/_N=180^@-109^@-27^@=44^@$

According to sine formula , we will have here,

$n/sinN=p/sinP=q/sinQ$

Hence, $p/(sin109^@)=q/(sin27^@)=22/sin44^@=22/0.6947$

Hence $p=22/0.6947xxsin109^@=22/0.6947xx0.9455=29.94$

and $q=22/0.6947xxsin27^@=22/0.6947xx0.454=14.38$

How do you solve the triangle NPQtriangle NPQ given P=109^circ, Q=57^circ, n=22P=109^circ, Q=57^circ, n=22?