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#