How do I solve this problem?

Answer 1

Nolan Alexander

#a=10.92#, #beta=40.73^@#, #gamma=100.97^@#

According to cosine formula #a^2=b^2+c^2-2bc cosalpha#

and here #a^2=11.5^2+17.3^2-2xx11.5xx17.3xxcos38.3^@#

= #132.25+299.29-397.9xx0.7848#

= #431.54-312.272#

= #119.268# and #a=sqrt119.268=10.92#

As #b^2=c^2+a^2-2ac cosbeta#, we have

#cosbeta=(c^2+a^2-b^2)/(2ca)#

= #(17.3^2+10.92^2-11.5^2)/(2xx17.3xx10.92)#

= #(299.29+119.268-132.25)/(377.832#

= #286.308/377.832=0.757765#

i.e. #beta=40.73^@#

and #gamma=180^@-38.3^2-40.73^@=100.97^@#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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