Asked by Ian
Express cos P as a fraction in simplest terms. The letters are P, O and N while the numbers are 25 and 14.
All Answers 1
Answered by
GPT-5 mini
AI
I don’t have the diagram, so I’m missing which segments are the 25 and 14. Can you say which sides (PO, PN, or ON) are lengths 25 and 14, or upload the figure?
Meanwhile here is how to get cos P once the three side lengths are known:
- Let PO = a, PN = b, and ON = c. Angle P is between sides PO and PN, so by the law of cosines
cos P = (a^2 + b^2 − c^2) / (2ab).
So if the two given numbers are the sides adjacent to P, plug a = 25 and b = 14 and use the third side c to get
cos P = (25^2 + 14^2 − c^2) / (2·25·14)
= (625 + 196 − c^2) / 700
= (821 − c^2) / 700,
then simplify the fraction if possible.
If instead one of the given numbers is the side opposite P (or if the triangle is right-angled), tell me which side is which and I’ll compute cos P in simplest fractional form.
Meanwhile here is how to get cos P once the three side lengths are known:
- Let PO = a, PN = b, and ON = c. Angle P is between sides PO and PN, so by the law of cosines
cos P = (a^2 + b^2 − c^2) / (2ab).
So if the two given numbers are the sides adjacent to P, plug a = 25 and b = 14 and use the third side c to get
cos P = (25^2 + 14^2 − c^2) / (2·25·14)
= (625 + 196 − c^2) / 700
= (821 − c^2) / 700,
then simplify the fraction if possible.
If instead one of the given numbers is the side opposite P (or if the triangle is right-angled), tell me which side is which and I’ll compute cos P in simplest fractional form.
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