Question

can someone just explain this please?

Answer 1

In △PQR, find the measure of ∡P.

Triangle PQR where angle Q is a right angle. PQ measures 33 point 8. PR measures 57 point 6. Measure of angle P is unknown

30.4°

59.6°

35.9°

54.1°

Answer 2

Do you know the trig function (sine, cosine, etc.) that relates the two known sides to your angle \( \angle P \) ?

Answer 3

Hint: think in terms of adjacent side, opposite side, and hypotenuse relative to angle P. The trig function that uses the pair you have is correct and we can set up the function equal to their ratio / take an inverse.

Answer 4

i know what adjacent is what hypotunes and opposite i also know cosine sine and the tan rules

Answer 5

Right. So in this case, we know what is adjacent to <P and what the hypotenuse is. These two are used in the cosine of <P, correct?
That allows us to make the following equation based on the definition of cosine:
\( \cos \angle P = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{33.8}{57.6} \)
That is clear so far?

Answer 6

yes it's clear

Answer 7

Then we would just need to apply the inverse cosine of both sides:
\( \cos ^{-1} \left( \cos \angle P \right) = \cos ^{-1} \left( \dfrac{33.8}{57.6} \right) \)
The inverse and original function cancel, leaving:
\( \angle P = \cos^{-1} \left ( \dfrac{33.8}{57.6} \right) \)
Which would require a calculator to evaluate.

Answer 8

i'm confused do you want me to simplify them?

Answer 9

You would enter \( \cos^{-1} \left( 33.8 / 57.6 \right)\) into a calculator (use degree mode) to get the degree measure of \( \angle P \). That is, if everything before that was clear.

Answer 10

is it ok if you explain it in a different way sir/ma'am (i don't know if your a boy or girl)

Answer 11

cos (alpha) = a --> alpha = arccos (a)
Dat sit

Answer 12

he uses cos^- = arccos.

Answer 13

so am i supposed to do it the way he showed me?

Answer 14

yes

Answer 15

so 54.07 but when you round you get 54.1?

Answer 16

yes

Answer 17

thank you @Loser66 and @ganeshie8