Question

Find the measure of angle P in the triangle below

Answer 1

`\displaystyle 53°`

Explanation:

We will be using the Law of Cosines to find the
`\displaystyle m∠P,` therefore we will use the given variables in the formulas:

Solving for Angles

`\displaystyle (p^2 + q^2 - r^2)/(2pq) = cos∠R \\ (p^2 - q^2 + r^2)/(2pr) = cos∠Q \\ (-p^2 + q^2 + r^2)/(2qr) = cos∠P`

**Use
`\displaystyle cos^(-1)` in your solution or it will be thrown off!

Solving for Sides

`\displaystyle p^2 + q^2 - 2pq\:cos∠R = r^2 \\ p^2 + r^2 - 2pr\:cos∠Q = q^2 \\ q^2 + r^2 - 2qr\:cos∠P = p^2`

**Perfourm the square root in your solution or it will be thrown off!

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`\displaystyle (-48^2 + 60^2 + 36^2)/(2[60][36]) = cos∠P → (-2304 + 3600 + 1296)/(4320) = cos∠P → (2592)/(4320) = cos∠P → (3)/(5) = cos∠P → 53,13010235° ≈ cos^(-1)\:(3)/(5) \\ \\ 53,13010235° ≈ m∠P → 53° ≈ m∠P`

I am delighted to assist you at any time. ☺️

Answer 2

`60^2=48^2+36^2\Rightarrow \triangle{PQR} \text{ is a right triangle}\Rightarrow \angle{Q}=90^o`


`\sin{\angle{P}}= (48)/(60) =0.8 \\ \\\angle{P}=\arcsin{0.8}=53.13^o`