Find the measure of angle P in the triangle below
Find the measure of angle P in the triangle below
\(60^2=48^2+36^2\Rightarrow \triangle{PQR} \text{ is a right triangle}\Rightarrow \angle{Q}=90^o\)
\(\sin{\angle{P}}= \frac{48}{60} =0.8
\\
\\\angle{P}=\arcsin{0.8}=53.13^o\)
Answer:
\(\displaystyle 53°\)
Step-by-step 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
\(\displaystyle \frac{p^2 + q^2 - r^2}{2pq} = cos∠R \\ \frac{p^2 - q^2 + r^2}{2pr} = cos∠Q \\ \frac{-p^2 + q^2 + r^2}{2qr} = cos∠P\)
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!
------------------------------------------------------------------------------------------
\(\displaystyle \frac{-48^2 + 60^2 + 36^2}{2[60][36]} = cos∠P → \frac{-2304 + 3600 + 1296}{4320} = cos∠P → \frac{2592}{4320} = cos∠P → \frac{3}{5} = cos∠P → 53,13010235° ≈ cos^{-1}\:\frac{3}{5} \\ \\ 53,13010235° ≈ m∠P → 53° ≈ m∠P\)
I am delighted to assist you at any time. ☺️