Question

Solve triangle PQR. Round intermediate results to 3 decimal places and final answers to 1 decimal place.

Answer 1

Answer:

$Q = 96.4^\circ$

$R = 58.4$

$P =25.2$

Step-by-step explanation:

Given

See attachment for PQR

Required

Solve the triangle

From the attached triangle, all sidea of the triangle are known.

i.e.

$PQ = 6\\QR=3\\PR=7$

So, we are to solve for $\angle P, Q \& R$

Using cosine rule:

$a^2 = b^2 + c^2 - 2bcCosA$

To solve for Q, we have:

$PR^2 = PQ^2 + QR^2 - 2 * PQ * QR Cos(Q)$

$7^2 = 6^2 + 3^2 - 2 * 6 * 3 * \cos(Q)$

$49 = 36 + 9 - 36 * \cos(Q)$

Collect like terms

$49 - 36 - 9 = - 36 * \cos(Q)$

$4 = - 36 * \cos(Q)$

Divide both sides by -36

$\cos(Q) = \frac{-4}{36}$

$\cos(Q) = -0.1111$

Take arc cos of both sides

$Q = cos^{-1}(0.1111)$

$Q = 96.4^\circ$

To solve for R, we make use of sine rule;

$\frac{a}{\sin A} = \frac{b}{\sin B}$

So, we have:

$\frac{PQ}{\sin R} = \frac{PR}{\sin Q}$

$\frac{6}{\sin R} = \frac{7}{\sin 96.4}$

Cross multiply

$\sin R * 7 = 6 * \sin 96.4$

Solve for sin R

$\sin R = \frac{6 * \sin 96.4}{7}$

$\sin R = \frac{6 * \0.9938}{7}$

$\sin R = 0.8518$

Take arc sin of both sides

$R = sin^{-1}(0.8518)$

$R = 58.4$

Solving P:

$P+ Q + R = 180$ --- angles in a triangle

Solve for P

$P =180-Q-R$

$P =180-96.4-58.4$

$P =25.2$