Question

use the figure below to answer the following questions. round all answers to two decimal places. enter deg after any degree values.

enter deg after any degree value.

a) what is ( mangle q )?

b) what is ( pr )?

c) what is ( qr )?

Explanation:

Part (a)

Step1: Recall triangle angle sum

In a triangle, the sum of angles is \(180^\circ\). This is a right triangle (\(\angle R = 90^\circ\)) and \(\angle P = 43^\circ\). Let \(\angle Q = x\). Then \(90^\circ + 43^\circ + x = 180^\circ\).

Step2: Solve for \(x\)

Simplify the equation: \(133^\circ + x = 180^\circ\). Subtract \(133^\circ\) from both sides: \(x = 180^\circ - 133^\circ = 47^\circ\). So \(m\angle Q = 47.00\) deg.

Part (b)

Step1: Identify trigonometric ratio

In right triangle \(PQR\), \(\sin(\angle Q)=\frac{PR}{PQ}\). We know \(PQ = 11\) and \(\angle Q = 47^\circ\). So \(\sin(47^\circ)=\frac{PR}{11}\).

Step2: Solve for \(PR\)

Multiply both sides by \(11\): \(PR = 11\times\sin(47^\circ)\). Calculate \(\sin(47^\circ)\approx0.7314\), so \(PR\approx11\times0.7314\approx8.05\).

Part (c)

Step1: Identify trigonometric ratio

In right triangle \(PQR\), \(\cos(\angle Q)=\frac{QR}{PQ}\). We know \(PQ = 11\) and \(\angle Q = 47^\circ\). So \(\cos(47^\circ)=\frac{QR}{11}\).

Step2: Solve for \(QR\)

Multiply both sides by \(11\): \(QR = 11\times\cos(47^\circ)\). Calculate \(\cos(47^\circ)\approx0.6820\), so \(QR\approx11\times0.6820\approx7.50\).

Answer:

s:
a) \(47.00\) deg
b) \(8.05\)
c) \(7.50\)