Question

a 15 - foot flagpole leans slightly, such that it makes an 80° angle with the ground. the shadow of the flagpole is 10 feet long when the sun has an unknown angle of elevation. how could the angle of elevation of the sun, x, be determined?
by determining the length of tv using ( tv^{2}=15^{2}+10^{2}-2(15)(10)cos80^{circ} ), and then determining the value of x using ( 15^{2}=tv^{2}+10^{2}-2(tv)(10)cos x )
by determining the value of x using ( x^{2}=15^{2}+10^{2}-2(15)(10)cos80^{circ} )
by determining the length of tv using ( tv^{2}=15^{2}+10^{2}+2(15)(10)cos80^{circ} ), and then determining the value of x using ( 15^{2}=tv^{2}+10^{2}+2(tv)(10)cos x )
by determining the value of x using ( x^{2}=15^{2}+10^{2}+2(15)(10)cos80^{circ} )

Explanation:

Step1: Use the Law of Cosines to find \(TV\)

The Law of Cosines formula is \(a^{2}=b^{2}+c^{2}-2bc\cos A\). In \(\triangle TUV\), for side \(TV\) (let \(TV = a\), \(TU = 15\) ( \(b = 15\) ), \(UV=10\) ( \(c = 10\) ), and \(A = 80^{\circ}\) ), we have \(TV^{2}=15^{2}+10^{2}-2(15)(10)\cos80^{\circ}\).

Step2: Use the Law of Cosines again to find \(x\)

Now, in \(\triangle TUV\), if we want to find \(x\), using the Law of Cosines formula \(a^{2}=b^{2}+c^{2}-2bc\cos A\) again. Let \(TU = 15\) ( \(a = 15\) ), \(TV\) (from step 1) as \(b\), \(UV = 10\) ( \(c = 10\) ), and \(A=x\). So \(15^{2}=TV^{2}+10^{2}-2(TV)(10)\cos x\)

Answer:

by determining the length of \(TV\) using \(TV^{2}=15^{2}+10^{2}-2(15)(10)\cos80^{\circ}\), and then determining the value of \(x\) using \(15^{2}=TV^{2}+10^{2}-2(TV)(10)\cos x\)