91Ó°ÊÓ

To solve an oblique triangle given the case ASA, the first step is to find the missing so that the law of sines can be used.

The opposite of a vector is a vector with the ______ magnitude and __________ direction. To obtain the opposite of a vector, multiply the vector by ____.

Find the semi perimeter of triangle \(A B C\). \(a=153 \mathrm{~cm}, b=174 \mathrm{~cm}, c=232 \mathrm{~cm}\)

Find all solutions to each of the following triangles: \(A=132.4^{\circ}, a=27.3 \mathrm{~cm}, b=50.2 \mathrm{~cm}\)

Draw the vector \(\mathbf{V}\) that goes from the origin to the given point. Then write \(\mathbf{V}\) in component form \(\langle a, b\rangle\). $$(-2,5)$$

Draw the vector \(\mathbf{V}\) that goes from the origin to the given point. Then write \(\mathbf{V}\) in component form \(\langle a, b\rangle\). $$(3,-3)$$

For each pair of vectors, find \(\mathbf{U} \cdot \mathbf{V}\). \(\mathbf{U}=5 \mathbf{i}+3 \mathbf{j} . \mathbf{V}=-5 \mathbf{i}+3 \mathbf{j}\)

In triangle \(A B C, A=40^{\circ}, b=19 \mathrm{ft}\), and \(a=18 \mathrm{ft}\). Use the law of sines to find \(\sin B\) and then give two possible values for \(B\).

Distance A hot-air balloon is held at a constant altitude by two ropes that are anchored to the ground. One rope is 120 feet long and makes an angle of \(65^{\circ}\) with the ground. The other rope is 115 feet long. What is the distance between the points on the ground at which the two ropes are anchored?

$$ \text { Solve each of the following triangles. } $$ $$ \text { Use the law of cosines to show that, if } A=90^{\circ} \text {, then } a^{2}=b^{2}+c^{2} \text {. } $$