Finding Sides

Help Questions

Math › Finding Sides

Questions 1 - 4

1

Solve for .

(Figure not drawn to scale).

There is not enough information

Explanation

The side-angle-side (SAS) postulate can be used to determine that the triangles are similar. Both triangles share the angle farthest to the right. In the smaller triangle, the upper edge has a length of , and in the larger triangle is has a length of . In the smaller triangle, the bottom edge has a length of , and in the larger triangle is has a length of . We can test for comparison.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}$

$\frac{8}{16}=\frac{6}{12}=\frac{1}{2}$

The statement is true, so the triangles must be similar.

We can use this ratio to solve for the missing side length.

$\frac{Upper\ edge_1}{Upper\ edge_2}=\frac{Lower\ edge_1}{Lower\ edge_2}=\frac{x}{Left\ edge_2}$

To simplify, we will only use the lower edge and left edge comparison.

$\small \frac{6}{12}=\frac{x}{5}$

Cross multiply.

$\small 12x=30$

$\small x=\frac{30}{12}=\frac{5}{2}=2.5$

2

Trig_id

If $\theta$ equals $55^{\circ}$ and is , how long is ?

Not enough information to solve

Explanation

This problem can be easily solved using trig identities. We are given the hypotenuse and $\theta (55^{\circ})$ . We can then calculate side using the $\sin$ .

$\sin=\frac{opposite}{hypotenuse}$

$\sin55^{\circ}=\frac{o}{15ft}$

Rearrange to solve for .

$o=\sin55^{\circ}*15ft$

$\dpi{100} \rightarrow 12.3ft$

If you calculated the side to equal then you utilized the $\cos$ function rather than the $\sin$ .

3

Solve for .

(Figure not drawn to scale).

Explanation

We can solve using the trigonometric definition of tangent.

$\small tan\Theta =\frac{opp}{adj}$

We are given the angle and the adjacent side.

$\small tan(22^o)=\frac{z}{18}$

We can find $\small tan(22^o)$ with a calculator.

$\small tan(22^o)=0.4$

$0.4=\frac{z}{18}$

$\small 18*0.4=z$

$\small z=7.3$

4

Triangle

What is the length of CB?

$\frac{AB}{\tan C}$

$\frac{AB}{\sin A}$

$\frac{\tan C}{AB}$

$\frac{\sin B}{AC}$

$\frac{AC}{\sin A}$

Explanation

$\tan X= \frac{OPPOSITE}{ADJACENT}$

$\tan C=\frac{AB}{CB}$

$CB= \frac{AB}{\tan C}$

Return to subject