Question

10TH GRADE SOLVING RIGHT TRIANGLES . PLEASE HELP!!! Even if you don't know all of them, any and all help is needed & appreciated!!

Answer 1

All of these questions test your familiarity on basic trigonometry involving right triangles. So, let's go to our old friend SOHCAHTOA, which tells us that
$\sin= \frac{opposite}{hypotenuse}$
$\cos= \frac{adjacent}{hypotenuse}$
$\tan= \frac{opposite}{adjacent}$

1) We know angle A and side c, and we're trying to find the length of side b. Which trig function relates all three of those variables? Well,

$\cos A = \frac{adj}{hyp}= \frac{b}{c}$

so it looks like cosine is the best choice here. Plugging in the values and solving for b, we see that

$\cos 72^{\circ} = \frac{b}{10}$
$b=10 \cos 72^{\circ} \approx \bf 3.1$, to the nearest tenth.
(Make sure your calculator is in degrees mode and not radians!)

2) This is really the same problem as #1, just in a word problem. We know angle $\angle ACB$ and length $AC$, and we're trying to find the width $w$. The function that relates all of these is tangent:

$\tan \angle ACB = \frac{opp}{adj} = \frac{w}{AC}$

So,

$\tan 61^{\circ} = \frac{w}{150 ft.}$
$w = (150 ft.) \tan 61^{\circ} \approx \bf 270.6 ft.$, to the nearest tenth.

3) This question is ambiguous to me, since it doesn't say which angle is the right angle. Does it include a diagram that we can't see? In any case, I will assume that C is the right angle, because that ends up giving nice integers for the side lengths. (Make sure to change your answers if I'm wrong about that.)

To find the length of side a, let's just use the Pythagorean theorem:

$a^2 + b^2 = c^2$
$a = \sqrt{c^2 - b^2} = \sqrt{17^2 - 15^2} = \bf 8$

We can then use the side lengths to find angles A and B in the right triangle. This time, we'll use the trig functions in "reverse" by using the inverse. I'll use the given sides b and c to solve for the angles:

$\cos A = \frac{adj}{hyp} = \frac{b}{c} = \frac{15}{17}$
$A = \cos^{-1} (\frac{15}{17}) \approx \bf 28.1^{\circ}$

$\sin B = \frac{opp}{hyp} = \frac{b}{c} = \frac{15}{17}$
$B = \sin^{-1} (\frac{15}{17}) \approx \bf 61.9^{\circ}$

Phew, hope all that helped!