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Precalc Chapter 6 formulas

Authored by Sheri White

Mathematics

11th - 12th Grade

Used 14+ times

Precalc Chapter 6 formulas
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10 questions

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1.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Given the following triangle, which formula would you use to find angle A?

7sin35=asin 105\frac{7}{\sin35}=\frac{a}{\sin\ 105}

sin105c = sin357\frac{\sin105}{c}\ =\ \frac{\sin35}{7}

A = sin1(7sin35sin 105)A\ =\ \sin^{-1}\left(\frac{7\sin35}{\sin\ 105}\right)

180 (35 + 105)180\ -\ \left(35\ +\ 105\right)

2.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Write the formula for finding angle R

arcsin(28sin39/41)\arcsin(28\sin39/41)

28sin39 = 41sin R\frac{28}{\sin39}\ =\ \frac{41}{\sin\ R}

282 + 412 2(28)(41)cos39°28^2\ +\ 41^{2\ }-2\left(28\right)\left(41\right)\cos39\degree

12(28)(41)Sin39°\frac{1}{2}\left(28\right)\left(41\right)Sin39\degree

3.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Write the formula for finding side c

4.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Which is not a valid Law of Cosine formula?

b = (a2 +c22ac CosB)b\ =\ \sqrt{\left(a^2\ +c^2-2ac\ CosB\right)}

cosAa=cosBb\frac{\cos A}{a}=\frac{\cos B}{b}

cos C=(b2 +a2 c2)2(a)(b) \cos\ C=\frac{\left(b^{2\ }+a^2\ -\ c^2\right)}{2\left(a\right)\left(b\right)}\

a2=b2+c2 2bc CosAa^2=b^2+c^2\ -2bc\ CosA

5.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

What formula would you use for finding the area of this triangle?

You need to first find the side p or angle P before finding the area.

You can't find the area of this triangle with the information given

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

 v =<v1, v2> and w = <w1, w2>v\ =<v_1,\ v_2>\ and\ w\ =\ <w_1,\ w_2>  



What is  vwv\cdot w  ?

 v1w1+v2w2v_1w_1+v_2w_2  

 <v1w1 , v2w2><v_1w_{1\ ,}\ v_2w_2>  

 <v1+w1 , v2+w2><v_1+w_1\ ,\ v_2+w_2>  

 v1w1 i + v2w2 jv_1w_1\ i\ +\ v_2w_2\ j  

7.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Given v = 52 and θ = 63\parallel v\parallel\ =\ 52\ and\ \theta\ =\ 63  find the linear combination for vector v.


 52 i + 63j52\ i\ +\ 63j  

 52cos63i +52 sin 63j52\cos63i\ +52\ \sin\ 63j  

 63 cos52 + 63 sin5263\ \cos52\ +\ 63\ \sin52  

 cos6352 i+sin6352j \frac{\cos63}{52}\ i+\frac{\sin63}{52}j\   

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