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OBLIQUE TRIANGLES

Authored by MARITES BERAN

Mathematics

9th - 10th Grade

CCSS covered

Used 14+ times

OBLIQUE TRIANGLES
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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

Media Image

Which law would you need to solve the given problem?

Law of Sines

Law of Cosines

Both

Neither

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

2.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

To solve triangle ABC, where a = 5, b = 6, c = 7. Which of the following would you need to use first?

Law of Sines

Law of Cosines

Pythagorean Theorem

None of the above

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

3.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

Media Image

Find the value of side AB.

27.99 units

21.74 units

18.84 units

13.59 units

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

4.

MULTIPLE CHOICE QUESTION

45 sec • 1 pt

If C = 135°, a = 52, and A = 30°, what is the value of c?

52

52252\sqrt{2}

5263\frac{52\sqrt{6}}{3}

26326\sqrt{3}

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

5.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

Media Image

Which of the following would you use to find the length of AC using Sine Rule?

sin⁡ 51.381°=sin⁡ AC85°\frac{\sin\ 51.3}{81\degree}=\frac{\sin\ AC}{85\degree}

51.3sin⁡ 81°=sin⁡ 85°AC\frac{51.3}{\sin\ 81\degree}=\frac{\sin\ 85\degree}{AC}

51.3sin⁡ 81°=ACsin⁡ 85°\frac{51.3}{\sin\ 81\degree}=\frac{AC}{\sin\ 85\degree}

51.3sin⁡ 85°=ACsin⁡ 81°\frac{51.3}{\sin\ 85\degree}=\frac{AC}{\sin\ 81\degree}

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

6.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

Media Image

Which of the following is the correct formula to find the measure of angle A?

 cos⁡−1 a2+b2−c22ab\cos^{-1}\ \frac{a^2+b^2-c^2}{2ab}  

 cos⁡−1 b2+c2−a22bc\cos^{-1}\ \frac{b^2+c^2-a^2}{2bc} 

 cos⁡−1 a2+b2−c22ab\cos^{-1}\ \frac{a^2+b^2-c^2}{2ab} 

 cos⁡−1 a2+c2−b22ac\cos^{-1}\ \frac{a^2+c^2-b^2}{2ac} 

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

7.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

Media Image

What is the length of PR?

PR2=9.2+8.5−(9.2)(8.5)cos⁡ 89°PR^2=9.2+8.5-\left(9.2\right)\left(8.5\right)\cos\ 89\degree

PR2=9.22+8.52−(9.2)2(8.5)2cos⁡ 89°PR^2=9.2^2+8.5^2-\left(9.2\right)^2\left(8.5\right)^2\cos\ 89\degree

PR2=9.22+8.52−2(9.2)(8.5)cos⁡ 89°PR^2=9.2^2+8.5^2-2\left(9.2\right)\left(8.5\right)\cos\ 89\degree

PR2=9.22+8.52+2(9.2)(8.5)cos⁡ 89°PR^2=9.2^2+8.5^2+2\left(9.2\right)\left(8.5\right)\cos\ 89\degree

Tags

CCSS.HSG.SRT.D.10

CCSS.HSG.SRT.D.11

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