The Student Room Group

This discussion is now closed.

Check out other Related discussions

Sine and Cosine rule question

Hello i have a question on Sine and Cosine rules...

The sides of a triangle are 7cm, 9am and 12cm. Find its angles and its area.

In a quadrilateral ABCD, AB = 4cm BC = 5cm CD = 7cm DA = 5cm and angle ABC is 87 degress. Find angel ADC.


Thanks a lot
From John
Reply 1
Avatar for Excalibur
Excalibur
15
For the triangle one, label each side a, b and c. The angle opposite each side, you call angle A, B and C respectively. You first need to find out the angle A.

cosA = (b^2 + c^2 - a^2)/2bc

a = 12, b = 9 and c = 7

Substitute to get

cosA = (9^2 + 7^2 - 12^2)/2 x 9 x 7)
cosA = -0.1111
A = cos inverse (-0.111)
A = 96.37 º

To work out angle B, use

cosB = (a^2 + c^2 - b^2)/2 x a x c

Then take away B and A from 180 to get C.
Reply 2
Avatar for Neo1
Neo1
OP
4
Ah right...

What about this question:

In a quadrilateral ABCD, AB = 4cm BC = 5cm CD = 7cm DA = 5cm and angle ABC is 87 degress. Find angel ADC.


Thanks a lot
From John
Reply 3
Avatar for Ana
Ana
19
Neo1
Ah right...

What about this question:

In a quadrilateral ABCD, AB = 4cm BC = 5cm CD = 7cm DA = 5cm and angle ABC is 87 degress. Find angel ADC.


Thanks a lot
From John


Split it up into two triangles - ABC and ACD. Use the cosine rule (c2 = a2 + b2 - 2abcosC) to find the length of the line AC, then the other form of it (cosC = (a2 + b2 - c2)/2ab) to find the angle ADC :smile:
Reply 4
Avatar for shizznitch
shizznitch
draw diagrams!!!!!!

they always help in this kind of question, you can 'break it down'! lol.

Related discussions

Latest

Trending

Trending

Articles for you