Radian Measures Question

The non-right angled triangle ABC is such that
AB = 12 mm,
AC = 14 mm,
BC = x mm.

The angle CAB has a value of 0.75 radians.
The point S lies on the line AB such that SC bisects the angle subtended by the lines AC and BC.
AS = y mm.

Find the area of triangle SCB.
Give your answer to the nearest integer.
(5)

(edited 1 week ago)

Reply 1

mqb2766

Original post by Uxie

The non-right angled triangle ABC is such that
AB = 12 mm,
AC = 14 mm,
BC = x cm.
The angle CAB has a value of 0.75 radians.
The point S lies on the line AB such that SC bisects the angle subtended by the lines AC and BC.
AS = y mm.
Find the area of triangle SCB.
Give your answer to the nearest integer.
(5)

What level is this and what have you tried / are you stuck on?

Reply 2

Nitrotoluene

Original post by Uxie

Hello Uxie!
Are you familiar with trigonometry? If you've studied it or are currently studying it, you're likely familiar with the laws of sines, cosines. Do you know the Pythagorean theorem?
If so, I'll provide guidance and support to help you overcome any challenges.

^IT ITALY FLAG.png

^IT
Talk to you soon.
Sandro

(edited 1 week ago)

Reply 3

Uxie

Hello! Thank for replying to my question :smile:

As a matter of fact I made the question a very long time ago and I simply posted it to see if anyone had any creative solutions!

Reply 4

mqb2766

Original post by Uxie

Hello! Thank for replying to my question :smile:

As a matter of fact I made the question a very long time ago and I simply posted it to see if anyone had any creative solutions!

How did you do it? Its "only" 5 marks and there is a fairly simple/elegant way to get the area (similar to the angle bisection theorem which isnt covered at a level, but its more for inspiration) once you get x.

(edited 1 week ago)

Reply 5

mqb2766

Original post by Uxie

Hello! Thank for replying to my question :smile:

As a matter of fact I made the question a very long time ago and I simply posted it to see if anyone had any creative solutions!

wbf, if you post your method (or at least a brief description), Ill fill in a bit more. There are a few ways you could do it from chasing sides and angles using elementary trig or the cos and sin rules, to using side or area ratios based on the angle bisection.

The side / angle chasing is a bit tedious/repetitive but it gets you there and would probably need a bit more than 5 marks, whereas using either ratio approach would really require a bit more of a hint in the question as its less standard.

Reply 6

Uxie

Hello. So here’s my general method:
(1) calculate the length x (which should say mm) - For this I simply used the cosine rule.
This got me 9.7 mm 1 d.p.

(2) Calculate the length AS by using the angle bisection theorem.
This gives (9.7/14) = (12-y)/y
Simplifying we obtain y = 7.1 mm 1 d.p.

(3) Calculate the length SC.
To do this I used my newly found value of y (7.1) and simply used the cosine rule again as triangle ASC is non-right angled.
This got me SC = 10.0 mm 1 d.p.

(4) Now use any valid method of quickly finding the area of a non-right angled triangle - I used Heron’s formula.
This got me an area of 23.33… mm^2

(5) Giving our answer to the stated accuracy gives us a final answer of 23 mm^2.

Reply 7

mqb2766

Original post by Uxie

Hello. So here’s my general method:
(1) calculate the length x (which should say mm) - For this I simply used the cosine rule.
This got me 9.7 mm 1 d.p.
(2) Calculate the length AS by using the angle bisection theorem.
This gives (9.7/14) = (12-y)/y
Simplifying we obtain y = 7.1 mm 1 d.p.
(3) Calculate the length SC.
To do this I used my newly found value of y (7.1) and simply used the cosine rule again as triangle ASC is non-right angled.
This got me SC = 10.0 mm 1 d.p.
(4) Now use any valid method of quickly finding the area of a non-right angled triangle - I used Heron’s formula.
This got me an area of 23.33… mm^2
(5) Giving our answer to the stated accuracy gives us a final answer of 23 mm^2.

Using angle bisection / heron is (slightly) beyond the usual syllabus (gcse/a level) but theyre not unusual so thats fine. A few ways are below.
1) Very elementary trig - drop an altitude from <B or <C onto the opposite side, then you can chase all the angles and sides using the basic trig functions sin(), cos() and tan() and calc the requried area as you have the necessary sides and angles.
2) Similar to 1) - instead of dropping an altitude, just chase the sides and angles using the cos and sin rules. Again fairly tedious, but it again uses standard formula.
3) Get x (cos rule) and area(ABC) (1/2 bc sin(A)), then as the bisected angle and side SC are common, area(ASC):area(BSC) = 14:x, so area(BSC) = area(ABC)*x/(14+x). Again uses standard formula, but its not that usual to think of area ratios at gcse/alevel so some sort of hint would be usual in a question. Also theres no need to calc y, but its kind of hinted at in the question. This is probably as simple as it gets.
4) similar to 3), use the angle bisection theorem to get AS: SB (which gives the same ratio/final calc as 3). Really the angle bisection theorem is unnecessary, but if youre asking for y, its kind of hinting at a side ratio rather than area ratio.

The question seems decent, but most kids (in an exam) would probably chase angles/sides as in 1) or 2) without any extra hint in the question and theres little in the question to hint at why one approach is going to be better than another. Labelling y in the question kind of suggests a side ratio/angle bisection approach (4), but then thinking you dont need to do that step and can simply do AC:BC as an area ratio (3),. Though as youve done, you can also use heron if you know the 3 sides.