Question
ABCD is a quadrilateral where \({\text{AB}} = 6.5,{\text{ BC}} = 9.1,{\text{ CD}} = 10.4,{\text{ DA}} = 7.8\) and \({\rm{C\hat DA}} = 90^\circ \). Find \({\rm{A\hat BC}}\), giving your answer correct to the nearest degree.
▶️Answer/Explanation
Markscheme
\({\text{A}}{{\text{C}}^2} = {7.8^2} + {10.4^2}\) (M1)
\({\text{AC}} = 13\) (A1)
use of cosine rule eg, \(\cos ({\rm{A\hat BC}}) = \frac{{{{6.5}^2} + {{9.1}^2} – {{13}^2}}}{{2(6.5)(9.1)}}\) M1
\({\rm{A\hat BC}} = 111.804 \ldots ^\circ {\text{ }}( = 1.95134 \ldots )\) (A1)
\( = 112^\circ \) A1
[5 marks]
Examiners report
Well done by most candidates. A small number of candidates did not express the required angle correct to the nearest degree.
Question
[Maximum mark: 7]
The diagram shows a triangle ABC with AB=9, AC = AD = 5, \( \hat{B}\)= 30°, C = 45°.
(a) Write down the value of the angle BÂ C. [1]
(b) Find the value of angle B \(\hat{D}\)A and hence the value of angle BÂ D. [2]
(c) Find BC. [2]
(d) Find BD.
▶️Answer/Explanation
Answer:
(a) B \(\hat{A}\) C = 105°
(b) B\(\hat{D}\) A = 135° , B \(\hat{A}\) D = 15°
(c) BC2 = 92 + 52 – 2×9×5×cos105° ⇒ BC = 11.4
BD2 = 92 + 52 – 2×9×5×cos15° ⇒ BD = 4.36
Question
[Maximum mark: 6]
In the triangle ABC, Â = 30°, BC = 3 and AB = 5. Find the two possible values of \(\hat{B}\).
▶️Answer/Explanation
Answer:
\(\sin C=\frac{c\sin A}{a}=\frac{5\times 0.5}{3}\)
\(\hat{C}=56.4^{o}\) and then \(\hat{B}=93.6^{o}\) or \(\hat{C}=123.6^{o}\) and then \(\hat{B}=26.24^{o}\)
Question
[Maximum mark: 7]
In a triangle ABC, Â = 35°, BC = 4 cm and AC = 6.5 cm. Find the possible values of \(\hat{B}\) and the corresponding values of AB.
▶️Answer/Explanation
Answer:
Sine rule: \(\frac{\sin 35}{4}=\frac{\sin \hat{B}}{6.5}\Rightarrow \sin \hat{B}=\frac{6.5\sin 35}{4}\Rightarrow \sin \hat{B}=0.932\)
Hence \(\hat{B}=68.8\) or \(\hat{B}=180-68.8 =111.2\)
If \(\hat{B}=68.8,\) then \(\hat{C}=180-35-68.8=76.2\)
Sine rule again: \(\frac{\sin 35}{4}=\frac{\sin 76.2}{AB}\Rightarrow AB=6.77\)
If \(\hat{B}=111.2,\) then \(\hat{C}=180-35-111.2=33.8 \)
Sine rule again: \(\frac{\sin 35}{4}=\frac{\sin 33.8}{AB}\Rightarrow AB=3.88\)
Question
[Maximum mark: 6]
In a triangle ABC, A\(\hat{B}\) C =30°, AB = 6cm , AC =\(3\sqrt{2}cm\). Find the possible lengths of [BC].
▶️Answer/Explanation
Answer:
METHOD 1:
Using the sine rule: \(\frac{\sin C}{6}=\frac{\sin 30^{o}}{3\sqrt{2}}\Rightarrow \sin C=\frac{1}{\sqrt{2}}\Rightarrow C=45^{o},135^{o}\)
If C = 45° , B = 105° \(\frac{3\sqrt{2}}{\sin 30^{o}}=\frac{BC}{\sin 15^{o}}\Rightarrow BC=2.20 cm\)
If \(C=135^{o}, B=15^{o}\) \(\frac{3\sqrt{2}}{\sin 30^{o}}=\frac{BC}{\sin 15^{o}}\Rightarrow BC=2.20 cm\)
Method 2:
Using the cosine rule: \(AC^{2}=6^{2}+BC^{2}-2(6)(BC)\cos 30^{o}\Rightarrow 18=36+BC^{2}-6\sqrt{3}BC\)
(In fact, this is a quadratic equation: \(BC^{2}-(6\sqrt{3})BC+18=0)\)
BC = 8.20 cm or BC = 2.20 cm.