(a)[3]
Show that $BD = 9.55\,\text{cm}$, when rounded to 2 decimal places.
(b(i))[1]
Show that $BCD = 112^{\circ}$.
(b(ii))[3]
Calculate the value of $CD$.
(c)[3]
Find the minimum distance from $D$ to $AB$.
Mathematics 0580 · IGCSE · Non-right-angled triangles
Non-right-angled triangles — practice question
ABCD is a quadrilateral in which $AB = 10.4\,\text{cm}$ and $AD = 6.5\,\text{cm}$. Angle $DAB = 64^{\circ}$, angle $BDC = 26^{\circ}$ and angle $DBC = 42^{\circ}$.
Worked solution & mark scheme
This 10-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Use the cosine rule: $\sqrt{10.4^2+6.5^2-2\times10.4\times6.5\cos64}$” …
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