Question

C
250
4.9 cm
80°
A
3.8 cm
D
B
ABC is a triangle.
D is a point on AB.
Work out the area of triangle BCD.
Give your answer correct to 3 significant figures.

Answer 1

Answer:

$10.4\: \sf cm^2\:(3\:sf)$

Step-by-step explanation:

$\Large\boxed{\sf Formulae}$

Sine Rule

$\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}$

(where A, B and C are the angles and a, b and c are the sides opposite the angles)

Cosine rule

$c^2=a^2+b^2-2ab\cos C$

Area of a triangle

$\textsf{Area of a triangle}=\dfrac12ab\sin C$

where:

• C is the angle
• c is the side opposite the angle
• a and b are the sides adjacent the angle

$\Large\boxed{\sf Calculation}$

To find the area of ΔBCD, find:

• length of CD
• angle CDB
• length of BD

Then use the sine rule for area of a triangle

Calculate length CD using cosine rule:

$\implies CD^2=4.9^2+3.8^2-2(4.9)(3.8)\cos 80^{\circ}$

$\implies CD^2=31.98334186...$

$\implies CD=5.655381673...\:\sf cm$

Calculate ∠ADC using sine rule:

$\implies \dfrac{\sin 80}{CD}=\dfrac{\sin ADC}{4.9}$

$\implies ADC=\sin^{-1}\left(\dfrac{4.9 \sin 80}{CD}\right)=58.568949^{\circ}$

Therefore, ∠CDB = 180° - 58.568949° = 121.431051°

Use sine rule to calculate DB:

$\implies \dfrac{DB}{\sin 25}=\dfrac{CD}{\sin CBD}$

$\implies DB=\dfrac{5.655381673\sin25}{\sin 33.568949}=4.322471258..\: \sf cm$

$\Large\boxed{\sf Solution}$

Use the sine rule for area of a triangle to find area of ΔBCD:

$\implies A=\dfrac12(CD)(DB)\sin CDB$

$\implies A=\dfrac12(5.655381673)(4.322471258)\sin (121.431051)$

$\implies A=10.4\: \sf cm^2\:(3\:sf)$