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

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

Explanation:

`\Large\boxed{\sf Formulae}`

Sine Rule

`(\sin A)/(a)=(\sin B)/(b)=(\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}=\frac12ab\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 (\sin 80)/(CD)=(\sin ADC)/(4.9)`

`\implies ADC=\sin^(-1)\left((4.9 \sin 80)/(CD)\right)=58.568949^(\circ)`

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

Use sine rule to calculate DB:

`\implies (DB)/(\sin 25)=(CD)/(\sin CBD)`

`\implies DB=(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=\frac12(CD)(DB)\sin CDB`

`\implies A=\frac12(5.655381673)(4.322471258)\sin (121.431051)`

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