# Engineering

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Triangle ABD in the diagram has a right angle A and sides AD = 4.9cm and AB = 7.0cm. side AB is extended to C so that ABC is a straight line.
a) use pythagoras theorem in triangle ABD to find the length of BD.
b)use trigonometry in triangle ABD to find angle ABD. hence find angle DBC.
c)angle BCD=21* use the sine formula in triangle BCD to find the length of CD
d)use the cosine formula in triangle BCD to find the length of BC

• geometry/trigonometry -

a. Pythagoras theorem
BD²=AD²+AB²
AD and AB are known.
Solve for BD.

b. Tan(∠ABD) = opposite/adjacent = DA/AB
Solve for ∠ABD.
Since ∠ABD and ∠DBC form a straight line, they are supplementary (add up to 180°).
Solve for ∠DBC:
∠DBC = 180 - ∠ABD

c. Sine rule
The sine rule applies when all angles are known plus one side, or two sides are known and one angle which is not the included angle.
The given triangle applies in to the first case.
BD, ∠DBC, ∠BCD are known, so apply the sine rule:
a/sin(A)=b/sin(B)=c/sin(C)
where A is the angle opposite side a, etc.
BD/sin(∠BCD)=DC/sin(∠DBC)
Solve for DC.

d. The cosine rule applies when two adjacent sides and the included angle are known.
By now, BD, DC and ∠BDC are known.
Note: ∠BDC = 180-(∠DBC+∠BCD)
So apply the cosine rule to solve for c
c²=a²+a²-2abcos(C)
or
BC²=BD²+CD²-2 BD CD cos(∠BDC)

Post to check your results if you wish.

• Engineering -

thank u

• geometry/trigonometry -

Triangle ABC is similar to triangle APQ with angle CBA equal to angle QPA and angle A common to both triangles. the length of sides AB, AP and PQ are 90cm,60cm and 30 cm. calculate the length of the side BC.

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