Question

Solve for BC without working out the size of angle A

Answer 1

BC = 6 units

Given following:

• AC = 4 units

• tan(A) = 3/2

Solve for BC:

Formula:

`\rightarrow \sf tan(A) =(opposite)/(adjacent)`

insert values and make them proportional

`\rightarrow \sf tan(A) =(BC)/(AC) = (3)/(2) = (6)/(4)`

So, BC is 6 units.

Answer 2

1) BC = 6

2) B = 33.7° (nearest tenth)

3)
`\sf AB=2√(13)`

Explanation:

Trigonometric ratios

`\sf \sin(\theta)=(O)/(H)\quad\cos(\theta)=(A)/(H)\quad\tan(\theta)=(O)/(A)`

where:

• 
  `\theta` is the angle
• O is the side opposite the angle
• A is the side adjacent the angle
• H is the hypotenuse (the side opposite the right angle)

Question 1

Given:

• Angle = A
• Side opposite angle = BC
• Side adjacent angle = AC = 4

Substituting the values into the tan ratio:

`\implies \sf \tan A=(BC)/(AC)=(BC)/(4)`

Given:

`\sf \tan A=(3)/(2)`

Therefore:

`\implies \tan \sf A = \tan \sf A`

`\implies \sf (BC)/(4)=(3)/(2)`

`\implies \sf BC=(4 \cdot 3)/(2)=6`

Question 2

Given:

• Angle = B
• Side opposite angle = AC = 4
• Side adjacent angle = BC = 6

Substituting the values into the tan ratio and solving for B:

`\implies \sf \tan B=(AC)/(BC)`

`\implies \sf \tan B=(4)/(6)`

`\implies \sf B=\tan ^(-1)\left((4)/(6)\right)`

`\implies \sf B=33.69006753...^(\circ)`

`\implies \sf B=33.7^(\circ)\:(nearest\:tenth)`

Question 3

Pythagoras’ Theorem

`a^2+b^2=c^2`

(where a and b are the legs, and c is the hypotenuse, of a right triangle)

Given:

• a = AC = 4
• b = BC = 6
• c = AB

Substituting the values into the formula and solving for AB:

`\implies \sf AC^2+BC^2=AB^2`

`\implies \sf 4^2+6^2=AB^2`

`\implies \sf 16+36=AB^2`

`\implies \sf AB^2=52`

`\implies \sf AB=√(52)`

`\implies \sf AB=2√(13)`