Solved W WINTON TRIGONOMETRY - WORKSHEET AC = 2.995-5 TASK B ... - Free Printable

Let’s solve each triangle one by one. We’re told not to use Pythagoras — so we’ll use trig ratios: sine, cosine, and tangent.

Remember for a right-angled triangle:

- sin(θ) = opposite / hypotenuse
- cos(θ) = adjacent / hypotenuse
- tan(θ) = opposite / adjacent

We’ll label the sides relative to the given angle (not the right angle).

---

Triangle 1:

Angle at A = 30°, hypotenuse AC = 6 cm
Right angle at B → side BC is opposite to angle A, AB is adjacent.

We want to find AB and BC.

→ Use sin(30°) = opposite/hypotenuse = BC / 6
sin(30°) = 0.5 → BC = 6 × 0.5 = 3 cm

→ Use cos(30°) = adjacent/hypotenuse = AB / 6
cos(30°) ≈ 0.866 → AB = 6 × 0.866 ≈ 5.196 cm (we can round to 5.2 cm if needed, but let’s keep precise for now)

But wait — maybe they expect exact values? Since it’s 30°, we know:

In a 30-60-90 triangle, sides are in ratio 1 : √3 : 2
Hypotenuse = 6 → shortest side (opposite 30°) = 3 → other leg = 3√3 ≈ 5.196

So:
BC = 3 cm
AB = 3√3 cm ≈ 5.2 cm

But since the problem says “without using Pythagoras”, and doesn’t specify exact or decimal, I’ll give both where helpful. But for consistency, let’s use calculator values as per worksheet style.

Actually, looking at the top of the page: they computed AC = √24.995 = 5 — so they’re using decimals. So we’ll use decimals.

So Triangle 1:
BC = 6 × sin(30°) = 6 × 0.5 = 3.0 cm
AB = 6 × cos(30°) = 6 × 0.8660 ≈ 5.196 cm → let’s say 5.20 cm (to 2 dp)

Wait — actually, let me check with more precision.

cos(30°) = √3/2 ≈ 0.8660254 → 6 × that = 5.196152 → so 5.20 cm is fine.

But maybe they want exact? The note says “triangles are not to scale” and no instruction on rounding. Let’s go with 2 decimal places unless it’s exact.

Actually, sin(30°) is exactly 0.5, so BC = 3.00 cm.

AB = 6 * cos(30°) = 6 * (√3/2) = 3√3 ≈ 5.196 → I’ll write 5.20 cm.

But let’s see what the worksheet expects. In problem 6, they have 45°, which gives nice numbers. Maybe for 30°, they accept 3 and 5.2.

I’ll proceed with decimals to 2 places.

---

Triangle 2:

Angle at A = 20°, hypotenuse AC = 9 cm
Right angle at B → BC opposite, AB adjacent.

BC = 9 × sin(20°)
sin(20°) ≈ 0.3420 → BC ≈ 9 × 0.3420 = 3.078 cm → 3.08 cm

AB = 9 × cos(20°)
cos(20°) ≈ 0.9397 → AB ≈ 9 × 0.9397 = 8.4573 cm → 8.46 cm

---

Triangle 3:

Angle at C = 25°, side CB = 12 cm (adjacent to angle C), right angle at B.

So, angle at C = 25°, adjacent side = CB = 12 cm, hypotenuse = AC, opposite = AB.

We need AB and AC.

→ tan(25°) = opposite / adjacent = AB / 12
tan(25°) ≈ 0.4663 → AB = 12 × 0.4663 ≈ 5.5956 cm → 5.60 cm

→ cos(25°) = adjacent / hypotenuse = 12 / AC
So AC = 12 / cos(25°)
cos(25°) ≈ 0.9063 → AC ≈ 12 / 0.9063 ≈ 13.240 cm → 13.24 cm

Alternatively, sin(25°) = AB / AC → but we already have AB, so better to use cos.

Yes.

---

Triangle 4:

Angle at C = 40°, side CB = 10 cm (adjacent to angle C), right angle at B.

So, adjacent = 10 cm, angle = 40°, need AB (opposite) and AC (hypotenuse).

→ tan(40°) = opposite / adjacent = AB / 10
tan(40°) ≈ 0.8391 → AB = 10 × 0.8391 = 8.391 cm → 8.39 cm

→ cos(40°) = adjacent / hypotenuse = 10 / AC
AC = 10 / cos(40°)
cos(40°) ≈ 0.7660 → AC ≈ 10 / 0.7660 ≈ 13.055 cm → 13.06 cm

---

Triangle 5:

Angle at A = 30°, side AB = 2 cm (adjacent to angle A), right angle at B.

Need BC (opposite) and AC (hypotenuse).

→ tan(30°) = opposite / adjacent = BC / 2
tan(30°) ≈ 0.5774 → BC = 2 × 0.5774 = 1.1548 cm → 1.15 cm

→ cos(30°) = adjacent / hypotenuse = 2 / AC
AC = 2 / cos(30°) ≈ 2 / 0.8660 ≈ 2.309 cm → 2.31 cm

Note: tan(30°) = 1/√3 ≈ 0.577, so BC = 2/√3 ≈ 1.1547

---

Triangle 6:

Angle at A = 45°, side AB = 3 cm (adjacent), right angle at B.

Since it’s 45°, it’s isosceles right triangle → legs equal.

So BC = AB = 3.00 cm

Hypotenuse AC = AB × √2 = 3 × 1.4142 ≈ 4.2426 cm → 4.24 cm

Using trig:

tan(45°) = 1 = BC / 3 → BC = 3 cm

cos(45°) = 3 / AC → AC = 3 / cos(45°) = 3 / (√2/2) = 3 × 2/√2 = 3√2 ≈ 4.2426 → same.

---

Now, let’s compile all answers clearly.

I’ll list each triangle with missing sides.

Triangle 1:
Given: angle A=30°, hypotenuse=6cm
Find: AB (adjacent), BC (opposite)
AB = 6 × cos(30°) ≈ 6 × 0.8660 = 5.196 → 5.20 cm
BC = 6 × sin(30°) = 6 × 0.5 = 3.00 cm

Triangle 2:
angle A=20°, hyp=9cm
AB = 9 × cos(20°) ≈ 9 × 0.9397 = 8.457 → 8.46 cm
BC = 9 × sin(20°) ≈ 9 × 0.3420 = 3.078 → 3.08 cm

Triangle 3:
angle C=25°, adjacent CB=12cm
AB (opposite) = 12 × tan(25°) ≈ 12 × 0.4663 = 5.5956 → 5.60 cm
AC (hyp) = 12 / cos(25°) ≈ 12 / 0.9063 = 13.240 → 13.24 cm

Triangle 4:
angle C=40°, adjacent CB=10cm
AB (opposite) = 10 × tan(40°) ≈ 10 × 0.8391 = 8.391 → 8.39 cm
AC (hyp) = 10 / cos(40°) ≈ 10 / 0.7660 = 13.055 → 13.06 cm

Triangle 5:
angle A=30°, adjacent AB=2cm
BC (opposite) = 2 × tan(30°) ≈ 2 × 0.5774 = 1.1548 → 1.15 cm
AC (hyp) = 2 / cos(30°) ≈ 2 / 0.8660 = 2.309 → 2.31 cm

Triangle 6:
angle A=45°, adjacent AB=3cm
BC (opposite) = 3 × tan(45°) = 3 × 1 = 3.00 cm
AC (hyp) = 3 / cos(45°) ≈ 3 / 0.7071 = 4.2426 → 4.24 cm

All calculated without Pythagoras, using trig functions only.

Final Answer:

Triangle 1: AB = 5.20 cm, BC = 3.00 cm
Triangle 2: AB = 8.46 cm, BC = 3.08 cm
Triangle 3: AB = 5.60 cm, AC = 13.24 cm
Triangle 4: AB = 8.39 cm, AC = 13.06 cm
Triangle 5: BC = 1.15 cm, AC = 2.31 cm
Triangle 6: BC = 3.00 cm, AC = 4.24 cm