SME
Grade 10 Math · Geometry
Right-triangle trigonometry: sine, cosine and tangent
Use this for “A right triangle has legs 5 and 12 and hypotenuse 13. Find sin A, cos A, tan A and the angle A.”
Diagram
sin A = a/c = 5/13 ≈ 0.3846
cos A = b/c = 12/13 ≈ 0.9231
tan A = a/b = 5/12 ≈ 0.4167
This is an example. Type your own values to start a new problem.
sin(
A
°)
=
a
c
cos(
A
°)
=
b
c
tan(
A
°)
=
a
b
s
Sine of A
worked out (not typed)
k
Cosine of A
worked out (not typed)
t
Tangent of A
worked out (not typed)
Formulas
a
2
+
b
2
=
c
2
5
2
+
12
2
=
13
2
tan(
A
°)
=
a
b
tan(22.6199°)
=
5
12
sin(
A
°)
=
a
c
sin(22.6199°)
=
5
13
cos(
A
°)
=
b
c
cos(22.6199°)
=
12
13
s
=
a
c
5
13
=
5
13
k
=
b
c
12
13
=
12
13
t
=
a
b
5
12
=
5
12
Assumptions
•
The right angle is at C. From angle A: a is the opposite side, b the adjacent side, c the hypotenuse.
•
sin A = opposite ÷ hypotenuse, cos A = adjacent ÷ hypotenuse, tan A = opposite ÷ adjacent.
•
The ratios depend only on the angle, not on the triangle’s size.
Step-by-step
WE KNOW
a = 5, b = 12
FIND
hypotenuse c (c), angle A (A), sine of A (s), cosine of A (k), tangent of A (t)
Step 1 · Find hypotenuse c (c)
Use a² + b² = c²
The
legs
and
the
hypotenuse
of
a
right
triangle:
a
2
+
b
2
=
c
2
.
c
=
√
a
2
+
b
2
c
=
√
5
2
+
12
2
c
=
√
25 + 144
c
=
√
169
c
=
13
Step 2 · Find angle A (A)
Use tan(A°) = a ÷ b
The inverse tangent turns the ratio back into the angle.
A
=
tan⁻¹(
a
b
)
A
=
tan⁻¹(
5
12
)
A
=
22.6199°
Step 3 · Find sine of A (s)
Use s = a ÷ c
The
sine
of
A
:
the
opposite
side
over
the
hypotenuse.
s
=
5
13
s
=
5
13
Step 4 · Find cosine of A (k)
Use k = b ÷ c
The
cosine
of
A
:
the
adjacent
side
over
the
hypotenuse.
k
=
12
13
k
=
12
13
Step 5 · Find tangent of A (t)
Use t = a ÷ b
The
tangent
of
A
:
the
opposite
side
over
the
adjacent
side.
t
=
5
12
t
=
5
12
Check
Put the numbers back in. Both sides should match.
5
2
+
12
2
=
13
2
✓
tan(22.6199°)
=
5
12
✓
sin(22.6199°)
=
5
13
✓
cos(22.6199°)
=
12
13
✓
5
13
=
5
13
✓
12
13
=
12
13
✓
5
12
=
5
12
✓
Topic check
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