FIND THE MISSING SIDE OF THE TRIANGLE INSCRIBED IN A SEMICIRCLE

Problem 2 :

Solution :

∠BAC = 90

Using Pythagorean theorem :

AB² + AC² = BC²

5² + 12² = x²

25 + 144 = x²

x² = 144 + 25

x² = 169

x = 13

So, the value of x is 13 cm.

Problem 3 :

Solution :

AC² + BC² = AB²

6² + 6² = x²

36 + 36 = x²

72 = x²

x = 8.48

So, the value of x is 8.48 cm.

Calculate the length of sides labelled in the circles below

Problem 4 :

Solution :

In the right triangle ABC,

AB = Opposite side, AC = Adjacent side and BC = Hypotenuse

cos θ = Adjacent side / Hypotenuse

cos 28 = AC/BC

cos 28^º = x/12

0.882 = x/12

x = 0.882(12)

x = 10.58

So, the value of x is 10.58.

Problem 5 :

Solution :

In the right triangle ABC,

AB = Hypotenuse, AC = Opposite side and BC = Adjacent side.

sin 55^º = opposite/hypotenuse

= AC/AB

= x/9

0.82 = x/9

x = 0.82 × 9

x = 7.38

So, the value x is 7.38.

Problem 6 :

Solution :

In the right triangle ABC, AB = Hypotenuse, BC = Opposite side and AC = Adjacent side

tan 49^º = opposite/adjacent

tan 49 = BC/AC

1.15 = 8/AC

AC = 8/1.15

AC = 6.95

AC = 7 cm

So, the value of x is 7 cm.