Question

What is the length of side AD?. . . 6 cm. 8.5 cm. 10.4 cm. 12 cm

Answer 1

The length of side AD is 6 cm.

$\boxed{ \ The \ Answer \ is \ A \ }$

Further explanation

We solve this problem using the principle of trigonometric ratio.

Look at the figure.  

Given that the triangle ABC is right-angled at B

• Length of AB = 12 cm
• ∠ C = 30°
• The segment of BD is perpendicular to AC side.

We start from this statement:

“sum of measures of angles in a triangle is 180°.”

In the triangle ABC:

∠A + ∠B + ∠C = 180°

∠A + 90° + 30° = 180°

∠A + 120° = 180°

∠A = 180° - 120°

∠A = 60°

See the picture again. The triangle ABD is a right angle at D.  

About angle D:

• AB is the hypotenuse
• AD is the adjacent side

Based on trigonometric ratios, the relationship between AD, AB, and angle A is as follows:

$\boxed{ \ cos \ A = \frac{adjacent}{hypotenuse} \ }$

$\boxed{ \ cos \ A = \frac{AD}{AB} \ }$

$\boxed{ \ cos \ 60^0 = \frac{AD}{12} \ }$

$\boxed{ \ \frac{1}{2} = \frac{AD}{12} \ }$

$\boxed{ \ AD = 12 \times \frac{1}{2} \ }$

We get the length of AD is 6 cm.

Note:

$\boxed{ \ sin \theta = \frac{opposite}{hypotenuse} \ } \ \boxed{ \ S = O / H \ }$

$\boxed{ \ cos \theta = \frac{adjacent}{hypotenuse} \ } \ \boxed{ \ C = A / H \ }$

$\boxed{ \ tan \theta = \frac{opposite}{adjacent} \ } \ \boxed{ \ T = O / A \ }$

$\boxed{ \ sin \ 30^0 = cos \ 60^0 = \frac{1}{2} \ }$

$\boxed{ \ sin \ 45^0 = cos \ 45^0 = \frac{1}{2} \sqrt{2} \ }$

$\boxed{ \ sin \ 60^0 = cos \ 30^0 = \frac{1}{2} \sqrt{3} \ }$

$\boxed{ \ tan \ 30^0 = \frac{1}{\sqrt{3}} = \frac{1}{3} \sqrt{3} \ }$

$\boxed{ \ tan \ 45^0 = 1 \ }$

$\boxed{ \ tan \ 60^0 = \sqrt{3} \ }$

Learn more

1. Finding the value of sine of angle brainly.com/question/9610296
2. Multiple choice about trigonometric ratios brainly.com/question/1445273
3. Trigonometric identities brainly.com/question/1980819

Keywords: the length, side AD, right-angled, adjacent, hypotenuse, opposite, sin, cos, tan, sum of measures of angles in triangle are 180°, 30°, 60°

Answer 2

In triangle ABC, AC = 12/ (sin30) = 12 / (1/2) = 24
DC = 24-x
DB = DC tan 30 = (24-x)  tan30 =(24−x)/√3


In triangle ADB using Pythagorean Theoremx2+((24−x)/√3)2=12^2x2+(24−x)^2/3=12^23x2+(24−x)^2=4324x2−48x+576=4324x2−48x+144=0x2−12x+36=0
x1 = x2 =6

AD = AC - DC = 24- (24-x) = 6