Question

HELPPPP please help me answer this question-

Answer 1

Answer:

$\displaystyle GH = 2 \sqrt{30}$

$\displaystyle FH = 2 \sqrt{10}$

$\displaystyle m \angle G = {30}^{ \circ}$

Step-by-step explanation:

QUESTION-2:

we are given a right angle triangle

it's a 30-60-90 triangle of which FH is the shortest side

remember that,in case of 30-60-90 triangle the the longest side is twice as much as the shortest side thus

our equation is

$\displaystyle 2FH=4\sqrt{10}$

divide both sides by 2

$\displaystyle\frac{2FH}{2}= \frac{ 4 \sqrt{10} }{2}$

$\displaystyle FH =2 \sqrt{10}$

Question-1:

in order to figure out GH we can use Trigonometry because the given triangle is a right angle triangle

as we want to figure out GH we'll use sin function

remember that,

$\displaystyle \sin( \theta) = \frac{opp}{hypo}$

let our opp, hypo and $\theta$ be GH, 4√10 and 60° respectively

thus substitute:

$\displaystyle \sin( {60}^{ \circ} ) = \frac{GH}{4 \sqrt{10} }$

recall unit circle:

$\displaystyle \frac{ \sqrt{3} }{2} = \frac{GH}{4 \sqrt{10} }$

cross multiplication:

$\displaystyle 2 GH = 4 \sqrt{10} \times \sqrt{3}$

simplify multiplication:

$\displaystyle 2 GH = 4 \sqrt{30}$

divide both sides by 2:

$\displaystyle GH = 2 \sqrt{30}$

QUESTION-3:

Recall that, the sum of the interior angles of a triangle is 180°

therefore,

$\displaystyle m \angle G + {60}^{ \circ} + {90}^{ \circ} = {180}^{ \circ}$

simplify addition:

$\displaystyle m \angle G + {150}^{ \circ} = {180}^{ \circ}$

cancel 150° from both sides

$\displaystyle m \angle G = {30}^{ \circ}$

Answer 2

ANSWER:

It is a 30-60-90 right angle triangle with base FH.

So, 2FH = 4√10

FH = 2√10.

Now, sin60° = GH/4√10

√3/2 = GH/4√10

GH = √3/2 × 4√10

GH = 2√30.

By angle sum property,

m∠G = 180° - 150°

m∠G = 30°.