Help please!
find the area of the isoceles triangle

This is a long answer. I got 82.5

Step-by-step explanation:

Area of isoceles triangle is

where b is the base and h is height.

Let draw a altuide going through Point D that split side FE into 2 equal lines. Let call that point that is equidistant from FE, H.

Since it is a altitude,it forms a right angle. So angle H=90.

Angle H is equidistant from F and E so

FH=11

EH=11.

The height is still unknown.

We can use pythagorean theorem to find side h but we need to know the slanted side or side DF to use the theorem.

Using triangle DFH, we know that angle H is 90 and angle F is 34. So using triangle interior rule,Angle D equal 56.

We can use law of sines to find side DF

Plug in the numbers

sin of 90 =1 so

Side df is about 13.3 inches.

Since we know our slanted side is 13.3 we can set up our pythagorean theorem equation,

is approximately 7.5 so dh=7.5 approximately.

Now using base times height times 1/2 multiply them out

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this