In ΔABC, and m∠ABC = 90°. D and E are the midpoints of and , respectively. If the length of is 9 units, the length of is units and m∠CAB is °.

Applying the midsegment theorem and the definition of isosceles triangle:

DE = 4.5 units

m∠CAB = 45°

The image that shows ΔABC is attached below.

Since AB = BC, therefore, ΔABC is an isosceles triangle.

This implies that, the base angles will be equal.

Thus:

If m∠ABC = 90°, therefore,

m∠CAB = ½(180 - 90)

m∠CAB = 45°.

DE is the midsegment of the triangle, and is parallel to the third side, CA = 9 units.

Based on the midsegment theorem, we have the following equation:

DE = ½(9)

DE = 4.5 units.

Therefore, applying the midsegment theorem and the definition of isosceles triangle:

DE = 4.5 units

m∠CAB = 45°

Learn more about midsegment theorem on:

link

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°

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