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artcher [175]
3 years ago
10

Find the altitude of an isosceles triangle with a vertex angle (that is, the non-base angle) of 70° and a base of 246. Draw a di

agram and find the length of the altitude

Mathematics
2 answers:
ki77a [65]3 years ago
7 0
Altitude = 175.6 units

mestny [16]3 years ago
4 0

Answer:

AD=175.66 units

Step-by-step explanation:

It is given that ABC is an isosceles triangle in which base is 246 and the non base angle is 70°.

Thus, applying the angle sum property in ΔABC, we get

∠A+∠B+∠C=180°

⇒70°+2∠B=180°

⇒∠B=55°

Thus, ∠B=∠C=55°.

Now, Draw AD perpendicular to BC such that BD=DC=123, then applying trigonometry in ΔADC, we have

\frac{AD}{DC}=tan55^{\circ}

⇒\frac{AD}{123}=1.428

⇒AD=123(1.428)

⇒AD=175.66 units

Thus, the length of the altitude is 175.66.

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A football is thrown from the top of the stands, 50 feet above the ground at an initial velocity of 62 ft/sec and at an angle of
Anvisha [2.4K]

a. i. The parametric equation for the horizontal movement is x = 43.84t

ii. The parametric equation for the vertical movement is y = 50 + 43.84t

b. the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

<h2>a. Parametric equations</h2>

A parametric equation is an equation that defines a set of quantities a functions of one or more independent variables called parameters.

<h3>i. Parametric equation for the horizontal movement</h3>

The parametric equation for the horizontal movement is x = 43.84t

Since

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  • the initial velocity, v = 62 ft/s,

the horizontal component of the velocity is v' = vcosФ.

So, the horizontal distance the football moves in time, t is x = vcosФt

= vtcosФ

= 62tcos45°

= 62t × 0.7071

= 43.84t

So, the parametric equation for the horizontal movement is x = 43.84t

<h3>ii Parametric equation for the vertical movement</h3>

The parametric equation for the vertical movement is y = 50 + 43.84t

Also, since

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  • the initial velocity, v = 62 ft/s,

the vertical component of the velocity is v" = vsinФ.

Since the football is initially at a height of h = 50 feet, the vertical distance the football moves in time, t relative to the ground is y = 50 + vsinФt

= 50 + vtcosФ

= 50 + 62tsin45°

= 50 + 62t × 0.7071

= 50 + 43.84t

<h3>b. Location of football at maximum height relative to starting point</h3>

The location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Since the football reaches maximum height at t = 1.37 s

The x coordinate of its location at maximum height is gotten by substituting t = 1.37 into x = 48.84t

So, x = 43.84t

x = 43.84 × 1.37

x = 60.0608

x ≅ 60.1 ft

The y coordinate of the football's location at maximum height relative to the ground is y = 50 + 48.84t

The y coordinate of the football's location at maximum height relative to the starting point is y - 50 = 48.84t

So,  y - 50 = 48.84t

y - 50 = 43.84 × 1.37

y - 50 = 60.0608

y - 50 ≅ 60.1 ft

So, the location of the football at its maximum height relative to the starting point is (60.1 ft, 60.1 ft)

Learn ore about parametric equations here:

brainly.com/question/8674159

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The process:

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