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ASHA 777 [7]
3 years ago
14

An obtuse triangle with area 12 has two sides of lengths 4 and 10. Find the length of the third side. (There are two answers.) (

Use law of cosines) ...?
Mathematics
1 answer:
mart [117]3 years ago
3 0
This question can be solved using the Herons equation where 
A = SQRT [s*(s-a)*(s-b)*(s-c)]

A = area of the triangle
a, b and c are the sides of the triangle

s = (a+b+c)/2

Since we are given the area, we can express "s" as a function of the third side c. This can be substituted in the original equation so as to obtain an expression to solve for the third side c

s = (4+10+c)/2 = (14+c)/2

using the solver function of the calculator or MS Excel, the third side is 7.21 units
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3 years ago
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romanna [79]

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Step-by-step

7 0
4 years ago
Find the angle between the given vectors to the nearest tenth of a degree. u = , v = (2 points)
Liono4ka [1.6K]

Answer:

<h2>3.6°</h2>

Step-by-step explanation:

The question is incomplete. Here is the complete question.

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

we will be using the formula below to calculate the angle between the two vectors;

u*v = |u||v| cos \theta

\theta is the angle between the two vectors.

u = 8i + 7j and v = 9i+7j

u*v = (8i + 7j )*(9i + 7j )

u*v = 8(9) + 7(7)

u*v = 72+49

u*v = 121

|u| = √8²+7²

|u| = √64+49

|u| = √113

|v| = √9²+7²

|v| = √81+49

|v| = √130

Substituting the values into the formula;

121= √113*√130 cos θ

cos θ = 121/121.20

cos θ = 0.998

θ = cos⁻¹0.998

θ = 3.6° (to nearest tenth)

Hence, the angle between the given vectors is 3.6°

5 0
3 years ago
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