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3 years ago

How to solve for area of triangle where A=67 degrees, b=11yd, and c=24yd

Mathematics

1 answer:

skad [1K]3 years ago

4 0

Use the formula of an area of a tringle:

$A=\dfrac{1}{2}bc\sin A$

We have b = 11 yd, c = 24 yd and m∠A = 67°

$A=\dfrac{1}{2}(11)(24)\sin67^o$

$\sin67^o\approx0.9205\\\A=(11)(12)(0.9205)=121.506\ yd^2$

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longer leg: 20 in

shorter leg: 15 in

hypotenuse: 25 in

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<u>Pythagorean Theorem</u>

$c^2 = a^2 + b^2$

c ... hypotenuse

a ... one leg

b ... another leg

Pythagorean theorem is used in right triangles (triangles in which one angle is 90°).

Step-by-step explanation:

longer leg: x

shorter leg: x - 5

hypotenuse: x + 5

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