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3 years ago

What is the measure of ∠S ?

Mathematics

2 answers:

zloy xaker [14]3 years ago

6 0

55 degrees that is the answer.

jeka943 years ago

4 0

Alright, lets get started.

We could use sine law to find the remaining angle.

$\frac{sinA}{a} = \frac{sinB}{b}$

$\frac{sinS}{63.6} = \frac{sin64}{69.8}$

Plugging the value of sin 64

$\frac{sinS}{63.6} = \frac{0.8987}{69.8}$

$\frac{sinS}{63.6} = 0.01287$

Cross multiplying

sin S = 0.8189

Taking inverse on both side

$S = sin^{-1}(0.8189)$

S = 55° : Answer

Hope it will help :)

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A. 4".

B. Approximately 9.54".

C. Approximately 13.54".

Step-by-step explanation:

Please find the attachment.

Let x be the distance from the peak of the kite to the intersection of the diagonals and y be the distance from the peak of the kite to the intersection of the diagonals.

We have been given that two sides of a kite are 10 inches and two sides are 5 inches. The shorter diagonal is 6 inches.

A. Since we know that the diagonals of a kite are perpendicular and one diagonal (the main diagonal) is the perpendicular bisector of the shorter diagonal.

We can see from our attachment that point O is the intersection of both diagonals. In triangle AOD the side length AD will be hypotenuse and side length DO will be one leg.

We can find the value of x using Pythagorean theorem as:

$(AO)^2=(AD)^2-(DO)^2$

$x^{2}=5^2-3^2$

$x^{2}=25-9$

$x^{2}=16$

Upon taking square root of both sides of our equation we will get,

$x=\sqrt{16}$

$x=\pm 4$

Since distance can not be negative, therefore, the distance from the peak of the kite to the intersection of the diagonals is 4 inches.

B. We can see from our attachment that point O is the intersection of both diagonals. In triangle DOC the side length DC will be hypotenuse and side length DO will be one leg.

We can find the value of y using Pythagorean theorem as:

$(OC)^2=(DC)^2-(DO)^2$

Upon substituting our given values we will get,

$y^2=10^2-3^2$

$y^2=100-9$

$y^2=91$

Upon taking square root of both sides of our equation we will get,

$y=\sqrt{91}$

$y\pm 9.539392$

$y\pm\approx 9.54$

Since distance can not be negative, therefore, the distance from intersection of the diagonals to the top of the tail is approximately 9.54 inches.

C. We can see from our diagram that the length of longer diagram will be the sum of x and y.

$\text{The length of the longer diagonal}=x+y$

$\text{The length of the longer diagonal}=4+9.54$

$\text{The length of the longer diagonal}=13.54$

Therefore, the length of longer diagonal is approximately 13.54 inches.

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