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Varvara68 [4.7K]

4 years ago

5

The light from the lamp casts a shadow of a man standing 15ft away from the lamppost, the shadow is 5ft long. The angle of eleva

tion from the tip of the shadow to the lamp is 50degrees. To the nearest foot, the lamppost is how tall?

1 answer:

Kipish [7]4 years ago

6 0

Check the picture below

make sure your calculator is in Degree mode.

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Prove that the triangle EDF is isosceles. Give reasons for your answer.

Gekata [30.6K]

Answer:

$\triangle EDF$ is isosceles.

Step-by-step explanation:

Please have a look at the attached figure.

We are <u>given</u> the following things:

$\angle EDF = y$

$\text{External }\angle DFG = 90 +\dfrac{y}{2}$

Let us try to find out $\angle E$ and $\angle DFE$ . After that we will compare them.

<u>Finding </u> $\angle DFE$ <u>:</u>

Side EG is a straight line so $\angle GFE = 180$

$\angle GFE$ is sum of internal $\angle DFE$ and external $\angle DFG$

$\angle GFE = 180 = \angle DFE + \angle DFG\\\Rightarrow 180 = \angle DFE + (90+\dfrac{y}{2})\\\Rightarrow \angle DFE = 180 - 90 - \dfrac{y}{2}\\\Rightarrow \angle DFE = 90 - \dfrac{y}{2} ....... (1)$

<u>Finding </u> $\angle E$ <u>:</u>

<u>Property of external angle:</u> External angle in a triangle is equal to the sum of two opposite internal angles of a triangle.

i.e. external $\angle DFG$ = $\angle E + \angle EDF$

$\Rightarrow 90+\dfrac{y}{2} = \angle E + y\\\Rightarrow \angle E = 90+\dfrac{y}{2} -y\\\Rightarrow \angle E = 90-\dfrac{y}{2} ....... (2)$

Comparing equations (1) and (2):

It can be clearly seen that:

$\angle DFE = \angle E =90-\dfrac{y}{2}$

The two angles of $\triangle EDF$ are equal hence $\triangle EDF$ is isosceles.

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Step-by-step explanation:

5'm

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