Question : The Square based pyramid depicted below has averticle height of 13cm. K is the mid point of CD and angle AKL =50 degrees. To the nearest tenth of degree, what is the measure of angle ACK?
Solution :
Length of Square be x
Tan 50º = 1.19175
= 13 / (x/2)
x = 21.81659041
x = 21.82
Sin 50º = (x/2) / AK
AK = (x/2) / 0.766
AK = 14.24
Tan ∟AKC = AK /(x/2) = 14.24/ 10.91 = 1.31
∟AKC = tanˉ¹(1.31)
∟AKC = 53º
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