Sin x = cos y
Angles x and y a complementary.
Sine and cosine are co-functions.
sin x = cos (90 - x)
sin x = cos y
Answer:
7/16.
Step-by-step explanation:
The first triangle has no shading.
The second triangle has 1/4 shaded.
The third has 3/9.
The fourth has 6/16.
Based on this pattern, we can assume that the number of small triangles in each triangle are going to be squared of numbers, since the first had 1, the second 4, the third 9, the fourth 16. So, the 8th triangle would have 8^2 small triangles, or 64 triangles in total.
The first triangle has no shaded triangles. The second has 1. The third has 3. The fourth has 6. If you study the pattern, the second triangle has 1 more than the previous, the third has 2 more, the fourth has three more. And so, the fifth triangle would have 6 + 4 = 10 triangles, the sixth would have 10 + 5 = 15 triangles, the seventh would have 15 + 6 = 21 triangles, and the eighth would have 21 + 7 = 28 shaded triangles.
So, the fraction of shaded triangles would be 28 / 64 = 14 / 32 = 7 / 16.
Hope this helps!
Answer:
The height of the pole is 167 m
Step-by-step explanation:
The given parameters are;
Increase in the length of the shadow = 90 m
Initial angle of elevation of the Sun = 58°
Final angle of elevation of the Sun = 36°
We have a triangle formed by the change in the length of the shadow and the rays from the two angle of elevation to the top of the pole giving an angle 22° opposite to the increase in the length of the shadow
We have by sin rule;
90/(sin (22°) = (Initial ray from the top of the pole to the end of the shadow's length)/(sin(122°)
Let the initial ray from the top of the pole to the end of the shadow's length = l₁
90/(sin (22°) = l₁/(sin(122°)
l₁ = 90/(sin (22°) ×(sin(122°) = 283.3 m
Therefore;
The height of the pole = 283.3 m × sin(36°) = 166.52 m
The height of the pole= 167 m to three significant figures.
Answer:
false is the correct answer of it if the answer going wrong sorry
12/10, 1.2, and any of the infinite multiples of the fraction such as 24/20, 120/100 etc...
